IV 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 191 X X X X X X X X X X X2 X X X X X X X X3 X X X X X X X4 X X X X5 X X X X6 X X X X7 X X X X X X X X X X X8 X X X X X X X X

1

I.1.1Identify and sketch the set of points satisfying.(a) jz � 1� ij = 1 (f) 0 < Im z < �(b) 1 < j2z � 6j < 2 (g) �� < Re z < �(c) jz � 1j2 + jz + 1j2 < 8 (h) jRe zj < jzj(d) jz � 1j+ jz + 1j � 2 (i) Re (iz + 2) > 0

(e) jz � 1j < jzj (j) jz � ij2 + jz + ij2 < 2

SolutionLet z = x+ iy, where x; y 2 R.(a) Circle, centre 1 + i, radius 1.

jz � 1� ij = 1, j(x� 1) + i (y � 1)j = 1, (x� 1)2 + (y � 1)2 = 12

(b) Annulus with centre 3, inner radius 1=2, outer radius 1.

1 < j2z � 6j < 2, 1 < 2 jz � 3j < 2,, 1=2 < jz � 3j < 1, (1=2)2 < (x� 3)2 + y2 < 12

(c) Disk, centre 0, radiusp3.

jx+ iy � 1j2 + jx+ iy + 1j2 < 8,

, (x� 1)2 + y2 + (x+ 1)2 + y2 < 8, x2 + y2 <�p3�2

(d) Interval [�1; 1].

jz � 1j+ jz + 1j � 2,q(x� 1)2 + y2 � 2�

q(x+ 1)2 + y2 ,

,�q

(x� 1)2 + y2�2��2�

q(x+ 1)2 + y2

�2,

,q(x+ 1)2 + y2 � x+ 1,

�q(x+ 1)2 + y2

�2� (x+ 1)2 , y = 0

Now, take y = 0 in the inequality, and compute the three intervals

2

x < �1; then jx� 1j+ jx+ 1j = � (x� 1)� (x+ 1) = �2x � 2;�1 � x � 1 then jx� 1j+ jx+ 1j = � (x� 1) + (x+ 1) = 2 � 2x > 1; then jx� 1j+ jx+ 1j = (x� 1) + (x+ 1) = 2x � 2:

(e) Half�plane x > 1=2.

jz � 1j < jzj , jz � 1j2 < jzj2 , jx+ iy � 1j2 < jx+ iyj2 ,, (x� 1)2 + y2 < x2 + y2 , x > 1=2

(f) Horizontal strip, 0 < y < �.(g) Vertical strip, �� < x < �.(h) CnR.

jRe zj < jzj , jRe (x+ iy)j2 < jx+ iyj2 , x2 < x2 + y2 , jyj > 0(i) Half plane y < 2.

Re (iz + 2) > 0, Re (i (x+ iy) + 2) > 0, �y + 2 > 0, y < 2

(j) Empty set.

jz � ij2 + jz + ij2 < 2, jx+ iy � ij2 + jx+ iy + ij2 < 2,, x2 + (y � 1)2 + x2 + (y + 1)2 < 2, x2 + y2 < 0

3

I.1.2Verify from the de�nitions each of the identities(a) z + w = �z + �w (b) zw = �z �w (c) j�zj = jzj (d) jzj2 = z�zDraw sketches to illustrate (a) and (c).

SolutionSubstitute z = x+ iy and w = u+ iv, and use the de�nitions.(a)

z + w = (x+ iy) + (u+ iv) = (x+ u) + (y + v) i =

= (x+ u)� (y + v) i = (x� iy) + (u� iv) = �z + �w:

(b)

zw = (x+ iy) (u+ iv) = (xu� yv) + (xv + yu) i == (xu� yv)� (xv + yu) i = (x� iy) (u� iv) = �z �w:

(c)

j�zj =��x+ iy�� = jx� iyj =qx2 + (�y)2 =px2 + y2 = jx+ iyj = jzj :

(d)

jzj2 = jx+ iyj2 =�p

x2 + y2�2= x2 + y2 =

= x2 � i2y2 = (x+ iy) (x� iy) = z�z:

4

I.1.3Show that the equation jzj2 � 2Re (�az) + jaj2 = �2 represents a circlecentered at a with radius �.

SolutionLet z = x+ iy and a = �+ i�, we have

jzj2 � 2Re (�az) + jaj2 == x2 + y2 � 2Re ((�� i�) (x+ iy)) + �2 + �2 =

= x2 + y2 � 2Re ((�x+ �y) + i (�y � �x)) + �2 + �2 == x2 + y2 � 2 (�x+ �y) + �2 + �2 =

= (x� �)2 + (y � �)2 :

Thus the equation jzj2 � 2Re (�az) + jaj2 = �2, becomes

(x� �)2 + (y � �)2 = �2;which is the equation for a circle of radius � centered at (�; �), which incomplex notation is the point a = �+ i�.

5

I.1.4Show that jzj � jRe zj+ jIm zj, and sketch the set of points for whichequality holds.

SolutionApply triangle inequality to z = Re z+ i Im z, to obtain jzj 6 jRe zj+ jIm zj.Now set z = x+ iy, and see then equality holds

jzj = jRe zj+jIm zj ,px2 + y2 =

px2+

py2 ,

�px2 + y2

�2=�px2 +

py2�2,

, x2+y2 =�px2�2+�px2�2+2px2py2 , x2y2 = 0, x = 0 or y = 0:

Equality holds only when z is real or pure imaginary, which are all the pointson the real and imaginary axis.

6

I.1.5Show that

jRe zj � jzj ; jIm zj � jzj :Show that

jz + wj2 = jzj2 + jwj2 + 2Re (z �w) :Use this to prove the triangle inequality jz + wj � jzj+ jwj.

SolutionLet z = x+ iy. Then since the square root function is monotone, we have

jRe zj = jxj =px2 6

px2 + y2 = jzj ;

jIm zj = jyj =py2 6

px2 + y2 = jzj :

Now for z; w we have

jz + wj2 == (z + w) (z + w) = (z + w) (�z + �w) = z�z + z �w + w�z + w �w =

= jzj2 + 2Re (z �w) + jwj2 ;

where we have used the fact that 2Re (z �w) = z �w+z �w = z �w+ �wz = z �w+w�z.We use both of the above facts and the trivial identities jzj = j�zj and jzwj =jzj jwj to prove the triangle inequality for z; w. We have

jz + wj2 = jzj2 + 2Re (z �w) + jwj2 � jzj2 + 2 jRe (z �w)j+ jwj2 �� jzj2 + 2 jz �wj+ jwj2 = jzj2 + 2 jzj jwj+ jwj2 = (jzj+ jwj)2 :

The desired inequality now follows by taking square root of both sides.

7

I.1.6For �xed a 2 C, show that jz � aj = j1� �azj = 1 if jzj = 1 and 1��az 6= 0.

SolutionIf jzj = 1, then j�zj = 1 and z�z = 1. Use this and get

jz � aj = jz � aj j�zj = jz�z � a�zj = j1� a�zj = j1� �azj :We have

jz � ajj1� �azj = 1;

as was to be shown.

8

I.1.7Fix � > 0, � 6= 1, and �x z0; z1 2 C. Show that the set of z satisfyingjz � z0j = � jz � z1j is a circle. Sketch it for � = 1

2and � = 2, with

z0 = 0 and z1 = 1. What happens when � = 1?

SolutionRecall that a circle in R2 centered at (a; b) with radius r is given by theequation

(x� a)2 + (y � b)2 = r2:We manipulate the equation

jz � z0j = � jz � z1j :The solutions set of the equation above remains the same if we square bothsides,

jz � z0j2 = �2 jz � z1j2 :Let z = x+ iy, z0 = x0 + iy0 and z1 = x1 + iy1. Thus our equation becomes

(x� x0)2 + (y � y0)2 = �2�(x� x1)2 + (y � y1)2

�:

Expanding the squares, and grouping terms, we have

�1� �2

�x2�2

�x0 � �2x1

�x+�x20 � �2x21

�+�1� �2

�y2�2

�y0 � �2y1

�y+�y20 � �2y21

�= 0

Dividing both sides by (1� �2), we have

x2 � 2(x0 � �2x1)

1� �2 x+(x20 � �2x21)1� �2 + y2 � 2(y0 � �

2y1)

1� �2 y +(y20 � �2y21)1� �2 = 0

Now complete the squares for both the x and y terms. Recall that

x2 � 2ax+ b = (x� a)2 � a2 + b:So we have

9

�x� x0 � �

2x11� �2

�2+(1� �2) (x20 � �2x21)� (x0 � �2x1)

2

(1� �2)2+

+

�y � y0 � �

2y11� �2

�2+(1� �2) (y20 � �2y21)� (y0 � �2y1)

2

(1� �2)2= 0:

This becomes

�x� x0 � �

2x11� �2

�2+

�y � y0 � �

2y11� �2

�2=

=(x0 � �2x1)2 + (y0 � �2y1)2 � (1� �2) (x20 + y20 � �2 (x21 + y21))

(1� �2)2=

=�2�(x1 � x0)2 + (y1 � y0)2

�(1� �2)2

;

which is the equation for a circle. If z0 = 0, and z1 = 1, we have�x� ��2

1� �2

�2+ y2 =

��

1� �2

�2:

When � = 12we have a circle of radius 2

3centered at

��13; 0�, and when � = 2,

we have a circle of radius 23centered at

�43; 0�. When � = 1, we have the

equation

jz � z0j = jz � z1j ;which is the line bisecting the two points. When z0 = 0; z1 = 1, this is theline Re z = 1

2.

10

­1 1 2

­2

­1

1

2

Re z

Im z

I.1.7

­1 1 2

­2

­1

1

2

Re z

Im z

I.1.7

­1 1 2

­2

­1

1

2

Re z

Im z

I.1.7

� = 12

� = 1 � = 2

11

I.1.8Let p (z) be a polynomial of degree n � 1 and let z0 2 C. Showthat there is a polynomial h (z) of degree n � 1 such that p (z) =(z � z0)h (z)+p (z0). In particular, if p (z0) = 0, then p (z) = (z � z0)h (z).

SolutionSet

p (z) = anzn + an�1z

n�1 + an�2zn�2 + � � �+ a2z2 + a1z + a0:

and

h (z) = bn�1zn�1 + bn�2z

n�2 + bn�3zn�3 + : : :+ b2z

2 + b1z + b0:

We equate coe¢ cients in the polynomial identity p (z) = (z � z0)h (z) +p (z0), and get

anzn + an�1z

n�1 + an�2zn�2 + � � �+ a2z2 + a1z + a0 =

= (z � z0)�bn�1z

n�1 + bn�2zn�2 + bn�3z

n�3 + : : :+ b2z2 + b1z + b0

�+p (z0) =

= bn�1zn + bn�2z

n�1 + bn�3zn�2 + : : :+ b2z

3 + b1z2 + b0z + p (z0)�

� bn�1z0zn�1 � bn�2z0zn�2 � bn�3z0zn�3 � : : :� b2z0z2 � b1z0z � b0z0 =bn�1z

n+(bn�2 � bn�1z0) zn�1+(bn�3 � bn�2z0) zn�2+(bn�4 � bn�3z0) zn�3+� � �� � �+ (b2 � b3z0) z3 + (b1 � b2z0) z2 + (b0 � b1z0) z + p (z0)� b0z0:

Equate and solve for the bj´s in terms of aj´s.

8<:an = bn�1ak = bk�1 � bkz0; 0 � k � n� 1a0 = p (z0)� b0z0

)

8>>>><>>>>:bn�1 = an

bk =n�k�1Pi=0

ak+1+izi0; 0 � k � n� 2

p (z0) =nPi=0

aizi0

Proof by induction on degree n of p (z), set

p (z) = anzn + an�1z

n�1 : : :+ a0;

where an 6= 0.

12

Fix z0 and write

p (z) = an (z � z0) zn�1 + r (z) ;where deg r (z) � n� 1.By using the induction hypothesis, we can assume that

r (z) = q (z) (z � z0) + c;where deg q (z) � deg r (z). Then

p (z) =�anz

n�1 + q (z)�(z � z0) + c = h (z) (z � z0) + c

Since deg q (z) � n� 2, deg r (z) � n� 1.Plug in z0, get p (z0) = c.

13

I.1.9Find the polynomial h (z) in the preceding exercise for the followingchoices of p (z) and z0(a) p (z) = z2 and z0 = i(b) p (z) = z3 + z2 + z and z0 = �1(c) p (z) = 1 + z + z2 + � � �+ zm and z0 = �1

SolutionFrom the preceding exercise we have

p (z) = (z � z0)h (z) + p (z0) :We solve for h (z) then

h (z) =p (z)� p (z0)z � z0

:

(a)We have that p (z) = z2 and z0 = i, thus p (z0) = p (i) = �1.Thus

h (z) =p (z)� p (z0)z � z0

=z2 + 1

z � i = z + i;

and

z2 = (z � i) (z + i)� 1:(b)We have that p (z) = z3 + z2 + z and z0 = �1, thus p (z0) = p (�1) = �1.Thus

h (z) =p (z)� p (z0)z � z0

=z3 + z2 + z + 1

z + 1=(z + 1) (z2 + 1)

z + 1= z2 + 1;

and

z3 + z2 + z = (z + 1)�z2 + 1

�� 1:

(c)We have that p (z) = 1 + z + z2 + � � �+ zm and z0 = �1, thus

14

p (z0) = p (�1) =�0; m odd1; m even

Thus

h (z) =p (z)� p (z0)z � z0

=

=

8<:(zm�1+zm�3���+z2+1)(z+1)

z+1= zm�1 + zm�3 + � � �+ z2 + 1 if m is odd

(zm�1+zm�3���+z3+z)(z+1)+1z+1

= zm�1 + zm�3 + : : :+ z3 + z if m is even

and

zm + zm�1 + � � � z2 + z + 1 =

=

�(z + 1) (zm�1 + zm�3 + � � �+ z2 + 1) if m is odd(z + 1) (zm�1 + zm�3 + � � �+ z3 + z) + 1 if m is even

15

I.1.10Let q (z) be a polynomial of degreem � 1. Show that any polynomialp (z) can be expressed in the form

p (z) = h (z) q (z) + r (z) ;

where h (z) and r (z) are polynomials and the degree of the remain-der r (z) is strictly less than m.Hint. Proceed by induction on the degree of p (z). The resultingmethod is called the division algorithm.

SolutionFirst suppose p (z) is the zero polynomial. (So, the degree of p (z) is �1.)The degree of r (z) must be less than the degree of q (z). If h (z) 6= 0, itfollows that the degree of h (z) q (z) is greater than the degree of r (z). Thisthen implies that h (z) q (z)+r (z) 6= 0. So, h (z) q (z)+r (z) = 0 implies thath (z) = 0, and thus r (z) = 0. So, the polynomials are h (z) = 0 and r (z) = 0,and these polynomials are the only ones that satisfy both conditions.

Now assume that the division algorithm is true for all polynomials p (z) ofdegree less than n. (Where n � 0.) If the degree of q (z) is greater than thedegree of p (z), and h (z) is nonzero, then h (z) q (z)+r (z) has degree greaterthan p (z). So, if the degree of q (z) is greater than the degree of p (z), thenh (z) = 0 and thus r (z) = p (z). This proves both existence and uniquenessof h (z) and r (z), in this case.

Now, suppose that the degree of q (z) is less than or equal to the degree ofp (z). Set

p (z) = anzn + an�1z

n�1 + � � �+ a1z + a0and

q (z) = bmzm + bm�1z

m�1 + � � �+ b1z + b0;where an 6= 0, bm 6= 0 and m � n.Let

p1 (z) =anbmzn�mq (z)� p (z) ;

16

then

p1 (z) =anbmzn�m

�bmz

m + bm�1zm�1 + � � �+ b1z + b0

���anz

n + an�1zn�1 + � � �+ a1z + a0

�:

The monomials of degree n cancel, and therefore p1 (z) is a polynomial of de-gree at most n�1. It follows, by assumption, that p1 (z) = h1 (z) q (z)+r1 (z),where h1 (z) and r1 (z) are the unique polynomials satisfying the conditionsabove.Let

h (z) =anbmzn�m � h1 (z) ;

and let

r (z) = �r1 (z) :Then

h (z) q (z) + r (z) =

=

�anbmzn�m � h1 (z)

�q (z)� r1 (z) =

=anbmzn�mq (z)� (h1 (z) q (z) + r1 (z)) =

=anbmzn�mq (z)� p1 (z) =

= p (z) :

Thus given p (z) and q (z), there exist polynomials h (z) and r (z) satisfyingthe above two conditions.

17

I.1.11Find the polynomials h (z) and r (z) in the preceding exercise forp (z) = zn and q (z) = z2 � 1.

SolutionRequire

zn = h (z)�z2 � 1

�+ r (z) ; deg r (z) � 1:

If n is even

zn =�zn�2 + zn�4 + zn�6 + � � �+ z2 + 1

� �z2 � 1

�+ 1:

If n is odd

zn =�zn�2 + zn�4 + zn�6 + � � �+ z3 + z

� �z2 � 1

�+ z:

Thus

h (z) =

�zn�2 + zn�4 + zn�6 + � � �+ z2 + 1; if n is even,zn�2 + zn�4 + zn�6 + � � �+ z3 + z; if n is odd. ;

and

r (z) =

�1; if n is even,z; if n is odd.

:

18

I.2.1Express all values of the following expressions in both polar andCartesian coordinates, and plot them.(a)

pi (c) 4

p�1 (e) (�8)1=3 (g) (1 + i)8

(b)pi� 1 (d) 4

pi (f) (3� 4i)1=8 (h)

�1+ip2

�25Solution(a)

pi =

n�ei(�=2+2k�)

�1=2= ei(�=4+k�); k = 0; 1

o=�ei�=4; ei5�=4

=n� (1 + i) =

p2o:

(b)

pi� 1 =

��p2ei(3�=4+2k�)

�1=2= 21=4ei(3�=8+k�); k = 0; 1

�=

=�21=4ei3�=8; 21=4ei11�=8

=��21=4 (cos (3�=8) + i sin (3�=8))

:

(c)

4p�1 =

n�ei(�+2k�)

�1=4= ei(�=4+k�=2); k = 0; 1; 2; 3

o=

=�ei�=4; ei3�i=4; ei5�=4; ei7�=4

=n(1� i)=

p2; (�1� i) =

p2o:

(d)

4pi =

n�ei(�=2+2k�)

�1=4ei(�=8+k�=2); k = 0; 1; 2; 3 =

�ei�=8; ei5�=8; ei9�=8; ei13�=8

o=

= f� (cos (�=8) + i sin (�=8)) ;� (cos (5�=8) + i sin (5�=8))g :

(e)

(�8)1=3 =n�23ei(�+2k�)

�1=3= 2ei(�=3+2k�=3); k = 0; 1; 2

o=

=�2ei�=3; 2ei�; 2ei5�=3

=n1 + i

p3;�2; 1� i

p3o:

(f)

19

Draw �gure and get tan �0 = �4=3) �0 = tan�1 ��4

3

�.

(3� 4i)1=8 =n�5ei(�0+2k�)

�1=8= 51=8ei(�0=8+k�=4); k = 0; 1; : : : ; 7

o=

=��51=8ei(�0=8);�51=8ei(�0=8+�=4);�51=8ei(�0=8+�=2);�51=8ei(�0=8+3�=4)

=�

�51=8 (cos (�0=8) + i sin (�0=8)) ;�51=8 (cos (�0=8 + �=4) + i sin (�0=8 + �=4)) ;�51=8 (cos (�0=8 + �=2) + i sin (�0=8 + �=2)) ;�51=8 (cos (�0=8 + 3�=4) + i sin (�0=8 + 3�=4))

:

(g)

(1 + i)8 =�p2ei(�=4+2k�)

�8= 16ei(2�+16k�) = 16ei2� = 16:

(h)

�1 + ip2

�25=�ei(�=4+2k�)

�25= ei(25�=4+50k�) = ei(�=4+6�+50k�) = ei�=4 =

1 + ip2:

­1 1

­1

1

I.2.1a

­1 1

­1

1

I.2.1b

­1 1

­1

1

I.2.1c

­1 1

­1

1

I.2.1d

­2 2

­2

2

I.2.1e

­1 1

­1

1

I.2.1f

20

­20 20

­20

20

I.2.1g

­1 1

­1

1

I.2.1h

21

I.2.2Sketch the following sets(a) jarg zj < �=4 (c) jzj = arg z(b) 0 < arg (z � 1� i) < �=3 (d) log jzj = �2 arg z

Solutiona) Sector. b) Sector. c) Is a spiral curve starting at 0, spiraling to 1. (d) Isa spiral curve, spiraling to 0, and to 1.

π/4

π/4­5 5

­5

5

I.1.2a

π/3

­5 5

­5

5

I.1.2b

(π/2,0)

(−π,0)

(−3π/2,0)

(2π,0)

I.2.2c

(1,0)

I.2.2d  (­3 < x < 3) I.2.2d  (­30 < x < 30) I.2.2d  (­800 < x < 800)

22

I.2.3For a �xed complex number b, sketch the curve

�ei� + be�i� : 0 � � � 2�

.

Di¤erentiate between the cases jbj < 1; jbj = 1 and jbj < 1.Hint. First consider the case b > 0, and then reduce the generalcase to this case by a rotation.

SolutionFor 0 < b < 1, an ellipse x2= (1 + b)2 + y2= (1� b)2 = 1, traversed in positivedirection with increasing �. For b = 1, an interval [�2; 2]. For 1 < b < +1,an ellipse traversed in negative direction. For b = �ei', express equation asei'=2

�ei(��'=2) + �e�i(��'=2)

�to see that curve is rotate of ellipse or interval

by '=2.

­3 ­2 ­1 1 2 3

­3

­2

­1

1

2

3

x

y

I.2.7  (b = 0.5)

­3 ­2 ­1 1 2 3

­3

­2

­1

1

2

3

x

y

I.2.7  (b = 1)

­3 ­2 ­1 1 2 3

­3

­2

­1

1

2

3

x

y

I.2.7  (b = 1.5)

23

I.2.4For which n is i an n-th root of unity?

Solution.i is an nth root of unity for in = 1, n = 4k, k = 1; 2; 3; : : :

24

I.2.5For n � 1, show that(a) 1 + z + z2 + � � �+ zn = (1� zn+1) = (1� z), z 6= 1

(b) 1 + cos � + cos 2� + � � �+ cosn� = 12+

sin�n+

12

��

2 sin �=2.

Solution(a)Set

sn = 1 + z + z2 + � � �+ zn;

and multiply sn with z and have

zsn = z + z2 + z3 + � � �+ zn+1:

Now subtract zsn from the sum sn, and have

sn (1� z) = 1� zn+1:If z 6= 1 we have after division by 1� z;

sn =1� zn+11� z :

(b)Apply (a) to z = ei� and to z = e�i�;

1 + ei� + ei2� + � � �+ ein� = 1� ei(n+1)�1� ei� ;

1 + e�i� + e�i2� + � � �+ e�in� = 1� e�i(n+1)�1� e�i� :

Add the identities, and use the de�nitions of sine and cosine.

25

2 (1 + cos � + cos 2� + � � �+ cosn�) =

=1� ei(n+1)�1� ei� +

1� e�i(n+1)�1� e�i� =

=

�1� ei(n+1)�

�e�i�=2

e�i�=2 � ei�=2 +

�1� e�i(n+1)�

�ei�=2

ei�=2 � e�i�=2 =

= � 12i

e�i�=2�1� ei(n+1)�

�sin (�=2)

+1

2i

ei�=2�1� e�i(n+1)�

�sin (�=2)

=

=1

2i

"ei�=2 � e�i�=2 + ei(n+

12)� � e�i(n+

12)�

sin (�=2)

#=sin (�=2) + sin

�n+ 1

2

��

sin (�=2)=

= 1 +sin�n+ 1

2

��

sin (�=2):

Divide both sides with 2 and get

1 + cos � + cos 2� + � � �+ cosn� = 1

2+sin�n+ 1

2

��

2 sin �=2:

26

I.2.6Fix n � 1. Show that the n � th roots of unity w0; : : : ; wn�1 satisfy(a) (z � w0) (z � w1) : : : (z � wn�1) = zn � 1:(b) w0 + � � �+ wn�1 = 0 if n � 2:(c) w0 � � �wn�1 = (�1)n�1 :

(d)n�1Pj=0

wkj =

�0; 1 � k � n� 1;n; k = n:

Solution(a)Let w0; : : : ; wn�1 be the n � th roots of unity then wj = e2�ji=n, and wj area roots of zn � 1 since

wnj � 1 =�e2�ji=n

�n � 1 = e2�ji � 1 = 0:By the fundamental theorem of algebra, and since the root are simple andthe coe¢ cient for the zn term is 1 it follows that

zn � 1 = (z � w0) (z � w1) � � � (z � wn�1) :(b)Using exercise (a) and multiply the factors in the product we get

zn�1 = zn�(w0 + w1 + � � �wn�1) zn�1+c2zn�2+ � � �+(�1)nw0w1 � � � � �wn�1:

By identi�cation of the coe¢ cients we see that

w0 + w1 + � � �+ wn�1 = 0 if n � 2:(c)From (b) follows that

�1 = (�1)nw0w1 � � �wn�1 , w0w1 � � �wn�1 = (�1)n�1 :(d)If 1 � k � n� 1

n�1Xj=0

wkj =n�1Xj=0

�e2�ji=n

�k=

n�1Xj=0

�e2�ki=n

�j=1�

�e2�ki=n

�n1� e2�ki=n =

1� e2�ki1� e2�ki=n = 0:

27

If k = n

n�1Xj=0

wkj =n�1Xj=0

�e2�ji=n

�n=

n�1Xj=0

e2�ji =

n�1Xj=0

1 = n:

28

I.2.7Fix R > 1 and n � 1, m � 0. Show that���� zm

zn + 1

���� � Rm

Rn � 1 ; jzj = R:

Sketch the set where equality holds. Hint. See (1.1) p.2.

Solution

a

ba + b

I.2.7

­1 1

­1

1

I.2.7 (n = 1)

­1 1

­1

1

I.2.7 (n = 2)

­1 1

­1

1

I.2.7 (n = 3)

­1 1

­1

1

I.2.7 (n = 4)

­1 1

­1

1

I.2.7 (n = 5)

We use that jz � wj � jzj � jwj see (1.1) on page 2 in CA, and have that

jzn + 1j > jznj � 1 = Rn � 1;where the last equaltity is due to that jzj = R.Some rearrangement gives

1

jzn + 1j �1

(Rn � 1) ;

29

and mulitplication by jzmj = Rm gives���� zm

zn + 1

���� � Rm

Rn � 1 ; jzj = R:

For equality, we must have

jzn + 1j = Rn � 1;because jzjm = Rm.We rearrarange this equality to

j�1j+ jzn + 1j = jznj

and we conclude that �1 and zn+1must lie on the same ray. (We have usedthe fact that jaj+jbj = ja+ bj implies that a; b lie on the same ray, see �rst �gure I.2.7)If

z = Rei�

then

zn = Rnein�:

Since R > 1, we require that ein� = �1 thus

z = wkRe�i=n;

where wk is an n � th root of unity.If

z = wkRe�i=n;

then

zn + 1 = �Rn + 1;and ���� zm

zn + 1

���� = Rm

Rn � 1 :

30

I.2.8Show that cos 2� = cos2 � � sin2 � and sin 2� = 2 cos � sin � using deMoivre�s formulae. Find formulae for cos 4� and sin 4� in terms ofcos � and sin �.

SolutionLet � 2 R be given. Then by de Moivre�s formulae (on page 8 in CA) wehave for all n 2 Z

cosn� + i sinn� = (cos � + i sin �)n

Hence for n = 2 we get

cos 2� + i sin 2� =

= (cos � + i sin �)2 = cos2 � + i2 cos � sin � � sin2 � ==�cos2 � � sin2 �

�+ i (2 cos � sin �) ;

then, by setting the real and imaginary parts equal to each other we obtain

cos 2� = cos2 � � sin2 �sin 2� = 2 cos � sin �:

Similarly, applying de Moivre�s formulae for n = 4 we get (using the BinomialTheorem)

cos 4� + i sin 4� = (cos � + i sin �)4 =

= cos4 � + i4 cos3 � sin � � 6 cos2 � sin2 � � i4 cos � sin3 � + sin4 � == cos4 � � 6 cos2 � sin2 � + sin4 � + i

�4 cos3 � sin � � 4 cos � sin3 �

�;

then, by setting the real and imaginary parts equal to each other we obtain

cos 4� = cos4 � � 6 cos2 � sin2 � + sin4 �sin 4� = 4 cos3 � sin � � 4 cos � sin3 �:

as required.

31

I.3.1Sketch the image under the spherical projection of the followingsets on the sphere(a) the lower hemisphere Z � 0(b) the polar cap 3

4� Z � 1

(c) lines of lattitude X =p1� Z2 cos �, Y =

p1� Z2 sin �, for Z �xed

and 0 � � � 2�(d) lines of longitude X =

p1� Z2 cos �, Y =

p1� Z2 sin �, for � �xed

and �1 � Z � 1(e) the spherical cap A � X � 1, with center lying on the equator,for �xed A. Separate into cases, according to various ranges of A.

Solution

­2 ­1 1 2

­2

­1

1

2y

I.3.1a

­4 ­2 2 4

­4

­2

2

4y

I.3.1b

­2 ­1 1 2

­2

­1

1

2y

I.3.1c

­2 ­1 1 2

­2

­1

1

2y

θ

I.3.1d

­6 ­4 ­2 2

­4

­2

2

4y

I.3.1e (A = ­1/2)

­2 ­1 1 2

­2

­1

1

2y

I.3.1e (A = 0)

32

­2 2 4 6

­4

­2

2

4y

I.3.1e (A = 1/2)

(a)Image is the unit disk.(b)Set Y = 0 and Z = 3=4 in X2 + Y 2 + Z2 = 1, we have X =

p7=4. We have

the formula x = X= (1� Z), thus x =p7. Image is the exterior of a disk,

with centre 0 and radiusp7.

(c) Image is the disk, with centre 0 and radiusp1 + z=

p1� z. We have that

X =p1� Z2 cos � and Y =

p1� Z2 sin �, we can take the radius to image

for � = 0, thus r =p1�Z21�Z =

p1+Zp1�Z .

(d) Image is lines of longitude � issuing from 0.(e)Case 1: �1 < A < 0Image is the exterior of the disk centered at 1=A with radius

p1� A2= jAj.

Case 2: A = 0Image is the left half�plane.Case 3: 0 < A < 1Image is a disk centered at 1=A with radius

p1� A2= jAj.

Set Y = 0 then Z goes from �p1� A2 to

p1� A2. We have the formula

x = X= (1� Z) thus x goes from A=�1 +

p1� A2

�=�1�

p1� A2

�=A to

A=�1�

p1� A2

�=�1 +

p1� A2

�=A.

33

I.3.2If the point P on the sphere corresponds to z under the stereo-graphic projection, show that the antipodal point �P on the spherecorresponds to �1=�z.

SolutionFor z = x + iy the corresponding point on the sphere under transformationgiven on p. 12 in CA is given by (X;Y; Z) where

X =2x

jzj2 + 1;

Y =2y

jzj2 + 1;

Z =jzj2 � 1jzj2 + 1

:

For z 6= 0, we have

�1�z= � 1

x� iy = �x+ iy

(x� iy) (x+ iy) = �x+ iy

x2 + y2= � x

jzj2� i yjzj2

which corresponds to the point on the sphere given by (X 0; Y 0; Z 0) where

X 0 =

�2xjzj2��1

�z

��2 + 1 = �2xjzj2 + 1

= �X;

Y 0 =

�2yjzj2��1

�z

��2 + 1 = �2yjzj2 + 1

= �Y;

Z 0 =

��1�z

��2 � 1��1�z

��2 + 1 = �jzj2 � 1

jzj2 + 1= �Z:

Hence, the map (X; Y; Z) 7�! (�X;�Y;�Z) on the sphere corresponds toz 7�! �1=�z of C.

34

I.3.3Show that as z travers a small circle in the complex plane in thepositive (counterclockwise) direction, the corresponding point Pon the sphere traverses a small circle in the negative (clockwise)direction with respect to someone standing at the center of thecircle and with body outside the sphere. (Thus the stereographicprojection is orientation reversing, as a map from the sphere withorientation determined by the unit outer normal vector to the com-plex plane with the usual orientation.)

SolutionDraw the picture. Or argue as follows. The orientation of the image circleis the same for all circles on the sphere orientated so that N is outside thecircle. This can be seen by moving one circle continuously to the other, andseeing that the image circles moves continuously. Thus we need to shrink itonly for the equator of the sphere oriented as indicated (�), if the South Poleis inside it. And the image circle is the unity and positive direction of theunit circle () is the converse. So the orientation of the image is clockwise(negative).

35

I.3.4Show that a rotation of the sphere of 180� about the X-axis corre-sponds under stereographic projection to the inversion z 7�! 1=z ofC.

SolutionFor z = x + iy the corresponding point on the sphere under transformationgiven on p. 12 in CA is given by (X;Y; Z) where

X =2x

jzj2 + 1;

Y =2y

jzj2 + 1;

Z =jzj2 � 1jzj2 + 1

:

For z 6= 0, we have

1

z=

1

x+ iy=

x� iy(x+ iy) (x� iy) =

x� iyx2 + y2

=x

jzj2� i yjzj2

which corresponds to the point on the sphere given by (X 0; Y 0; Z 0) where

X 0 =

2xjzj2��1

z

��2 + 1 = 2x

jzj2 + 1= X;

Y 0 =

�2yjzj2��1

z

��2 + 1 = �2yjzj2 + 1

= �Y;

Z 0 =

1jzj2 � 11jzj2 + 1

= �jzj2 � 1

jzj2 + 1= �Z:

Hence, the map (X; Y; Z) 7�! (X;�Y;�Z) on the sphere corresponds toinversion z 7�! 1=z of C. Moreover, the map (X;Y; Z) 7�! (X;�Y;�Z) isgiven by rotation of the sphere by 180� about the X-axis.

36

I.3.5Suppose (x; y; 0) is the spherical projection of (X; Y; Z). Show thatthe product of the distances from the north pole N to (X; Y; Z) andfrom N to (x; y; 0) is 2. What is the situation when (X; Y; Z) lies onthe equator on the sphere?

SolutionThe distance from from the north pole N = (0; 0; 1) to (X; Y; Z) isq

X2 + Y 2 + (Z � 1)2;

and the distance from the north pole N = (0; 0; 1) to (x; y; 0) =�X1�Z ;

Y1�Z ; 0

�is s

X2

(1� Z)2+

Y 2

(1� Z)2+ 1:

The product of distances is

�X2 + Y 2 + (Z � 1)2

�1=2� X2

(1� Z)2+

Y 2

(1� Z)2+ 1

�1=2=X2 + Y 2 + (Z � 1)2

1� Z = 2:

When (X; Y; Z) lie on the equator, the product is simply the square of thedistance from N to a point on the equator. By the Pythagorean Law, this is1 + 1 = 2.

37

I.3.6We de�ne the chordal distance d (z; w) between two points z; w 2 C�to be the length of the straight line segment joining to the pointsP and Q on the unit sphere whose stereographic projections are zand w, respectively.(a) Show that the chordal distance is a metric, that is, it is sym-metric, d (z; w) = d (w; z); satis�es the triangle inequality d (z; w) �d (z; �) + d (�; w); and d (z; w) = 0 if and only if z = w.(b) Show that the chordal distance from z to w is given by

d (z; w) =2 jz � wjq

1 + jzj2q1 + jwj2

; z; w 2 C:

(c) What is d (z;1)? Remark. The expression for d (z; w) showsthat in�nitesimal arc length corresponding to the chordal metricis given by

d� (z) =2ds

1 + jzj2;

where ds = jdzj is the usual Euclidean in�nitesimal arc length. Thein�nitesimal arc length d� (z) determines another metric, the spher-ical metric � (z; w), on the extended complex plane. See SectionIX.3.

Solution(a)Follows from fact that Euclidean distance in R3 is a metric on the sphere.(b)Set z = x1 + iy1; w = x2 + iy2, we have

jz � wj2 == (z � w) (z � w) = (z � w) (z � w) =

= zz�zw � zw + ww = jzj2 + jwj2 � zw � zw == jzj2 + jwj2 � (x1 + iy1) (x2 � iy2)� (x1 � iy1) (x2 + iy2) =

= jzj2 + jwj2 � 2x1x2 � 2y1y2 : (1)

38

We take square of distance between P and Q

d (z; w)2 = (X �X 0)2+ (Y � Y 0)2 + (Z � Z 0)2 =

= X2 + Y 2 + Z2 +X 02 + Y 02 + Z 02 � 2 (XX 0 + Y Y 0 + ZZ 0) =

= 2� 2 4x1x2 + 4y1y2 +

�jzj2 � 1

� �jwj2 � 1

��jzj2 + 1

� �jwj2 + 1

� !=

= 2

�jzj2 + 1

� �jwj2 + 1

�� 4x1x2 � 4y1y2 �

�jzj2 � 1

� �jwj2 � 1

��jzj2 + 1

� �jwj2 + 1

� !=

= 2

2jzj2 + 2 jwj2 � 4x1x2 � 4y1y2�

jzj2 + 1� �jwj2 + 1

� !(1)=

4 jz � wj2�jzj2 + 1

� �jwj2 + 1

� :Taking the positive square root we have

d (z; w) =2 jz � wjq

1 + jzj2q1 + jwj2

; z; w 2 C:

(c)

d (z;1) = limw!1

2 jz � wjq1 + jzj2

q1 + jwj2

= limw!1

2 jz=w � 1jq1 + jzj2

q1= jwj2 + 1

=2q

1 + jzj2:

39

I.3.7Consider the sphere of radius 1

2in (X; Y; Z)� space, resting on the

(X;Y; 0)� plane, with south pole at the origin (0; 0; 0) and north poleat (0; 0; 1). We de�ne a stereographic projection of the sphere ontothe complex plane as before, so that corresponding points (X; Y; Z)and z � (x; y; 0) lie on the same line through the north pole. Findthe equations for z = x + iy in terms of X; Y; Z, and the equationsfor X; Y; Z in terms of z. What is the corresponding formula for thechordal distance? Note. In this case, the equation of the sphere isX2 + Y 2 +

�Z � 1

2

�2= 1

4.

SolutionThe line through P = (X; Y; Z) and N = (0; 0; 1) is on the form N +t (P �N), and it meats the xy � plane when

(x; y; 0) = (0; 0; 1) + t ((X; Y; Z)� (0; 0; 1)) = (tX; tY; 1 + t (Z � 1)) :

By simultaneous equations we have, and solve for X, Y and Z8<:x = tXy = tY1 + t (Z � 1) = 0

)

8<:X = x

t

Y = yt

Z = 1� 1t

We solve for the t that is a point on the sphere X2 + Y 2 +�Z � 1

2

�2= 1

4we

have

x2

t2+y2

t2+

�1

2� 1t

�2=1

4, jzj2 +

�t

2� 1�2=t2

4,

, jzj2 � t+ 1 = 0, t = jzj2 + 1

thus we have 8><>:X = x

1+jzj2

Y = y

1+jzj2

Z = jzj2

1+jzj2 :

40

Set z = x1 + iy1; w = x2 + iy2, we have

jz � wj2 == (z � w) (z � w) = (z � w) (z � w) =

= zz�zw � zw + ww = jzj2 + jwj2 � zw � zw == jzj2 + jwj2 � (x1 + iy1) (x2 � iy2)� (x1 � iy1) (x2 + iy2) =

= jzj2 + jwj2 � 2 (x1x2 + y1y2) : (1)

And the coordical distance

41

d (z; w)2 = (X1�X2)2 + (Y1�Y 2)2 + (Z1�Z2)2 =

=

�x1

1 + jzj2� x2

1 + jwj2�2+

�y1

1 + jzj2� y2

1 + jwj2�2+

jz1j2

1 + jzj2� jz2j2

1 + jwj2

!2=

=

�x1�1 + jwj2

�� x2

�1 + jzj2

��2+�y1�1 + jwj2

�� y2

�1 + jzj2

��2+�jzj2 � jwj2

�2�1 + jzj2

�2 �1 + jwj2

�2 =

=(x21 + y

21)�1 + jwj2

�2+ (x22 + y

22)�1 + jzj2

�2+ jzj4 � 2 jzj2 jwj2 + jwj4�

1 + jzj2�2 �1 + jwj2

�2 +

+�2x1x2

�1 + jwj2

� �1 + jzj2

�� 2y1y2

�1 + jwj2

� �1 + jzj2

��1 + jzj2

�2 �1 + jwj2

�2 =

=jzj2

�1 + jwj2

�2+ jwj2

�1 + jzj2

�2+ jzj4 � 2 jzj2 jwj2 + jwj4�

1 + jzj2�2 �1 + jwj2

�2 +

+�2 (x1x2 + y1y2)

�1 + jwj2

� �1 + jzj2

��1 + jzj2

�2 �1 + jwj2

�2 =

=jzj2 + 2 jzj2 jwj2 + jzj2 jwj4 + jwj2 + 2 jwj2 jzj2 + jwj2 jzj4 + jzj4 � 2 jzj2 jwj2 + jwj4�

1 + jzj2�2 �1 + jwj2

�2 +

+�2 (x1x2 + y1y2)

�1 + jwj2

� �1 + jzj2

��1 + jzj2

�2 �1 + jwj2

�2 =

=

�jzj2 + jwj2

� �1 + jwj2

� �1 + jzj2

�� 2 (x1x2 + y1y2)

�1 + jwj2

� �1 + jzj2

��1 + jzj2

�2 �1 + jwj2

�2 =

=

�jzj2 + jwj2 � 2 (x1x2 + y1y2)

� �1 + jwj2

� �1 + jzj2

��1 + jzj2

�2 �1 + jwj2

�2 =

=

�jzj2 + jwj2 � 2 (x1x2 + y1y2)

��1 + jzj2

� �1 + jwj2

� (1)=

(1)=

jz � wj2�jzj2 + 1

� �jwj2 + 1

� :This gives

42

d (z; w) =jz � wjq�

jzj2 + 1�q�

jwj2 + 1� :

43

I.4.1Sketch each curve and its image under w = z2.(a) jz � 1j = 1 (c) y = 1 (e) y2 = x2 � 1; x > 0(b) x = 1 (d) y = x+ 1 (f) y = 1=x; x 6= 0

Solution

­4 ­2 2 4

­4

­2

2

4

x

yI.4.1a z­plane

­4 ­2 2 4

­4

­2

2

4

x

yI.4.1b z­plane

­4 ­2 2 4

­4

­2

2

4

x

yI.4.1c z­plane

­4 ­2 2 4

­4

­2

2

4

x

yI.4.1a w­plane

­4 ­2 2 4

­4

­2

2

4

x

yI.4.1b w­plane

­4 ­2 2 4

­4

­2

2

4

x

yI.4.1c w­plane

­4 ­2 2 4

­4

­2

2

4

x

yI.4.1d z­plane

­4 ­2 2 4

­4

­2

2

4

x

yI.4.1e  z­plane

­4 ­2 2 4

­4

­2

2

4

x

yI.4.1f z­plane

44

­4 ­2 2 4

­4

­2

2

4

x

yI.4.1d w­plane

­4 ­2 2 4

­4

­2

2

4

x

yI.4.1e w­plane

­4 ­2 2 4

­4

­2

2

4

x

yI.4.1f w­plane

45

I.4.2Sketch the image of each curve in the preceding problem under theprincipal branch of w =

pz, and also sketch, on the same grid but

in a di¤erent color, the image of each curve under the other branchofpz.

Solution

w = z1=2 =�jzj ei argjzj

�1=2=�jzj eiArgjzj+i2�n

�1=2= �

pjzjeiArg z=2

­4 ­2 2 4

­4

­2

2

4

x

yI.4.2a z­plane

­4 ­2 2 4

­4

­2

2

4

x

yI.4.2b z­plane

­4 ­2 2 4

­4

­2

2

4

x

yI.4.2c z­plane

­4 ­2 2 4

­4

­2

2

4

x

yI.4.2a w­plane

­4 ­2 2 4

­4

­2

2

4

x

yI.4.2b w­plane

­4 ­2 2 4

­4

­2

2

4

x

yI.4.2c w­plane

46

­4 ­2 2 4

­4

­2

2

4

x

yI.4.2d z­plane

­4 ­2 2 4

­4

­2

2

4

x

yI.4.2e  z­plane

­4 ­2 2 4

­4

­2

2

4

x

yI.4.2f z­plane

­4 ­2 2 4

­4

­2

2

4

x

yI.4.2d w­plane

­4 ­2 2 4

­4

­2

2

4

x

yI.4.2e w­plane

­4 ­2 2 4

­4

­2

2

4

x

yI.4.2f w­plane

47

I.4.3(a) Give a brief description of the function z 7�! w = z3, consideredas a mapping from the z � plane to the w- plane. (Describe whathappens to w as z traverses a ray emanating from the origin, and asz traverses a ray a circle centered at the origin.) (b) Make branchcuts and de�ne explicitly three branches of the inverse mapping.(c) Describe the construction of the Riemann surface of z1=3.

Solution(a)The function w = f (z) = z3. For z = rei�, we have z3 = r3ei3�. The radicalrays at angle � are mapped to rays at angle 3�, that is, argw = 3arg z. Themagnitude of a complex number is cubed, jwj = jzj3. Circles, centered at theorigin of radius r, are mapped to cocentric circles with radius r3.(b)We make branch cuts at ( �1; 0],

z = w1=3 =�jwj ei argw

�1=3=�jwj eiArgw+i2�n

�1=3= ei2�n=3

pjwjeiArgw=3

we chooseg (z) = z1=3 = r1=3ei�=3, �� < � < �.Sheet 1 : Take f1 (z) = g (z),Sheet 2 : Take f2 (z) = e2�i=3g (z),Sheet 3 : Take f3 (z) = e4�i=3g (z).

(c) Top edge of cut on sheet 1 to bottom edge of cut on sheet 2. Top edge ofcut on sheet 2 to bottom edge of cut on sheet 3. Top edge of cut on sheet 3to bottom edge of cut on sheet 1. The endpoints for f (z) is continuous onsurface

Spara

48

± ± ± ± ± ± ±

± ±

± ±

± ±

±

± ± ± ± ± ± ±

I.4.3a

­1 1

­1

1

I.2.1b

­1 1

­1

1

± ± ± ± ± ±

± ± ± 

± ± ±

I.2.1c

Spara

49

I.4.4Describe how to construct the Riemann surfaces for the following functions(a) w = z1=4; (b) w =

pz � i; (c) w = (z � 1)2=5 :

Remark. To describe the Riemann surface of a multivalued function, beginwith one sheet for each branch of the function, make branch cuts so that thebranches are de�ned continuously on each sheet, and identify each edge ofa cut on one sheet to another appropriate edge so that the function valuesmatch up continuously.

Solution(a) Use four sheets, can make branch cuts along real axis from �1 to 0.(b) Use two sheets, can make branch cuts along horizontal line from �1+ ito i.(c) Use �ve sheets, can make branch cuts along real axis from �1 to 1.

50

I.5.1Calculate and plot for ez for the following points z.(a) 0 (c) � (i� 1) =3 (e) �i=m; m = 1; 2; 3; : : :(b) �i+ 1 (d) 37�i (f) m (i� 1) m = 1; 2; 3; : : :

Solution

­1 1

­1

1

I.5.1e

­0.4 ­0.2 0.2

­0.2

0.2

I.5.1f

(a)

e0 = 1:

(b)

e�i+1 = e�ie1 = �e:(c)

e�(i�1)3 = e��=3e�i=3 = e��=3

1

2+

p3

2i

!= 0:351 + 0:006i:

(d)

e37�i = e36�ie�i = �1:(e)We take the limit for the sequence e�i=m, m = 1; 2; 3; : : : as m ! 1, andhave

e�i=m ! 1;

51

as m ! 1. Because��e�i=m�� = 1 the sequence approaches its limit along a

circle with radius 1.(f)We take the limit for the sequence em(i�1) = e�memi, m = 1; 2; 3; : : : asm!1, and have

e�memi ! 0;

as m ! 1. Because je�memij = e�m the sequence spiraling to its limit inorigo.

52

I.5.2Sketch each of the following �gures and its image under the expo-nential map w = ez. Indicate the images of horizontal and verticallines in your sketch.(a) the vertical strip 0 < Re z < 1,(b) the horizontal strip 5�=3 < Im z < 8�=3,(c) the rectangle 0 < x < 1, 0 < y < �=4,(d) the disk jzj � �=2,(e) the disk jzj � �,(f) the disk jzj � 3�=2.

Solution

­4 ­2 2 4

­4

­2

2

4

I.5.2a z­plane

­4 ­2 0 2 4

2

4

6

8

I.5.2b z­plane

1 2 3

1

2

3

I.5.2c z­plane

­4 ­2 2 4

­4

­2

2

4

I.5.2a w­plane

­4 ­2 2 4

­4

­2

2

4

I.5.2b  w­plane

1 2 3

1

2

3

I.5.2c w­plane

53

­4 ­2 2 4

­4

­2

2

4

I..5.2d  z­plane

­4 ­2 2 4

­4

­2

2

4

I.5.2e  z­plane

­4 ­2 2 4

­4

­2

2

4

I.5.2f  z­plane

­4 ­2 2 4

­4

­2

2

4

I..5.2d w­plane

­10 10 20 30

­20

­10

10

20

I.5.2e w­plane

100

­100

100

I.5.2f w­plane

54

I.5.3Show that e�z = ez.

Solution

e�z = ex�iy = exe�iy = ex (cos (�y) + i sin (�y)) = ex (cos y � i sin y) == ex(cos y + i sin y) = ex (cos y + i sin y) = exeiy = ex+iy = ez:

55

I.5.4Show that the only periods of ez are the integral multiples of 2�i,that is, if ez+� = ez for all z, then � is an integer times 2�i.

SolutionShow that the only periods of ez are the integral muliples of 2�i that is, ifez+� = ez for all z, then � is an integer times 2�i.

ez = ez+� = eze� ) e� = 1) � = 2�mi

56

I.6.1Find and plot log z for the following complex numbers z. Specifythe principal value.(a) 2 (b) i (c) 1 + i (d)

�1 + i

p3�=2

Solution(a)Suppose that n = 0;�1;�2; : : :

log 2 = log j2j+ iArg 2 + 2�ni = log 2 + i2�n:(b)Suppose that n = 0;�1;�2; : : :

log i = log jij+ iArg i+ 2�ni = i�=2 + i2�n:(c)Suppose that n = 0;�1;�2; : : :

log (1 + i) =

= log j1 + ij+ iArg (1 + i) + 2�ni = logp2 + i�=4 + i2�n =

= log 2=2 + i�=4 + i2�n:

(d)Suppose that n = 0;�1;�2; : : :

log�1+i

p3

2

�=

= log���1+ip32

���+ iArg �1+ip32

�+ 2�ni =

= �i=3 + i2�n:

57

­1 1 2

­20

­10

10

20

I.6.1a

­1 1 2

­20

­10

10

20

I.6.1b

­1 1 2

­20

­10

10

20

I.6.1c

­1 1 2

­20

­10

10

20

I.6.1d

58

I.6.2Sketch the image under the map w = Log z of each of the following�gures.(a) the right halv-plane Re z > 0,(b) the half-disk jzj < 1, Re z > 0,(c) the unit circle jzj = 1,(d) the slit annulus

pe < jzj < e2, z 62 (�e2;�

pe),

(e) the horizontal line y = e,(f) the vertical line x = e.

Solution(b) We have a disk with radius less then 1, this means jzj < 1, thus log jzj< 0 and is unbounded, therefore, it goes from 0 to �1. Since Re (z) > 0,the polar angle is between ��

2and �

2.

(d) Here, we have an annulus, the polar angle is from �� to �, thus the argu-ment is in this range. Since, the log function is the inverse of the exponentialmap, circles in z � plane are stright lines in the w � plane (w = log z).Therefore, the image of this annulus under the log function is the rectangle,bounded by x = log j

pej = 1

2and x = log je2j = 2.

­3 ­2 ­1 1 2 3

­3

­2

­1

1

2

3

I.6.2a  z­plane

­2 2

­2

2

I.6.2b  z­plane

­4 ­2 2 4

­4

­2

2

4

I.6.2c  z­plane

59

­3 ­2 ­1 1 2 3

­3

­2

­1

1

2

3

I.6.2a  w­plane

π/2

−π/2

I.6.2b w­plane

π

−π

I.6.2c w­plane

­8 ­6 ­4 ­2 2 4 6 8

­8­6­4­2

2468

I.6.2d  z­plane

­4 ­2 2 4

­4

­2

2

4

I.6.2e  z­plane

­4 ­2 2 4

­4

­2

2

4

I.6.2f  z­plane

­4 ­2 2 4

­4

­2

2

4

I.6.2d  w­plane

­4 ­2 2 4

­4

­2

2

4

I.6.2e  w­plane

­4 ­2 2 4

­4

­2

2

4

I.6.2f  w­plane

60

I.6.3De�ne explicitly a continuous branch of log z in the complex planeslit along the negative imaginary axis, Cn [0;�i1).

Solution

I.6.3

We have log z = log rei� = log r + i�. To avoid the negative imaginary axiswe chose ��=2 < � < 3�=2. We use the branch

f�rei��= log r + i�; ��=2 < � < 3�=2;

of log z.

61

I.6.4How would you make a branch cut to de�ne a single-valued branchof the function log (z + i� 1)? How about log (z � z0)?

SolutionAny straight line cut from z0 to 1, in any direction, will do.

62

I.7.1Find all values and plot them.(a) (1 + i)i (b) (�i)1+i (c) 2�1=2 (d)

�1 + i

p3�(1�i)

Solution(a)Suppose that m;n = 0;�1;�2; : : :

(1 + i)i =

= ei log(1+i) = ei(logj1+ij+iArg(1+i)+i2�m) = ei(logp2+i�=4+i2�m) = e�2�me��=4ei log

p2 = [m = �n] =

= e2�ne��=4ei logp2:

(b)Suppose that m;n = 0;�1;�2; : : :

(�i)1+i == e(1+i) log(�i) = e(1+i)(logj�ij+iArg(�i)+i2�m) = e(1+i)(�i�=2+i2�m) =

= e�2�me�=2ei(��=2+2�m) = [m = �n] = e2�ne�=2ei(��=2�2�n) == �ie2�ne�=2:

(c)Suppose that m = 0;�1;�2; : : :

2�1=2 =

= e� log 2=2 = e�(logj2j+iArg 2+i2�m)=2 = e�(log 2+i2�m)=2 = e� log 2=2e�i�m =

= �1=p2:

(d)Suppose that n = 0;�1;�2; : : :

63

�1 + i

p3�1�i

=

= e(1�i) log(1+ip3) = e(1�i)(logj1+i

p3j+iArg(1+ip3)+i2�n) =

= e(1�i)(log 2+i�=3+i2�n) = e2�nelog 2+�=3ei( � log 2+�=3+2�n) =

= e2�nelog 2+�=3ei( � log 2+�=3):

64

I.7.2Compute and plot log

h(1 + i)2i

i.

SolutionWe rewrite the expression, suppose that k;m; n = 0;�1;�2; : : :

logh(1 + i)2i

i=

= logh�elog(1+i)

�2ii= log

�e2i log(1+i)

�= log

�e2i(logj1+ij+iArg(1+i)+i2�k)

�=

= loghe2i(log

p2+i�=4+i2�k)

i= log

�e�4�ke��=2ei log 2

�= [k = �m] = log

�e4�me��=2ei log 2

�=

= log��e4�me��=2ei log 2��+iArg �e4�me��=2ei log 2�+i2�n = 4�m��=2+i log 2+i2�n =

= ��=2 + i log 2 + 4�m+ 2�in:

Thus the set is a square lattice.

­10 10 20

­20

­10

10

20

I.7.2

65

I.7.3Sketch the image of the sector f0 < arg z < �=6g under the map w =za for(a) a = 3

2(b) a = i (c) a = i+ 2

Use only the principal branch of za.

Solution

­5 5

­5

5

I.7.3a  z­plane

­5 5

­5

5

I.7.3b  z­plane

­5 5

­5

5

I.7.3c  z­plane

­5 5

­5

5

I.7.3a  w­plane

­2 2

­2

2

I.7.3b w­plane

1

1

I.7.3c  w­plane

Some points on the y-axis is marked, and the point (1; 0) too.

66

I.7.4Show that (zw)a = zawa, where on the right we take all possibleproducts.

SolutionLet � 2 (zw)a, and k;m; n = 0;�1;�2; : : :

� =

= ea log(zw) = ea(logjzwj+iArg(zw)+i2�n) = ea(logjzj+logjwj+iArg z+iArgw+i2�m+i2�n) =

= ea(logjzj+iArg z+i2�(m+n))ea(logjwj+iArgw) 2 zawa:

Conversely, if � 2 zawa, say

� =

= ea log zea logw = ea(logjzj+iArg z+i2�m)ea(logjwj+iArgw+i2�n) =

= ea(logjzj+logjwj+iArg z+iArgw+i2�(m+n)) = ea(logjzwj+iArg(zw)+i2�k+i2�(m+n)) =

= ea(logjzwj+iArg(zw)+i2�(k+m+n)) =

= ea log(zw) 2 (zw)a

We have that (zw)a = zawa as sets

67

I.7.5Find iii. Show that is does not coincide with ii�i = i�1.

SolutionWe rewrite the expression ii, suppose that k = 0;�1;�2; : : :

ii =

= ei log i = ei(logjij+iArg(i)+i2�k) = ei(i�=2+i2�k) =

= e�2�k��=2:

Now we can �nd iiiwe use that we have ii, suppose that k;m = 0;�1;�2; : : :

iii

=

=�ii�i= ei log(i

i) = ei(log(e�2�k��=2)) = ei(logje�2�k��=2j+iArg(e�2�k��=2)+i2�m) =

= ei(�2�k��=2+i2�m) = e�2�mei(�2�k��=2) = �ie�2�m == �ie2�m:

Which not coincide with ii�i = i�1 = 1=i = �i which is smaller thanf�ie2�m;m = 0;�1;�2; : : :g.

68

I.7.6Determine the phase factors of the function za (1� z)b at the branchpoints z = 0 and z = 1. What conditions on a and b guarantee thatza (1� z)b can be de�ned as a (continuous) single-valued functionon Cn [0; 1]?

SolutionPhase factors is e2�ia at 0 and e2�ib at 1. Require e2�iae2�ib = 1, or a+ b = n,where n 2 Z , to have a continuous single�valued determination of za (1� z)bon Cn [0; 1].

69

I.7.7Let x1 < x2 < � � � < xn be n consecutive points on the real axis.Describe the Riemann surface of

p(z � x1) � � � (z � xn). Show that

for n = 1 and n = 2 the surface is topologically a sphere with certainpunctures corresponding to the branch points and 1. What is itwhen n = 3 or n = 4? Can you say anything for general n? (Anycompact Riemann surface is topologically a sphere with handles.Thus a torus is topologically a sphere with one handle. For a givenn, how many handles are there, and where do they come from?)

SolutionFor n = 1 and n = 2, the functions f1 =

pz � x1 and f2 =

p(z � x1) (z � x2),

and for a general n we have fn (x) =p(z � x1) (z � x2) : : : (z � xn). To con-

stuct the Riemann Suface we use two sheets with slits [x1; x2] ; [x3; x4] ; : : :.If n is odd, we also need a slit [xn;+1) . Identify top edge of slit on onesheet with bottom edge of slit on other sheet. If n = 1 and n = 2 the surfaceis topologically a sphere with certain punctures corresponding to the branchpoints and 1. If n = 3 or n = 4, surface is a torus. For a general evenn, fn (x) will be continuous on Cn ([x1; x2] [ [x3; x4] [ : : : [ [xn�1; xn]), andits Riemann surface will be a sphere with n

2� 1 holes. For a general odd n

though, fn (x) will be continuous on Cn ([x1; x2] [ [x3; x4] [ : : : [ [xn;+1]),its Riemann surface will be a sphere with n�1

2holes.

70

I.7.8Show that

pz2 � 1=z can be de�ned as a (single-valued) continuous

function outside the unit disk, that is, for jzj > 1. Draw branchcuts so that the function can be de�ned continuous o¤ the branchcuts. Describe the Riemann surface of the function.

SolutionThe function is z

p1� 1=z3. If jzj > 1, can use the principal value of the

square root to de�ne a branch of the function. There are branch points atz = 0 and z3 = 1, that is at 0; 1; e2�i=3 and e�2�i=3. Make two branch cuts byconnecting any two pairs of point by curves; for instance, connect 0 to 1 bya straight line, and the other cube roots of unity by a straight line or arc ofunit circle. The resulting two-sheeted surface with identi�cation of cuts andwith points at in�nity is a torus.

71

I.7.9Consider the branch of the function

qz (z3 � 1) (z + 1)3 that is posi-

tive at z = 2. Draw branch cuts so that this branch of the functioncan be de�ned continuously o¤ the branch cuts. Describe the Rie-mann surface of the function. To what value at z = 2 does thisbranch return if it is continued continuously once counterclockwisearound the circle fjzj = 2g?

SolutionWe have that the function is

(z + 1) z2pzp(1� 1=z3) (1 + 1=z):

If jzj > 1, the second square root can be de�ned to be single-valued forjzj > 1. The value of the function at z = 2 returns to the negative of theirinitial value then we travers the circle jzj = 2 , because

pz does. We can

construct a Riemann Surface by making cuts at �1; 0; 1; e2�i=3; e�2�i=3 and1. The function for the Riemann Surface is

� (z + 1)qz (z � 1) (z + 1) (z � e2�i=3) (z � e�2�i=3):

72

I.7.10Consider the branch of the function

qz (z3 � 1) (z + 1)3 (z � 1) that is

positive at z = 2. Draw branch cut so that this branch of the func-tion can be de�ned continuously o¤ the branch cuts. Describe theRiemann surface of the function. To what value at z = 2 does thisbranch return if it is continued continuously once counterclockwisearound the circle fjzj = 2g?

SolutionWe have that the function is

z4q(1� 1=z3) (1 + 1=z)3 (1� 1=z)

The branch that is positive for z = 2 is, it is de�ned continuously for jzj > 1.Branch return to original value around circle jzj = 2. We can construct aRiemann Surface by making cuts at 0;�1; e2�i=3 and e�2�i=3.The function for the Riemann Surface is

� (z � 1) (z + 1)qz (z + 1) (z � e2�i=3) (z � e�2�i=3):

73

I.7.11Find the branch points of 3

pz3 � 1 and describe the Riemann surface

of the function.

Solution

­1 1

­1

1

I.7.11a

­1 1

­1

1

I.7.11b

­1 1

­1

1

I.7.11c

We rewrite as follows

3pz3 � 1 = 3

q(z � 1) (z � e2�i=3) (z � e�2�i=3):

This equation will have 3 branch points, which are the cube root of unity.So the phase factor ise2�i=3. The Riemann surface is obtained ty pasting three sheets with thecorresponding branch cuts, we end up with a one hole torus.Make two cuts, from 1 to e�2�i=3, on each sheet. In this case the cuts sharecommon endpoint.We nee three shets where f0 (z), f1 (z) = e2�i=3f0 (z) and f2 (z) = e�2�i=3f0 (z)Note: Can use Riemann�formula to see that surface is a torus.Cheek by going around each little tip what phase change to, which of foryour sheet.

74

I.8.1Establish the following addition formulae(a) cos (z + w) = cos z cosw � sin z sinw;(b) sin (z + w) = sin z cosw + cos z sinw;(c) cosh (z + w) = cosh z sinhw + sinh z sinhw;(d) sinh (z + w) = sinh z coshw + cosh z sinhw;

Solution(a)Using the de�nitions of sine and cosine functions given on page 29 in CA wehave,

cos (z + w) =

=ei(z+w) + e�i(z+w)

2=1

4

�2ei(z+w) + 2e�i(z+w)

�=

=1

4

��ei(z+w) + ei(z�w) + e�i(z�w) + e�i(z+w)

�+

+�ei(z+w) � ei(z�w) � e�i(z�w) + e�i(z+w)

��=

=eiz + e�iz

2

eiw + e�iw

2� e

iz � e�iz2i

eiw � e�iw2i

=

= cos z cosw � sin z sinw:

(b)Using the addition formula (a), and using that sin z = cos (z � �=2) andcos z = � sin (z � �=2) we have,

sin (z + w) = cos�z + w � �

2

�=

= cos z cos�w � �

2

�� sin z sin

�w � �

2

�=

= cos z sinw + sin z cosw = sin z cosw + cos z sinw:

(c)Using the addition formula (a) and the formulas cos (iz) = cosh z and sin (iz) =i sinh z given on page 30 in CA we have,

75

cosh (z + w) = cos (i (z + w)) = cos (iz) cos (iw)� sin (iz) sin (iw) == cosh z � coshw � i sinh z � i sinhw = cosh z coshw + sinh z sinhw:

(d)Using the addition formula (b) and the formulas cos (iz) = cosh z and sin (iz) =i sinh z given on page 30 in CA we have,

sinh (z + w) = �i sin (i (z + w)) = �i [sin (iz) cos (iw) + cos (iz) sin (iw)] == �i [i sinh z � coshw + cosh z � i sinhw] = sinh z coshw + cosh z sinhw:

76

I.8.2Show that jcos zj2 = cos2 x + sinh2 y, where z = x + iy. Find all zerosand periods of cos z.

Solution.Use trigonometric formulas from page 29 and 30 in CA we have,

cos z = cos (x+ iy) = cosx cos (iy)�sin x sin (iy) = cosx cosh y�i sin x sinh y:

Now take the modulus squared, and use cosh2 y = 1 + sinh2 y,

jcos zj2 = cos2 x cosh2 y + sin2 x sinh2 y == cos2 x

�1 + sinh2 y

�+ sin2 x sinh2 y =

= cos2 x+�cos2 x+ sin2 x

�sinh2 y = cos2 x+ sinh2 y:

The identity for jcos zj2 shows that the only zeros of cos z are the zeros ofcos z on the real axis, because

cos z = 0,�cosx = 0 , x = �

2+m�; m = 0;�1;�2; : : : ;

sinh y = 0 , y = 0:

Translation by any period � of cos z sends zeros to zeros. Thus any periodis an integral multiple of �, and since odd integral multiples are not periods,the only periods of cos z are 2�n; �1 < n <1.

77

I.8.3Find all zeros and periods of cosh z and sinh z.

SolutionWe have that cosh z = cos (iz), thus the zeros of cosh z are at z = i�=2 +im�,m = 0;�1;�2; : : :, and the periods of cosh z are 2�mi, m = 0;�1;�2; : : :.

We have that sinh z = �i sin (iz), thus the zeros of sinh z are at z = m�i,m = 0;�1;�2; : : :, and the periods of cosh z are 2�mi, m = 0;�1;�2; : : :.

78

I.8.4Show that

tan�1 z =1

2ilog

�1 + iz

1� iz

�;

where both sides of the identity are to be interpreted as subsets ofthe complex plane. In other words, show that tanw = z if and onlyif 2iw is one of the values of the logarithm featured on the right.

SolutionSet z = tanw, we have that

z = tanw =sinw

cosw=

eiw � e�iwi (eiw + e�iw)

=e2iw � 1i (e2iw + 1)

:

Solve for e2iw

e2iw =1 + iz

1� iz ;

and take logarithm and solve for w

w =1

2ilog

�1 + iz

1� iz

�:

Because z = tanw then w 2 tan�1 z and we have the identity

tan�1 z =1

2ilog

�1 + iz

1� iz

�:

79

I.8.5Let S denote the two slits along the imaginary axis in the complexplane, one running from i to +i1, the other from �i to �i1. Showthat (1 + iz) = (1� iz) lies on the negative real axis (�1; 0] if andonly if z 2 S. Show that the principal branch

Tan�1 z =1

2iLog

�1 + iz

1� iz

�maps the slit plane CnS one-to-one onto the vertical strip fjRewj < �=2g.

Solution

i

­i

I.8.5 z­plane

­1

I.8.5

−π

π

I.8.5

w1 =1+iz1�iz w2 = Logw1

−π/2 π/2

I.8.5 w­plane

w = 12iw2 =

12iLog

�1+iz1�iz

�We begin to show that 1+iz

1�iz 2 (�1; 0] if and only if z 2 (�i1;�i][ [i; i1).Set

1 + iz

1� iz = w1

and solve for z, thus

80

z =1� w11 + w1

i

As w1 goes from �1 to �1, 1�w11+w1

i goes from �i to �i1 along iR, and whenw1 goes from �1 to 0, 1�w1

1+w1i goes from i1 to i along iR.

Since we have that

1 + iz

1� iz = w1 ) z =1� w11 + w1

i

the map is one to one.Remark that the map makes correspondence between the interval [�i:� i1]in z � plane and [�1;�1] in w1 � plane, and between interval [i; i1] in z� plane and [�1; 0] in w1 � plane.We have that w2 = Logw1 maps Cn(�1; 0) onto fjImwj < �g, thus w = w2

2i

maps jImw2j < � onto�jRewj < �

2

, se �gures.

We have that the function

Tan�1 z =1

2iLog

�1 + iz

1� iz

�maps CnS onto

�jRewj < �

2

, where Tan�1 z is the principal branch for

tan�1 z. The other branches are given by fn (z) = Tan�1 z + n�, �1 < n <1.

81

I.8.6Describe the Riemann surface for tan�1 z.

Solution

i

­i

+ + 

+ +

+ + 

+ +

_ _ 

_ _

_ _ 

_ _

I.8.6 z­plane

­1+ + + + + +_ _ _ _ _ _

I.8.6

−π

π+ + + + + + + + + + +

_ _ _ _ _ _ _ _ _ _ _

I.8.5

w1 =1+iz1�iz w2 = Logw1

−π/2 π/2

+ + 

+ + 

+ + 

+ + 

+ + 

+

_ _ 

_ _ 

_ _ 

_ _ 

_ _ 

_

I.8.6 w­plane

w = 12iw2 =

12iLog

�1+iz1�iz

�We have that

f0 (z) = Tan�1 z =

1

2iLog

�1 + iz

1� iz

�;

other branches of tan�1 z are fn (z) = f0 (z) + n�, where �1 < n <1.Use one copy of the double slit plane S for each integer n, and de�ne fn (z) =Tan�1 z + n� on the nth sheet to the (n+ 1) th sheet along one of the cuts,so that fn (z) and fn+1 (z) have same value at the junction.

82

Make in�nite many copies of S and call them Sn. De�ne fn (z) = Tan�1 z+n�on Sn. Identify "+" side of cut on Sn to "�" side of cut on Sn+1. Thenfn on Sn continuous to fn+1 on Sn+1, and fn maps Sn onto vertical strip�n� 1

2< Re z < n+ 1

2

. Note that composite function is not de�ned at

�i and �1 (endpoints of slits), and its image omits the sequence 12+ n,

�1 < n <1.

83

I.8.7Set w = cos z and � = eiz. Show that � = w �

pw2 � 1. Show that

cos�1w = �i loghw �

pw2 � 1

i;

where both sides of the identity are to be interpreted as subsets ofthe complex plane.

SolutionSet w = cos z and � = eiz, we have that

w = cos z =eiz + e�iz

2=� + 1=�

2:

Solve for �

� = w �pw2 � 1;

and set � = eiz and take the logarithm and solve for z

z = �i log�w �

pw2 � 1

�:

Because w = cos z then z 2 cos�1w we have the identity

cos�1w = �i log�w �

pw2 � 1

�:

84

I.8.8Show that the vertical strip jRe (w)j < �=2 is mapped by the functionz (w) = sinw one-to-one onto the complex z� plane with two slits(�1;�1] and [+1;+1) on the real axis. Show that the inversefunction is the branch of sin�1 z = �iLog

�iz +

p1� z2

�obtained by

taking the principal value of the square root. Hint. First showthat the function 1 � z2 on the slit plane omits the negative realaxis, so that the principal value of the square root is de�ned andcontinuous on the slit plane, with argument in the open intervalbetween ��=2 and �=2.

SolutionSet w = x+ iy, by formulas on page 30 in CA we have that

sinw = sin (x+ iy) = sin (x) cos (iy) + cos (x) sin (iy) =

= sinx cosh (y) + i cos (x) sinh y:

If jRewj = jxj < �=2 then sinw is mapped on Cn (�1;�1] [ [+1;+1),because

Im (sinw) = 0) y = 0) sinw = sin (x)

and

�1 < sin (x) < 1for these x.Let now z = sinw. The value of w is referred to as sin�1 z, that is, thecomplex number whose sine is z. Note that

z =eiw � e�iw

2i: (1)

Now with � = eiw and 1=� = e�iw in (1), we have

z =� � 1=�2i

:

Mulitplying the above by 2i� and doing some rearranging, we �nd that

85

2iz� = �2 � 1 or �2 � 2iz� � 1 = 0:With the quadratic formula, we solve this equation for � and �nd

� = iz +p1� z2 or eiw = iz +

p1� z2

We now take the logarithm of both sides of the last equation and divide theresult by i to obtain

w =1

ilog�iz +

p1� z2

�and, since w 2 sin�1 z,

sin�1 z = �i log�iz �

p1� z2

�:

86

II 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 191 X X X X X X X X X X X X X X X X2 X X X X X X3 X X X X X4 X X X X X5 X X X X X X X X678

1

II.1.1Establish the following:(a) lim

n!1nn+1

= 1 (c) limn!1

2np+5n+1np+3n+1

= 2; p > 1

(b) limn!1

nn2+1

= 0 (d) limn!1

zn

n!= 0; z 2 C:

Solution(a)

limn!1

n

n+ 1= lim

n!1

1

1 + 1=n= 1:

(b)

limn!1

n

n2 + 1= lim

n!1

1=n

1 + 1=n2= 0:

(c)

limn!1

2np + 5n+ 1

np + 3n+ 1= lim

n!1

2 + 5n1�p + n�p

1 + 3n1�p + n�p= 2:

(d)Let m be a positive integer so that m > 2 jzj and let n > m, then

0 �����znn!���� = jzjm jzjn�m

m! � (m+ 1) � (m+ 2) � � � � � n =

=jzjm

m!� jzjm+ 1

� jzjm+ 2

� � � � � jzjn� jzj

m

m!� jzjm� jzjm� � � � � jzj

m�

� jzjm

m!� 12� 12� � � � � 1

2=jzjm

m!� 1

2n�m! 0

as n ! 1, it follows by instängning that�� znn!

�� ! 0 as n ! 1 as was to beshown.

2

II.1.2For which values of z is the sequence fzng1n=1 bounded? For whichvalues of z does the sequence converge to 0?

SolutionThe sequence is bounded for jzj � 1, and the series converge to 0 for jzj < 1.

3

II.1.3Show that fnnzng converges only for z = 0.

SolutionIf we choose

n >2

jzjthen

2 < n � jzjand

jnnznj = jn � zjn > 2n !1when n!1 for z 6= 0.

4

II.1.4Show that lim

N!1N !

Nk(N�k)! = 1; k � 0.

SolutionWe have that

N !

Nk (N � k)! =N (N � 1) : : : (N � k + 1)

N �N � : : : �N =

�1� 1

N

��1� 2

N

�: : :

�1� k � 1

N

�! 1

as N !1 (k is �xed).

5

II.1.5Show that the sequence

bn = 1 +1

2+1

3+ � � �+ 1

n� log n; n � 1;

is decreasing, while the sequence an = bn � 1=n is increasing. Showthat the sequences both converge to the same limit . Show that12< < 3

5. Remark. The limit of the sequence is called Euler�s

constant. It is not known whether Euler�s constant is a rationalnumber or an irrational number.

SolutionWe have that

bn = 1 +1

2+1

3+ � � �+ 1

n� 1 +1

n� log n;

an = 1 +1

2+1

3+ � � �+ 1

n� 1 � log n:

We trivially have that

log n =

Z n

1

1

tdt ((1))

An elementary estimation gives

1

k + 1<

k+1Zk

1

tdt <

1

k((2))

We take the di¤erence an+1 � an and by (1) and (2) we have that

an+1 � an =1

n� log (n+ 1) + log n = 1

n�

n+1Zn

dt

t> 0;

thus an+1 > an, so the sequence an is increasing.We take the di¤erence bn+1 � bn and by (1) and (2) we have that

6

bn+1 � bn =1

n+ 1� log (n+ 1) + log n = 1

n+ 1�

n+1Zn

dt

t< 0;

thus bn+1 < bn, so the sequence bn is decreasing.We have after some investigation on the increasing sequence an that

an � a7 =49

20� log 7 > 1

2; n � 7;

thus 12< an, if n � 7.

We have after some investigation on the decreasing sequence bn that

bn � b22 =19 093 197

5173 168� log 22 < 3

5; n � 22

thus bn � 3=5, if n � 22.The both sequences converge to the same limit because bn�an = 1=n havethe limit 0 as n!1. And this limit is in the interval 1

2< < 3

5.

7

II.1.6For a complex number �, we de�ne the binomial coe¢ cient "�choose n" by�

�0

�= 1;

��n

�=� (�� 1) � � � (�� n+ 1)

n!; n � 1:

Show the following

(a) The sequence��n

�is bounded if and only if Re� � �1.

(b)��n

�! 0 if and only if Re� > �1.

(c) If � 6= 0; 1; 2; : : :, then�

�n+ 1

����n

�! �1.

(d) If Re� � �1, � 6= �1, then����� �n+ 1

����� > ����� �n

����� for all n � 0.(e) If Re a > �1 and � is not an integer, then

����� �n+ 1

����� < ����� �n

�����for n large

Solution(a)We have that

����� �n

����� = nYk=1

����1� �+ 1k���� = 1Y

k=1

=

s1� 2 (1 + Re�)

k+(1 + Re�)2

k2+(1 + Im�)2

k2:

Let

tn =(1 + Re�)2

k2+(1 + Im�)2

k2� 2 (1 + Re�)

k

If Re� < �1, we have that tk > 0 and����n

���!1 then n!1 becausePtk

diverge is��n

�unlimited.

If Re� = �1, we have that tk = (Im�)2

k2and

����n

��� ! A then n ! 1, andA 6= 0 becauxe for all k we have that

��1� �+1k

�� > 1 andP tk =P (Im�)2

k2<1

so��n

�is limited.

8

If Re� > �1 we have that it exists k 2 N so that �1 < tk < 0 for all k > K.This gives that

����n

��� is a decreasing series with respect to n, and so limited.The conclution is that

��n

�is limited if and only if Re� � �1.

b)We have that

��n

�! 0 if and only if

����n

��� ! 0. From (a) we see thatthe only possibility for this is Re� > �1. For Re� > �1 we can see that����n

��� � 0 is decreasing. LetM = limn!1

��n

�. Then we hav thatM = lim

n!1

��n

�=

limn!1

��n

����nn+1

�= �M so M = �M () M = 0 so

��n

�! 0, then n ! 1 if

and only if Re� > �1.

c)If � 6= 0; 1; 2; : : :, then

��n

�6= 0 and we can divide, to get�

�n+1

���n

� =

�(��1):::(��(n+1)+1)(n+1)!

�(��1):::(��n+1)n!

=�� nn+ 1

=�n� 1

1 + 1n

! �1

as n!1.

d)j( �n+1)jj(�n)j

= j��njn+1

= distance(�;n)distance(�1;n) > 1, if Re� � �1, � 6= �1

e)Let � be �xed and Re� > �1. Then

j( �n+1)jj(�n)j

=q

(Re��n)2+(Im�)2

(n+1)2< 1 for n > j�j2�1

2(1+Re�):

Parts (a) and (b) seems to require some facts about products:

nQk=1

�1 + �

k

�!1 as n!1, � > 0

nQk=1

�1� �

k

�!1 as n!1, � > 0

nQk=1

�1� �

k2

�! A 6= 0 as n!1

9

II.1.7De�ne x0 = 0, and de�ne by induction xn+1 = x2n+

14for n � 0. Show

that xn ! 12.

Hint. Show that the sequence is bounded and monotone, and that any limitsatis�es x = x2 + 1

4.

SolutionWe �rst prove by induction that for every integer n � 0 we have

0 � xn �1

2and xn � xn+1:

For n = 0, we have x0 = 0 and x1 = x20 +14= 1

4and so

0 � x0 �1

2and x0 = 0 �

1

4= x1:

Hence, the assertion holds for n = 0. Now suppose that the assertion holdsfor n = k, that is,

0 � xk �1

2and xk � xk+1:

For 0 � x � 12, it is straightforward to show that x � x2 + 1

4� 1

2. Since

0 � xk � 12and xk+1 = x2k +

14we have

0 � xk � xk+1 = x2k +1

4��1

2

�2+1

4=1

2

and hence 0 � xk+1 � 12, and

xk+1 � x2k+1 +1

4= xk+2:

Hence, the inductive step holds and so the sequence fxngn is bounded andincreasing. It follows by Bounded Monotone Sequence Theorem that itconverges, say xn ! x. Note that the real-valued function f given byf (t) = t2 + 1

4for t 2 R is continuous and therefore we have,

x = limn!1

xn+1 = limn!1

f (xn) = f (x) = x2 +

1

4:

It follows that x = x2 + 14and so x = 1

2. Hence, xn ! 1

2as required.

10

II.1.8Show that if sn ! s, then jsn � sn�1j ! 0.

SolutionWe use standard "=2 proof and let sn ! s.Let " > 0, choose N such that

jsn � sj < "=2for n > N .If n > N , then we have

jsn+1 � snj = jsn+1 � s+ s� snj � jsn+1 � sj+ jsn � sj < "=2 + "=2 = ":

Thus

jsn+1 � snj ! 0

as n!1.

11

II.1.9Plot each sequence and determine its lim inf and lim sup.(a) sn = 1 +

1n+ (�1)n (c) sn = sin (�n=4)

(b) sn = (�n)n (d) sn = xn (x 2 R �xed)

Solution(a)lim sup sn = 2 lim inf sn = 0(b)lim sup sn =1 lim inf sn = �1(c)lim sup sn = 1 lim inf sn = 1(d)

lim sup sn =

8<:+1; jxj > 1;1; jxj = 1;0; jxj < 1:

and

lim inf sn =

8>>>><>>>>:�1; �1 < x < �1;�1; x = �1;0; jxj < 1;1; x = 1;+1; x > 1:

12

II.1.10At what points are the following function continuous? Justify youranswer.(a) z (c) z2= jzj(b) z= jzj (d) z2= jzj3

Solution(a)continuous everywhere(b)Continuous except at z = 0, where it is not de�ned(c)Continuous for jzj 6= 0. It is not de�ned at z = 0, but it has a limit 0 asz ! 0. If we de�ne the function to be 0 at z = 0, it is contionuous there.(d)Contionuous for jzj 6= 0. It has no limit as z ! 0

13

II.1.11At what points does the function Arg z have a limit? Where is Arg zcontinuous? Justify your answer.

SolutionArg z have a limit at each point of Cn (�1; 0]. It is continuous in Cn (�1; 0].It is discontinuous at each point of (�1; 0]. Values of function ! f��g atpoints of (�1; 0], the function may take any value in intervall [��; �] asz ! 0. The value is depending in that direktion we approach z = 0.

14

II.1.12Let h (z) be the restriction of the function Arg z to the lower half-plane fIm z < 0g. At what points does h (z) have a limit? What isthe limit?

SolutionNo limit at 0, has limit at all other points of closed lower half-plane, limit= �� at points of (�1; 0).

15

II.1.13For which complex values of � does the principal value of z� havea limit as z tends to 0? Justify your answer.

SolutionSet z = rei� = eLog rei� = eLog r+i� and � = Re�+ i Im�, then

z� =�eLog r+i�

�Re�+i Im�= eLog rRe�+iLog r Im�+i�Re��� Im� =

= eLog rRe�e�� Im�ei(Log r Im�+�Re�) = rRe�e�� Im�ei(Log r Im�+�Re�);

thus

jz�j = rRe�e�� Im�:Because e�� Im� is bounded we look at rRe� in three intervalsIf Re� < 0, rRe� ! 1 as r ! 0 thus jz�j have no limit as z ! 0, and notz� either.If Re� = 0, we get z� = e�� Im�, have no limit unless Im� = 0, thus we havelimit if � = 0.If Re� > 0, rRe� ! 0 as r ! 0 thus jz�j have limit 0 as z ! 0, and z� ! 0as z ! 0.The conclution is z� ! 0 if Re� > 0, and z� ! 1 if � = 0, otherwise wehave no limit at 0.

16

II.1.14Let h (t) be a continuous complex-valued function on the unit in-terval [0; 1], and consider

H (z) =

Z 1

0

h (t)

t� zdt:

Where is H (z) de�ned? Where is H (z) continuous? Justify youranswer. Hint. Use the fact if jf (t)� g (t)j < " for 0 � t � 1, thenR 10jf (t)� g (t)j dt < ".

SolutionWe have that H (z) is de�ned for z 2 Cn [0; 1]. If h (z1) = 0 for somez1 2 [0; 1]. Then H (z) is also de�ned for z = z1.

Then H(z) =1R0

h(t)t�z dz is continuous for z 2 Cn [0; 1]. If zn ! z 2 Cn [0; 1],

then h(t)t�zn !

h(t)t�z uniformly for 0 � t � 1, so the used proof shows that

H (zn)! H (z).

17

II.1.15Which of the following sets are open subsets of C?Which are close?Sketch the sets.(a) The punctured plane, Cn f0g .(b) The exterior of the open unitdisk in the plane, fjzj � 1g .(c) The exterior of the closed unitdisk in the plane, fjzj > 1g .(d) The plane with the open unitinterval removed, Cn (0; 1) .(e) The plane with the closed unitinterval removed, Cn [0; 1] .(f) The semidisk, fjzj < 1; Im (z) � 0g .(g) The complex plane, C.

Solution(a) open (c) open (e) open (g) both open and closed(b) closed (d) neither (f) neitherNote: Sketches are di¢ cult to do reasonably.

18

II.1.16Show that the slit plane Cn (�1; 0] is star-shaped but not convex.Show that the slit plane Cn [�1; 1] is not star-shaped. Show that apunctured disk is not star-shaped.

Solution(a) Cn (�1; 0] is star�shaped, because we can see every point from z = 1.To show that Cn (�1; 0] is not convex we may choose any two points suchthat the stright line segment joining them contains a point in (�1; 0]. Forexample take �1 + i and �1 � i. These two points are joined by a verticalline which contain �1. From this follows that Cn (�1; 0] is not convex.(b) Cn [�1; 1] is not star�shaped. Given any z we have that �z can not beeseen from a stright line between z and �z that must contain 0. We also havethat if z belong to the domain then �z belongs also to the domain.(c) Same argument as in (b) shows that Cn f0g is not star�shaped.

19

II.1.17Show that a set is convex if and only if it is star-shaped with respectto each of its points.

Solution (A. Kumjian)Let D � C, since the empty set trivially satis�es both conditions we willassume that D 6= ?.Suppose �rst that D is convex. We must show that D is star-shaped withrespect to each of its points. So �x z0 2 D, to show that D is star-shapedwith respect to z0. We must show that for every point z 2 D the line segmentconnecting z0 to z is contained in D. Let z 2 D be given, then since D isconvex it contains the line segment joining z0 and z. Hence, D is star-shapedwith respect to z0, then, since z0 was chosen arbitrarily, D is star-shapedwith respect to each of its points.

Conversely, suppose that D is star-shaped with respect to each of its points.To show that D is convex we must show that given two points in D, theline segment joining them is also contained in D. So let z0; z1 2 D be given.Then since D is star-shaped with respect to z0, D contains the line segmentjoining z0 and z1. Hence, D is convex.

20

II.1.18Show that the following are equivalent for an open subset U of thecomplex plane.(a)Any two points of U can be joined by a path consisting of straightlinesegments parallel to the coordinate axis.(b)Any continuously di¤erentiable function h (x; y) on U such thatrh =0 is constant.(c)If V and W are disjoint open subsets of U such that U = V [W ,then either U = V or U = W . Remark. In the context of topologicalspaces, this latter property is taken as de�nition of connectedness.

SolutionShow that the following equivalent for an open subset U of the complex plane.(a) ) (b) Suppose rh = 0 in D. Fix z0 2 D. If z1 2 D, join by polygonalcurve. rh = 0) h is constant on each segment of curve ) h (z0) = h (z1).) h is constant in D.(b) ) (c) Suppose rh = 0, h not constant, say h (z0) = 0 for some z0. LetW = fh = 0g, V = fh 6= 0g. Evidently V is open. Since rh = 0, h isconstant in a neighborhood of each point, so fh = 0g is open andW is open.W 6= ?, V 6= ?, W \ V = ?, W [ V = U(c) ) (a) Suppose z0 2 U . Let D = points that can be joined to z0 by apolygonal curves, intervals parallel to coordinate axis. Evidently D is open.Also if can connected points near z0 2 U , then can connect to U , so UnD isopen. By (c), UnD must be empty, some D 6= ?:

21

II.1.19Give a proof of the fundamental theorem of algebra along the fol-lowing lines. Show that if p (z) is a nonconstant polynomial, thenjp (z)j attains its minimum at some point z0 2 C. Assume thatthe minimum is attained at z0 = 0, and that p (z) = 1 + azm + � � � ,where m � 1 and a 6= 0. Contradict the minimality by showing that��P �"ei�0��� < 1 for an appropriate choice of �0.Solution (K. Seip)Set

p (z) = anzn + an�1z

n�1 + � � �+ a0;where we assume an 6= 0. Since

limjzj!1

jp (z)jjzjn = janj ;

9 R > 0 such that jp (z)j > ja0j for all jzj > R. Thus jp (z)j attainsits minimum in jzj � R. We may assume it is attained at z0 = 0.Suppose a0 6= 0, say a0 = 1. Then p (z) = 1 + azm+ higher order

terms, a 6= 0. Choose z = "�� 1a

� 1m (any m � th root does the job).

Then

p (z) = 1� "m +O�"m+1

�:

Thus for " su¢ ciently small " we have jp (z)j < 1, which is a contra-diction.

22

Follow the suggestion.Let

p (z) = a0 + a1z + am+1zm+1 + :::+ aNz

N ; aN 6= 0:Choose R so that

jaN jRN > ja0j+ ja1jR + jam+1jRm+1 + :::+ jaN�1jRN�1:Then p (z) 6= 0 for jzj = R, the term aNz

N dominates and further jp (z)j >jp (0)j for jzj = R.Let z0 be a point in the disk jzj 6 R at which the continuous function forjp (z)j attains a minimum. Then jz0j < R, so that is a disk centred at z0 sothat jp (z)j > jp (z0)j on the disk. Suppose p (z0) 6= 0. We can assume thatp (z0) = 1.Write

p (z) = 1 + amP (z � z0)m +O�(z � z0)m+1

�;

where am 6= 0. Suppose am = r0ei�0.Conside

zm = z0 + "e�i�0=me�i�=m; " > 0:

We have

p (zm) = 1 + r0ei�0eme�i�0e�i� +O

�"m+1

�= 1� r0"m +O

�"m+1

�:

Have p (zm) � 1 � r0"m < 1 for " > 0 small. Contradiction! We concludethat p (z0) = 0.

23

II.2.1Find the derivatives of the following function.(a) z2 � 1 (c) (z2 � 1)n (e) 1= (z2 + 3) (g) (az + b) = (cz + d)

(b) zn � 1 (d) 1= (1� z) (f) z= (z3 � 5) (h) 1= (cz + d)2

Solution(a) 2z (c) n (z2 � 1)n�1 2z (e) �2z= (z2 + 3)2 (g) (ad� bc) = (cz + d)2

(b) nzn�1 (d) 1= (1� z)2 (f) (�2z3 � 5) = (z3 � 5)2 (h) �2c= (cz + d)3

24

II.2.2Show that

1 + 2z + 3z2 + � � �+ nzn�1 = 1� zn

(1� z)2� nzn

1� z :

SolutionUse the geometric sum

1 + z + z2 + z3 + � � �+ zn = 1� zn+11� z ; z 6= 1:

Di¤erentiate both sides

1 + 2z + 3z2 + � � �+ nzn�1 =

=(n+ 1) (�1) zn (1� z) + (1� zn+1)

(1� z)2=

=nzn+1 � nzn + zn+1 � zn + 1� zn+1

(1� z)2=

=1� zn

(1� z)2+nzn+1 � nzn

(1� z)2=1� zn

(1� z)2� nzn

(1� z) :

25

II.2.3Show from the de�nition that the functions x = Re z and y = Im zare not complex di¤erentiable at any point.

SolutionDi¤erentiation f (z) = x = Re z (from the de�nition)

lim�z!0

f (z +�z)� f (z)�z

= lim�z!0

Re (z +�z)� Re (z)�z

=

= lim�z!0

Re�z

�z=

�1; if �z = �x0; if �z = i�y

:

Di¤erentiation g (z) = y = Im z (from the de�nition)

lim�z!0

g (z +�z)� g (z)�z

= lim�z!0

Im (z +�z)� Im (z)�z

=

= lim�z!0

Im�z

�z=

�0; if �z = �x�i; if �z = i�y :

The functions x = Re z and y = Im z are not complex di¤erentiable at anypoint, because the functions have di¤erent limit as �z ! 0 through real andimaginary axis.

26

II.2.4 (+ IV.8.2)Suppose f (z) = az2 + bz�z + c�z2, where a, b, and c are �xed complexnumbers. By di¤erentiating f (z) by hand, show that f (z) is com-plex di¤erentiable at z if and only if bz + 2c�z = 0. Where is f (z)analytic?

SolutionWe will �nd d

dzf (z) using the limit de�nition

f (z +�z)� f (z)�z

=

=a (z +�z)2 + b (z +�z) (z +�z) + c(z +�z)

2 � (az2 + bz�z + c�z2)�z

=

=a�z2+2z�z +�z2

�+ b�z�z + z�z+�z�z +�z�z

�+ c��z2 + 2�z�z +�z

2�� (az2 + bz�z + c�z2)

�z=

=2az�z + a�z2 + bz�z + b�z�z + b�z�z + 2c�z�z + c�z

2

�z=

= 2az + a�z + bz�z

�z+ b�z + b�z + 2c�z

�z

�z+ c

�z2

�z=

= 2az + a�z + b�z + b�z + c�z�z

�z+ (bz + 2c�z)

�z

�z

We know �z is not analytic at any open set of C and lim4z!0

��z�zdoes not ex-

ist. Therefore, unless bz + 2c�z = 0, the derivative ddzf (z) would not exist.

Therefore f (z) is analytic on C if b = c = 0, otherwise is the function notanalytic on any open set.

27

II.2.5Show that if f is analytic on D, then g (z) = f (�z) is analytic on there�ected domain D� = f�z : z 2 Dg, and g0 (z) = f 0 (�z).

Solution (A. Kumjian)Let f be analytic on the domain D and de�ne g on D� as above (note thatD� is also a domain). For a complex-valued function ' de�ned near a pointz0 it is easy to show that

limz!z0

' (z) exists i¤ limz!z0

' (z) exists;

and if either exists, then the two limits are complex conjugates of each other.Let z 2 D� be given. We �rst show that g is di¤erentiable at z and thatg0 (z) = f 0 (�z).

limh!0

1

h(g (z + h)� g (z)) = lim

h!0

1

h

�f�z + h

�� f (�z)

��= lim

k!0

1

k(f (�z + k)� f (�z)) = f 0 (�z)

where equation (�) follows using the substitution k = �h and basic algebraicproperties of conjugation, note that conjugation is a continuous function soh ! 0 i¤ �h ! 0. Since f is analytic on the domain D, its derivative f 0

is continuous on D. Further, since g0 (z) = f 0 (�z) for all z 2 D�, g0 is thecomposite of continuous functions and therefore is itself continuous. Hence,g is analytic on D�.

28

II.2.6Let h (t) be a continuous complex-valued function on the unit in-terval [0; 1], and de�ne

H (z) =

Z 1

0

h (t)

t� zdt; z 2 Cn [0; 1] :

Show that H (z) is analytic and compute its derivative. Hint. Dif-ferentiate by hand, that is, use the de�ning identity (2.4) of com-plex derivative.

SolutionWe use de�nition (2.4) of derivative to �nd H 0 (z).

H 0 (z) =

= lim�z!0

H (z +�z)�H (z)�z

= lim�z!0

1

�z

�Z 1

0

h (t)

t� z ��zdt�Z 1

0

h (t)

t� zdt�=

= lim�z!0

1

�z

�Z 1

0

h (t)

�1

t� z ��z �1

t� z

��=

= lim�z!0

�Z 1

0

h (t)

(t� z ��z) (t� z)

�=

Z 1

0

h (t)

(t� z) (t� z)dt =

=

Z 1

0

h (t)

(t� z)2dt:

H 0 (z) is continuous in z, also by uniform convergence of integrand H (z) isanalytic for z 2 Cn [0; 1].

29

II.3.1Find the derivatives of the following functions.(a) tan z = sin z

cos z(b) tanh z = sinh z

cosh z(c) sec z = 1= cos z

Solution(a)

d

dztan z =

d

dz

sin z

cos z=cos z � cos z � sin z � (� sin z)

cos2 z=cos2 z + sin2 z

cos2 z=

1

cos2 z:

(b)

d

dztanh z =

d

dz

sinh z

cosh z=cosh z � cosh z � sinh z � sinh z

cosh2 z=cosh2 z � sinh2 z

cosh2 z=

1

cosh2 z:

(c)

d

dzsec z =

d

dz

1

cos z= � 1

cos2 z(� sin z) = sin z

cos2 z= tan z sec z:

30

II.3.2Show that u = sinx sinh y and v = cos x cosh y satisfy the Cauchy-Riemann equations. Do you recognize the analytic function f =u+ iv? (Determine its complex form)

SolutionWe have

@u

@x=

@

@xsin x sinh y = cos x sinh y =

@

@ycosx cosh y =

@v

@y;

@u

@y=

@

@ysin x sinh y = sinx cosh y = � @

@xcosx cosh y = �@v

dx:

Hence, u and v satisfy Cauchy-Riemann equations, and now we calculatef (z).

f (z) =

= u+ iv =

�eix � e�ix

2i

��ey � e�y

2

�+ i

�eix + e�ix

2

��ey + e�y

2

�=

= �i�eix+y � eix�y � e�ix+y + e�ix�iy

4

�+i

�eix+y + eix�y + e�ix+y + e�ix�y

4

�=

= ieix�y + e�ix+y

2= iei(x+iy) + e�i(x+iy)

2= ieiz + e�iz

2=

= i cos z:

31

II.3.31 2 3 P L K

LLLShow that if f and �f are both analytic on a domain D, then f isconstant.

Solution (A. Kumjian)Let D be a domain and let f be a complex valued function such that both fand �f are analytic on D. Then both Re f = 1

2

�f + �f

�and Im f = 1

2i

�f � �f

�must also be analytic on D. Since Re f and Im f are also both real-valued,it follows by the theorem on page 50, that both are constant. Hence, f =Re f + i Im f is constant.

32

II.3.4Show that if f is analytic on a domain D, and if jf j is constant,then f is constant.Hint. Write �f = jf j2 =f .

Solution (with use of hint)If f (z) = 0 for some z 2 D. Then f � 0 on D, since jf j is constant. Assumethat f (z) 6= 0 for every z 2 D. Since f is analytic in D we have that1=f and �f = jf j2 =f (where jf j is constant) are analytic. From this followsthat Re f and Im f are analytic and real-valued and therefore constant. Sof = Re f + i Im f is constant.

33

II.3.4Show that if f is analytic on a domain D, and if jf j is constant,then f is constant.Hint. Write �f = jf j2 =f .

SolutionSet f = u + iv, where u and v are realvalued functions. Now suppose thatjf j = k is constant. We have that u2+ v2 = k2. Di¤erentiate both sides withrespect to x respectively y �

uu0x + vv

0x = 0;

uu0y + vv

0y = 0:

(1)

We use Cauchy-Riemanns equations u0x = v

0y and u

0y = �v

0x in (1) and have�

uu0x � vu

0y = 0;

uu0y + vu

0x = 0:

(2)

In the simultaneusly systems of equations in (2) we �rst �rst multiply the�rst row with u and second row with v in (2) and add them together, andhave (u2 + v2)u

0x = 0. And if we in the same system multiply the �rst row

with �v and the second row with u and add them we have (u2 + v2)u0y = 0.

Thus �(u2 + v2)u

0x = 0;

(u2 + v2)u0y = 0:

(3)

If in (3) u2 + v2 = 0, and we have that u = v = 0 and thus f is constant.Suppose that u2 + v2 6= 0. We have that (3) gives that u

0x = u

0y = 0, with

says that u is constant. Cauchy-Riemanns equations gives�0 = u

0x = v

0y;

0 = u0y = �v

0y;

(4)

thus by (4) f is constant.

34

II.3.5If f = u+ iv is analytic, then jruj = jrvj = jf 0j.

Solution (K. Seip)Because f = u+ iv is analytic. Then

ru =�@u

@x;@u

@y

�=

�@u

@x;�@v@x

�=

�@v

@y;�@v@x

�from which it follows that

jruj = jf 0 (z)j = jrvj :

35

II.3.6If f = u + iv is analytic on D, then rv is obtained by rotating ruby 90�. In Particular, ru and rv are orthogonal.

Solution (A. Kumjian)Let � 2 R then rotation of a vector (x; y) 2 R2 by the angle � (about theorigin) is given by matrix multiplication:

�xy

�7! A�

�xy

�=

�cos � � sin �sin � cos �

��xy

�=

�x cos � � y sin �x sin � + y cos �

�:

Hence, rotation by 90� is given by the transformation (x; y) 7! (�y; x). Bythe Cauchy-Riemann equations we have

rv =�@v

@x;@v

@y

�=

��@u@y;@u

@x

�= A�=2ru:

Thus, rv is obtained by rotating ru by 90�. It follows that ru and rv areorthogonal.

Solution (D. Jakobsson)Given since f is analytic, then

(0.6)@u

@x=@v

@yand

@u

@y= �@v

@x

Since ru =�@u@x; @u@y

�and rv =

�@v@x; @v@y

�all we need to check is whether

rv � rv = 0. To prove that we use (0.6).

ru � rv = @v

@x

@u

@x+@v

@y

@u

@y=@v

@x

@v

@y� @v@y

@v

@x= 0

We conclude that ru and rv are orthogonal.Notice that the operation that sends ru! rv is given by the counter-clock-wise rotation matrix with � = �

2. Keeping in mind (0.6), one can write the

following.

rv =�vxvy

�=

�0 �11 0

��vy�vx

�=

�cos �

2� sin �

2

sin �2cos �

2

��vy�vx

�= R

��2

�ru

36

Remark.We have that

vx =@v

@x; vy =

@v

@y; ux =

@u

@x; uy =

@u

@y;

and

R (�=2) = A�=2:

37

II.3.7Sketch the vector �elds ru and rv for the following functions f = u+ iv.(a) iz (b) z2 (c) 1=z

Solution

­4 ­2 2 4

­4

­2

2

4

x

y

II.3.7a

­4 ­2 2 4

­4

­2

2

4

x

y

II.3.7a

­4 ­2 2 4

­4

­2

2

4

x

yII.3.7a

(a)

f (z) = iz = �y + ix)�u = �yv = x

)�ru = (0;�1)rv = (1; 0)

(b)

f (z) = z2 = x2 � y2 + 2ixy )�u = x2 � y2v = 2xy

)�ru = (2x;�2y)rv = (2y; 2x)

(c)

f (z) =1

z=

z

jzj2=x� iyx2 + y2

)�u = x

x2+y2= cos �

r

v = � yx2+y2

= � sin �r8>>>><>>>>:

@u@x= y2�x2

(x2+y2)2= sin2 ��cos2 �

r2= � cos 2�

r2

@u@y= �2xy

(x2+y2)2= �2 cos � sin �

r2= � sin 2�

r2

@v@x= 2xy

(x2+y2)2= 2 cos � sin �

r2= sin 2�

r2

@v@y= y2�x2

(x2+y2)2= sin2 ��cos2 �

r2= � cos 2�

r2�ru = 1

r2(� cos 2�;� sin 2�)

rv = 1r2(� sin 2�;� cos 2�)

38

II.3.8 (see III.4.2) (36)Derive the polar form of the Cauchy-Riemann equations for u andv,

@u

@r= 1

r@v@�; @u

@�= �r @v

@r:

Check that for any integer m, the functions u�rei��= rm cos (m�)

and v�rei��= rm sin (m�) satisfy the Cauchy-Riemann equations.

SolutionSet x = r cos � and y = r sin �, then take the partial derivative,

@x@r= cos � @x

@�= �r sin �

@y@r= sin � @y

@�= r cos �

Use Cauchy-Riemanns equations and the derivative to get the �rst equation

@u

@r=@u

@x

@x

@r+@u

@y

@y

@r=@u

@xcos � +

@u

@ysin � =

@v

@ycos � � @v

@xsin � =

=1

r

��r @v@xsin � + r

@v

@ycos �

�=1

r

�@v

@x

@x

@�+@v

@y

@y

@�

�=1

r

@v

@�;

Use Cauchy-Riemanns equations and the derivative to get the second equa-tion

@u

@�=@u

@x

@x

@�+@u

@y

@y

@�= �r@u

@xsin �+r

@u

@ycos � = �r@v

@ysin ��r @v

@xcos � =

= �r�@v

@xcos � +

@v

@ysin �

�= �r

�@v

@x

@x

@r+@v

@y

@y

@r

�= �r@v

@r:

we get �@u@r= 1

r@v@�;

@u@�= �r @v

@r:

Set u�rei��= rm cos (m�) and v

�rei��= rm sin (m�) and use it in the

Cauchy-Riemann�s equations,

39

@u

@r=

@

@r(rm cos (m�)) = mrm�1 cos (m�) =

1

rrmm cos (m�) =

1

r

@

@�(rm sinm�) =

1

r

@v

@�;

@u

@�=

@

@�(rm cos (m�)) = �mrm sin (m�) = �rmrm�1 sin (m�) = �r @

@r(rm sin (m�)) = �r@v

@r:

40

II.4.1Sketch the gradient vector �elds ru and rv for(a) u+ iv = ez (b) u+ iv = Log z

Solutiona)We have

u+ iv = ez = ex cos y + iex sin y

and get

ru = (ex cos y;�ex sin y) = ex (cos y;� sin y)(in black), and

rv = (ex sin y; ex cos y) = ex (sin y; cos y)(in red).b)We have

u+ iv = Log z = log r + i�

and get

ru =1

r�!ur ;

rv =1

r�!u� :

FIGURE II.4.1a NF FIGURE II.4.1b NF

41

II.4.2Let a be a complex number a 6= 0, and let f (z) be an analytic branchof za on Cn (�1; 0]. Show that f 0 (z) = af (z) =z. (Thus f 0 (z) = aza�1,where we pick the branch of za�1 that corresponds to the originalbranch of za divided by z.)

SolutionWe may write the branch as follows

f (z) = za = ea log z = ea(Log z+2�im) = eaLog z+2�iam = ceaLog z;

where c = e2�iam for some m 2 Z. It follows by the chain rule that

f 0 (z) = ceaLog za

z=af (z)

z:

So, the desired result follows.

42

II.4.3Consider the branch of f (z) =

pz (1� z) on Cn [0; 1] that has posi-

tive imaginary part at z = 2. What is f 0 (z)? Be sure to specify thebranch of the expression for f 0 (z).

SolutionSet f (z) =

pz (1� z) (principal branch) and di¤erentiate

f 0 (z) =1

2

1pz (1� z)

� ddz(z (1� z)) = 1

2

1� 2zpz (1� z)

:

Branch of f (z) is i� increasing for x > 0 large, thus f 0 (z) is i� positive forx > 0 large.Use branch of f 0 (z) =

pz (1� z) that it is i� positive, i.e., same on f (z).

f 0 (z) =1

2

1� 2zf (z)

:

Solution (D. Jakobsson)Given f (z) =

pz (1� z), we can de�ne w = f (z) and get w2 = z (1� z).

d

dz(1� z) z = d

dzw2

1� 2z = 2wdwdz

One gets

d

dz

pz (1� z) = 1� 2z

2pz (1� z)

and is de�ned on Cn [0; 1]. Take the principal branch with the positive imag-inary part.

43

II.4.4Recall that the principal branch of the inverse tangent functionwas de�ned on the complex plane with two slits on the imaginaryaxis by

Tan�1 = 12iLog

�1+iz1�iz

�; z 62 (�i1;�i] [ [i; i1) :

Find the derivative of Tan�1 z. Find the derivative of tan�1 z forany analytic branch of the function de�ned on a domain D.

SolutionWe have that

d

dzTan�1 z =

1

2i

�1

1 + izi� 1

1� iz (�i)�=1

2

�1� iz + 1 + iz(1 + iz) (1� iz)

�=

1

1 + z2:

Any two branches of tan�1 z di¤er by a constant, so derivatives are same.

44

II.4.5Recall that cos�1 (z) = �i log

�z �pz2 � 1

�. Suppose g (z) is an ana-

lytic branch of cos�1 (z), de�ned on a domain D. Find g0 (z). Dodi¤erent branches of cos�1 (z) have the same derivative?

SolutionWe have that

d

dzcos�1 z =

=�i

z �pz2 � 1

�1� 1

2

�z2 � 1

��1=2 � 2z� ==

�iz �pz2 � 1

�1� zp

z2 � 1

�=

�ipz2 � 1

=

=�1p1� z2

:

Derivatives of branches of cos�1 z are not always the same.

45

II.4.6Suppose h (z) is an analytic branch of sin�1 (z), de�ned on a domainD. Find h0 (z). Do di¤erent branches of sin�1 (z) have the samederivative?

Solution (A. Kumjian)Regard h (z) as a local inverse of f (z) where f (z) = sin z. Then f 0 (z) =cos z, so using the formula given in the statement of theorem on page 51 wehave

h0 (z) =1

f 0 (h (z))=

1

cosh (z)

if cosh (z) 6= 0.One local inverse of f satis�es h1 (0) = 0, while another local inverse of fsatis�es h2 (0) = �. Using the formula above we see

h01 (0) =

1

cos 0= 1 and h

02 (0) =

1

cos �= �1:

Hence, di¤erent branches of sin�1 (z) do not necessarily have the same deriv-ative.

46

II.4.7Let f (z) be a bounded analytic function, de�ned on a boundeddomain D in the complex plane, and suppose that f (z) is one-one-one. Show that the area of f (D) is given by

Area (f (D)) =

ZZD

jf 0 (z)j2 dxdy:

Solution (K. Seip)We have f : D ! C, jf (z)j � M for some M < 1 when z 2 D. Since f isassumed to be one-to-one, we may compute

A (f (D)) =

ZZf(D)

dudv

by the change of variables (u (x; y) ; v (x; y)) ! (x; y), and since det Jf =jf 0 (z)j2, we get

Area (f (D)) =

ZZD

jf 0 (z)j2 dxdy:

47

II.4.8Sketch the image of the circle fjz � 1j � 1g under the map w = z2.Compute the area of the image.

Solution

­4 ­2 2 4

­4

­2

2

4

x

yII.4.8 z­plane

­4 ­2 2 4

­4

­2

2

4

x

yII.4.8 w­plane

We have that f (z) = z2, that gives f 0 (z) = 2z and Jf (z) = 4 jzj2.The area of the image isZZ

Jfdxdy = 4

ZZ(x�1)2+y261

�x2 + y2

�dxdy:

Set t = x� 1, get

4

ZZt2+y261

�t2 + 2t+ 1 + y2

�dtdy =

= 4

264 ZZt2+y261

�t2 + y2

�dtdy + 0 +

ZZt2+y261

dtdy

375 == 4

h�2+ �i= 6�:

Solution (D. Jakobsson)Given w = f (z) = z2, one can di¤erentiate f (z) and get d

dzf (z) = 2z =

2 (x+ iy), we can use maple to integrate and get the following

48

Area (f (D)) =

=

ZZD

jf 0 (z)j2 dxdy =Z 2

0

Z p1�(x�1)2�p1�(x�1)2

4 jx+ iyj2 dx dy =

=

Z 2

0

Z p1�(x�1)2�p1�(x�1)2

4�x2 + y2

�dx dy =

Z 2

0

8x2p2x� x2+8=3

�2x� x2

�3=2dx =

= 6�

Change of variables �x = 1 + sin tdx = cos tdt

Z �=2

��=28 (1 + sin t)2

p1� sin2 t+ 8

3

�1� sin2

�3=2cos tdt =

=

Z �=2

��=28

1 + 2sin t|{z}

odd

+ sin2 t

!cos2+

8

3cos4 tdtdt =

= 16

Z �=2

0

2 cos2�23cos4 tdt =

= 8

Z �=2

0

2 (1 + cos 2t)� 13

�1 + 2 cos 2t+ cos2 2t

�dt =

= 8

Z �=2

0

5

3+4

3cos 2t� 1

3(1 + cos 4t) dt =

= 8 � �2

�5

3� 16

�= 4�

9

6= 6�:

49

II.4.9Compute ZZ

D

jf 0 (z)j2 dxdy;

for f (z) = z2 and D the open unit disk fjzj < 1g. Interpret youranswer in terms of areas.

SolutionBy the result aboveZZ

jzj<1x>0

jf 0 (z)j2 dxdy =ZZjzj<1x<0

jf 0 (z)j2 dx dy = �;

thus ZZjzj<1

jf 0 (z)j2 dx dy = 2�:

Remark.The function maps the top and bottom halves of the unit disk one-to-oneonto the unit disk, so by Exersice II.4.7 the integrals on top and bottomhalves are each �.

50

II.4.101 2 3 P L K

For smooth functions g and h de�ned on a bounded domain U , we de�ne theDirichlet form DU (g; h) by

DU (g; h) =

ZZU

�@g

@x

@h

@x+@g

@y

@h

@y

�dx dy

Show that if z = f (�) is a one-to-one analytic function from the boundeddomain V onto U , then

DU (g; h) = DV (g � f; h � f) :Remark. This shows that the Dirichlet form is a "conformal invariant".

SolutionDu (g; h) =

RRU

rgrhdxdy Assume h; g ????

Du (g; h) =RRU

r (g � f)r (h � f)dudv =RRU

g (u (x; y) ; v (x; y))h (u (x; y) ; v (x; y)) dudv

@g(u(x;y);v(x;y))@x

@h(u(x;y);v(x;y))@x

+@g(u(x;y);v(x;y))

@y@h(u(x;y);v(x;y))

@y=�

@g@u

@u@x+ @g

@v@v@x

� �@h@u

@u@x+ @h

@v@v@x

�+�@g@u

@u@y+ @g

@v@v@y

��@h@u

@u@y+ @h

@v@v@y

�=

@g@u

@h@u

�@u@x

�2+ @g

@u@u@x

@h@v

@v@x+ @g

@v@v@x

@h@u

@u@x+ @g

@v@h@v

�@v@x

�2+

@g@u

@h@u

�@u@y

�2+ @g

@u@u@y

@h@v

@v@y+ @g

@v@v@y@h@u

@u@y+ @g

@v@h@v

�@v@y

�2=

@g@u

@h@u

�@u@x

�2+ @g

@v@h@v

�@v@x

�2+ @g

@u@h@u

�@u@y

�2+ @g

@v@h@v

�@v@y

�2=

@g@u

@h@u

�@u

@x

�2+

�@u

@y

�2!| {z }

jf 0(z)j2

+ @g@v@h@v

�@v

@x

�2+

�@v

@y

�2!| {z }

jf 0(z)j2

+

@g

@u

@u

@x

@h

@v

@v

@x+@g

@v

@v

@x

@h

@u

@u

@x+@g

@u

@u

@y

@h

@v

@v

@y+@g

@v

@v

@y

@h

@u

@u

@y| {z }Use C-R equationsRR

U

rgrh jf 0 (z)j2dxdy =RRU

rgrhdudv

51

II.5.1Show that the following functions are harmonic, and �nd its har-monic conjugates:(a) x2 � y2 (c) sinh x sin y (e) tan�1 (y=x) ; x > 0

(b) xy + 3x2y � y3 (d) ex2�y2 cos (2xy) (f) x= (x2 + y2)

Solution(a)Set

u (x; y) = x2 � y2 )�u0x = 2xu0y = �2y

)�u00xx = 2u00yy = �2

)4u = 0;

thus is the function u (x; y) is harmonic. Cauchy Riemann give us�v0x = �u

0y = 2y ) v (x; y) = 2xy + g (y)) v

0y = 2x+ g

0 (y)

v0y = u

0x = 2x

Thus we have that

g0 (y) = 0) g (y) = C;

thus

v (x; y) = 2xy + C:

(b)Set

u (x; y) = xy+3x2y�y3 )�u0x = y + 6xyu0y = x+ 3x

2 � 3y2 )�u00xx = 6yu0yy = �6y

)4u = 0;

thus is the function u (x; y) is harmonic. Cauchy Riemann give us

�v0x = �u

0y = �x� 3x2 + 3y2 ) v (x; y) = �x2

2� x3 + 3xy2 + g (y)) v

0y = 6xy + g

0 (y)

v0y = u

0x = y + 6xy

Thus we have that

52

g0 (x) = y ) g (y) =y2

2+ C;

thus

v (x; y) =y2

2+ 3xy2 � x

2

2� x3 + C:

In paranthesis we remark that the analytic function f (z) = u + iv is givenby

f (z) = xy+3x2y�y3+i�y2

2+ 3xy2 � x

2

2� x3 + C

�= �iz2 (z + 1=2)+iC:

(c)Set

u (x; y) = sinhx sin y )�u0x = cosh x sin yu0y = sinhx cos y

)�u00xx = sinhx sin yu0yy = � sinh x sin y

)4u = 0;

thus the function u (x; y) is harmonic. Cauchy Riemann give us

�v0x = �u

0y = � sinh x cos y ) v (x; y) = � coshx cos y + g (y)) v

0y = cosh x sin y + g

0 (y)

v0y = u

0x = cosh x sin y

Thus we have that

g0 (y) = 0) g (y) = C;

thus

v (x; y) = � coshx cos y + CIn paranthesis we remark that the analytic function f (z) = u + iv is givenby

53

f (z) = sinhx sin y � i (coshx cos y + C) =

=

�ex � e�x

2

��eiy � e�iy

2i

�� i�ex + e�x

2

��eiy + e�iy

2

�+ iC =

= � i4

��ex � e�x

� �eiy � e�iy

�+�ex + e�x

� �eiy + e�iy

��+ iC =

= �i�ex+iy + e�(x+iy)

2

�+ iC = �i cosh z + iC:

(d)Set

u (x; y) = ex2�y2 cos (2xy))

�u0x = 2xe

x2�y2 cos (2xy)� 2yex2�y2 sin (2xy)u0y = �2yex

2�y2 cos (2xy)� 2xex2�y2 sin (2xy) )

)�u00xx = 2e

x2�y2 (cos 2xy + 2x2 cos 2xy � 2y2 cos 2xy � 4xy sin 2xy)u0yy = �2ex

2�y2 (cos 2xy + 2x2 cos 2xy � 2y2 cos 2xy � 4xy sin 2xy) )4u = 0;

thus the function u (x; y) is harmonic. Cauchy Riemann give us

8<:v0x = �u

0y = 2ye

x2�y2 cos (2xy) + 2xex2�y2 sin (2xy)) v (x; y) = ex

2�y2 sin (2xy) + g (y))) v

0y = �2yex

2�y2 sin (2xy) + 2xex2�y2 cos (2xy) + g0 (y)

v0y = u

0x = 2xe

x2�y2 cos (2xy)� 2yex2�y2 sin (2xy)

Thus we have that

g0 (y) = 0) g (y) = C;

thus

v (x; y) = ex2�y2 sin (2xy) + C

In paranthesis we remark that the analytic function f (z) = u + iv is givenby

f (z) = ex2�y2 cos (2xy) + i

�ex

2�y2 sin (2xy) + C�= ez

2

+ iC:

(e)

54

Set

u (x; y) = tan�1 (y=x) ; x > 0)

=)

8><>:u0x =

1

1 + (y=x)2

�� yx2

�=

�yx2 + y2

u0y =

x

x2 + y2

=)

)

8>><>>:u00xx =

2xy

(x2 + y2)2

u0yy =

�2xy(x2 + y2)2

)4u = 0;

thus the function u (x; y) is harmonic. Cauchy Riemann give us

8><>:v0x = �u

0y =

�xx2 + y2

) v (x; y) = a� 12log (x2 + y2) + g (y)) v

0y =

�yx2 + y2

+ g0 (y)

v0y = u

0x =

�yx2 + y2

Thus we have that

g0 (y) = 0) g (y) = C;

thus

v (x; y) = �12log�x2 + y2

�+ C:

(f)Set

u (x; y) =x

x2 + y2)

8>><>>:u0x = �

x2 � y2

(x2 + y2)2

u0y = �

2xy

(x2 + y2)2

)

8>><>>:u00xx = �2x

3y2 � x2

(x2 + y2)3

u0yy = 2x

3y2 � x2

(x2 + y2)3

)4u = 0;

thus is the function u (x; y) is harmonic. Cauchy Riemann give us

55

8>><>>:v0x = �u

0y =

2xy

(x2 + y2)2) v (x; y) =

�y(x2 + y2)

+ g (y)) v0y = �

x2 � y2

(x2 + y2)2+ g0 (y)

v0y = u

0x = �

x2 � y2

(x2 + y2)2

Thus we have that

g0 (y) = 0) g (y) = C;

thus

v (x; y) =�y

(x2 + y2)+ C

In paranthesis we remark that the analytic funktion f (z) = u + iv is givenby

f (z) =x

x2 + y2+ i

��y

(x2 + y2)+ C

�=

�z

jzj2+ iC =

�z

z�z+ iC =

1

z+ iC:

56

II.5.2Show that if v is a harmonic conjugate for u, then �u is a harmonicconjugate for v.

SolutionIf f (z) = u (x; y) + iv (x; y) is analytic then �if (z) = v (x; y) � iu (x; y) isanalytic. Thus �u is a harmonic conjugate of v.

57

II.5.3De�ne u (z) = Im (1=z2) for z 6= 0, and set u (0) = 0.(a) Show that all partial derivatives of u with respect to x existat all points of the plane C, as do all partial derivative of u withrespect to y.(b) Show that @2u

@x2+ @2u

@y2= 0.

(c) Show that u is not harmonic on C.(d) Show that @2u

@x@ydoes not exist at (0; 0).

Solution(a)We �rst de�ne

u (z) =

8<: Im1

z2= Im

�z2

jzj4= Im

(x� iy)2

(x2 + y2)2=

�2xy(x2 + y2)2

(x; y) 6= (0; 0)

0 (x; y) = (0; 0)

For all (x; y) 6= (0; 0), the n � th derivative of u (z) will be f(x;y)

(x2+y2)n+2with

f (x; y) a polynomial of degree n+2 in (x; y). At (0; 0), the function itself is0, thus u (x; y) 2 C1 (x). By symmetry u (x; y) 2 C1 (y).(b)We have

@2u

@x2+@2u

@y2=@u

@x

�2y3x2 � y2

(x2 + y2)3

�+@u

@y

�2x3y2 � x2

(x2 + y2)3

�=

= �24xy x2 � y2

(x2 + y2)4+ 24xy

x2 � y2

(x2 + y2)4= 0

(c)Notice that the function u (x; y) is not harmonic because the function itself,its �rst and second derivative are not continuous at the origin. The limit ofu (x; y) as (x; y)! (0; 0) does not exist, for instance u

�1n; 1n

�goes to �1 as

n!1 whereas we de�ned the function u to be 0 at the origin.

limn!1

u

�1

n;1

n

�= lim

n!1� n

2

2= �1:

58

(d)

@u

@y=�2x � (x2 + y2)2 + 2xy � 2 (x2 + y2) � 2y

(x2 + y2)4=�2x3 + 6xy2

(x2 + y2)3

thus

@2u

@x@y=(�6x2 + 6y2) � (x2 + y2)3 � (�2x3 + 6xy2) � 3 (x2 + y2)2 � 2x

(x2 + y2)6=6x4 � 36x2y2 + 6y4

(x2 + y2)4:

We have

@2u

@x@y=

(6x4�36x2y2+6y4

(x2+y2)4(x; y) 6= (0; 0)

0 (x; y) = (0; 0)

@2u

@x@y(x; 0) =

�6=x4; x 6= 00; x = 0

:

This partial derivative is not continuous at x = 0, so @2u@x@y

does not exist at(0; 0)..

59

II.5.4Show that if h (z) is a complex-valued harmonic function (solutionof Laplace�s equation) such that zh (z) is also harmonic, then h (z)is analytic.

SolutionFirst partial derivative for zh (z) is,

@ (zh)

@x=

@z

@xh+ z

@h

@x;

@ (zh)

@y=

@z

@yh+ z

@h

@y:

Second derivative for zh (z) is,

@2 (zh)

@x2=

@2z

@x2h+

@z

@x

@h

@x+@z

@x

@h

@x+ z

@2h

@x2;

@2 (zh)

@y2=

@2z

@y2h+

@z

@y

@h

@y+@z

@y

@h

@y+ z

@2h

@y2:

We have z = x+iy, thus @z=@x = 1, @z=@y = i, and @2z=@x2 = @2z=@y2 = 0,therefore

@2 (zh)

@x2= 2

@h

@x+ z

@2h

@x2;

@2 (zh)

@y2= i2

@h

@y+ z

@2h

@y2:

Add the equations

@2 (zh)

@x2+@2 (zh)

@y2= 2

@h

@x+ 2i

@h

@y+ z

�@2h

@x2+@2h

@y2

�:

Because h (z) and zh (z) is harmonic, then both @2 (zh) =@x2+@2 (zh) =@y2 =0 and @2 (h) =@x2 + @2 (h) =@y2 = 0, we have

@h

@x= �i@h

@y:

60

Write h = u+ iv, get

@u

@x+ i@v

@x= �i

�@u

@y+ i@v

@y

�=@v

@y� i@u@y:

Now we take real and imaginary parts and get

@u@x= @v

@yand @u

@y= � @v

@x:

Thus yields Cauchy-Riemann�s equations for u+ v, then h (z) is analytic aswas to be shown.

61

II.5.5Show that Laplace�s equation in polar coordinates is

@2u

@r2+1

r

@u

@r+1

r2@2u

@�2= 0:

SolutionLet x = r cos � and y = r sin �, and get partial derivatives

@x

@r= cos �;

@x

@�= �r sin �;

@y

@r= sin �;

@y

@�= r cos �:

Taking the �rst derivative of u with respect to r

@u

@r=@u

@x

@x

@r+@u

@y

@y

@r=@u

@xcos � +

@u

@ysin � = cos �

@u

@x+ sin �

@u

@y:

Taking the second derivative of u with respect to r

@2u

@r2=

=@2u

@x2@x

@rcos � +

@2u

@y@x

@y

@rcos � +

@2u

@x@y

@x

@rsin � +

@2u

@y2@y

@rsin � =

=@2u

@x2@x

@rcos � +

@2u

@y@xsin � cos � +

@2u

@x@ycos � sin � +

@2u

@y2@y

@rsin � =

= cos2 �@2u

@x2+ 2 sin � cos �

@2u

@x@y+ sin2 �

@2u

@y2:

Taking the �rst derivative of u with respect to �

@u

@�=@u

@x

@x

@�+@u

@y

@y

@�=@u

@x(�r sin �) + @u

@yr cos �:

Taking the second derivative of u with respect to �

62

@2u

@�2=

=@2u

@x2@x

@�(�r sin �) + @2u

@y@x

@y

@�(�r sin �) + @u

@x(�r cos �)+

+@2u

@x@y

@x

@�(r cos �) +

@2u

@y2@y

@�(r cos �) +

@u

@y(�r sin �) =

=@2u

@x2(�r sin �) (�r sin �) + @2u

@y@x(r cos �) (�r sin �) + @u

@x(�r cos �)+

+@2u

@x@y(�r sin �) (r cos �) + @

2u

@y2(r cos �) (r cos �) +

@u

@y(�r sin �) =

= r2 sin2 �@2u

@x2� 2r2 sin � cos � @

2u

@x@y+ r2 cos2 �

@2u

@y2� r cos �@u

@x� r sin �@u

@y:

Combining these partial derivatives, one gets

@2u

@r2+1

r

@u

@r+1

r2@2u

@�2=

= cos2 �@2u

@x2+ 2 sin � cos �

@2u

@x@y+ sin2 �

@2u

@y2+1

r

�cos �

@u

@x+ sin �

@u

@y

�+

+1

r2

�r2 sin2 �

@2u

@x2� 2r2 sin � cos � @

2u

@x@y+ r2 cos2 �

@2u

@y2� r cos �@u

@x� r sin �@u

@y

�=

= cos2 �@2u

@x2+ 2 sin � cos �

@2u

@x@y+ sin2 �

@2u

@y2+1

rcos �

@u

@x+1

rsin �

@u

@y+

+ sin2 �@2u

@x2� 2 sin � cos � @

2u

@x@y+ cos2 �

@2u

@y2� 1rcos �

@u

@x� 1rsin �

@u

@y=

=�sin2 � + cos2 �

� @2u@x2

+�sin2 � + cos2 �

� @2u@y2

=@2u

@x2+@2u

@y2:

Because u is a harmonic function we have @2u=@x2 + @2u=@y2 = 0 and wehave,

@2u

@r2+1

r

@u

@r+1

r2@2u

@�2=@2u

@x2+@2u

@y2= 0:

63

II.5.6Show using Laplace�s equation in polar coordinates that log jzj isharmonic on the punctured plane Cn f0g.

SolutionSet z = rei�, we have u (r; �) = log jzj = log r. We can compute the Laplacianof log jzj without worrying about the origin, i.e. r 6= 0 because this case istaken out, thus we have

� log r =@2u

@r2+1

r

@u

@r+1

r2@u2

@�2= � 1

r2+1

r

1

r+ 0 = 0:

64

II.5.7Show that log jzj has no conjugate harmonic function on the punc-tured plane Cn f0g, though it does have a conjugate harmonic func-tion on the slit plane Cn (�1; 0).

Solution (A. Kumjamin)Set D := Cn (�1; 0] and let u : D ! R be given by u (z) := log jzj, it iseasily veri�ed that u (z) = ReLog z and this shows that u is harmonic on Dby the main theorem in section II.5. Moreover, it follows by Cauchy-Riemannequations that a harmonic conjugate v is given by

v (z) := ImLog z = Arg z for z 2 D:

Any other harmonic conjugate di¤ers from v by a constant. Now if we set~D := Cn f0g and de�ne ~u : D ! R by ~u (z) := log jzj, it can easily beshown by an argument similar to the one above that ~u is harmonic on ~Deven though it cannot be expressed as the real part of an analytic functionon ~D (the property of being harmonic is local). Suppose that ~u does have aharmonic conjugate, say ~v, on ~D. Then its restriction to D would have tobe a harmonic conjugate of u and so it would be of the form z 7�! Arg z+ cfor some c 2 R. But such a function has no continuous extension to ~D.This contradicts the fact that ~v is harmonic on ~D (since this implies ~v iscontinuous on ~D). Therefore, log jzj has no conjugate harmonic function onthe punctured plane Cn f0g.

Solution (K. Seip)In Cn (�1; 0], we know that any harmonic conjugate of log jzj must havethe form

v (x; y) = Arg z + c:

Such a function can not be made continuous in C since for x < 0, we have

limy!00+

v (x; y) = � + c;

and

limy!00�

v (x; y) = �� + c:

65

II.5.8Show using Laplace�s equation in polar coordinates that u

�rei��=

� log r is harmonic. Use the polar form of the Cauchy-Riemannequations. (Exercise 3.8) to �nd a harmonic conjugate v for u.What is the analytic function u+ iv?

SolutionSet

u�rei��= � log r )

(u0r =

�

ru0� = log r

)(u00rr = �

�

r2u00�� = 0

The function u�rei��= � log r is harmonic because

�u = urr +1

rur +

1

r2u�� = �

�

r2+�

r2= 0:

Cauchy Riemann equations in polar form (page 50) gives us

8<: v0r = �

1

ru0� = �

log r

r) v (r; �) = �(log r)

2

2+ g (�)) v

0� = g

0 (�)

v0� = ru

0r = �

Thus we have that

g0 (�) = � ) g (�) =�2

2+ C;

thus

v (r; �) = �(log r)2

2+�2

2+ C:

We have that the analytic function f (z) = u+ iv is given by

f (z) = � log r+i

�2

2� (log r)

2

2+ C

!= � i

2(log r + i�)2+C = � i

2(log z)2+iC:

66

II.6.1Sketch the families of level curves of u and v for the followingfunctions f = u+ iv.(a) f (z) = 1=z (b) f (z) = 1=z2 (c) f (z) = z6

Determine where f (z) is conformal and where it is not conformal.

Solution(a)We have that

f (z) =1

z=

�z

jzj2=x� iyx2 + y2

)

8<: u =x

x2 + y2

v = � y

x2 + y2

Conformal except at z = 0, where it is not de�ned. (Conformal everywhereif we view it as a mapping of the sphere).We have

u =1

c;

1

c(x2 + y2) = x;

�x� c

2

�2+ y2 =

c2

4;

which is circles centred on x � axis and tangent to y � axis.We have

v =1

a;

1

a(x2 + y2) = �y; x2 +

�y +

a

2

�2=a2

4;

which is circles centred on y � axis and tangent to x � axis.(b)We have

f (z) =1

z2=z2

jzj4=(x� iy)2

(x2 + y2)2=x2 � y2 � 2ixy(x2 + y2)2

)

=)(u = x2�y2

(x2+y2)2= cos2 ��sin2 �

r2= cos(2�)

r2

v = �2xy(x2+y2)2

= �2 cos � sin �r2

= � sin(2�)r2

and

67

u =1

c; r2 = c cos (2�) ; v =

1

c; r2 = �c sin (2�) :

Conformal everywhere except at z = 0.(c)We have that

f (z) = z6 = r6 (cos (6�) + i sin (6�)))�u = r6 cos (6�)v = r6 sin (6�)

Figure repeats itself. Rotate of fu = 0g by �=12 is fv = 0g. Each grindrepeats itself every �=6.Conformal everywhere except at z = 0.

FIGURE II.6.1a NF FIGURE II.6.1b NF FIGURE II.6.1c NF

68

II.6.2Sketch the families of level curves of u and v for f (z) = Log z = u+iv.Relate your sketch to a �gure in Section I.6.

Solution

0

π/4π/2

3π/4

−3π/4−π/2

−π/4

II.6.2

We have

f (z) = Log z = log jzj+iArg z = log r+i� )�u = log jzjv = Arg z

)�u = log rv = �

:

If u = constant and r = constant we have circles centred at 0, and if v =constanat and � = constanat we have rays issuing from 0, �� < � < �.Refer to the Figure for the map w = Log z in section I.6. The sketch is ofthe inverse images of horizontal and vertical lines in that Figure.

69

II.6.3Sketch the families of level curves of u and v for the function f =u + iv given by (a) f (z) = ez, (b) f (z) = e�z, where � is complex.Determine where f (z) is conformal and where it is not conformal.

Solutionf (z) = ez = ex (cos y + i sin y)) u = ex cos yv = ex sin yu = const, x = � log (c cos y) = � log c� log (cos y)v = const, x = � log (c sin y) = � log c� log (sin y)Level curves are invariant under ????? in x�direction. Level curves are in-variant under ????? in y�direction by �. The function f (z) is conformaleverywhere.f (z) = e�z = e(a+ib)(x+iy) = eax�by (cos (ay + bx) + i sin (ay + bx)))�u = eax�by cos (ay + bx)v = eax�by sin (ay + bx)

Figure repeat themselvesu = 0 at ay + bx = n� + �

2, y = � b

ax + n�

a+ �

2a. v = 0 at ay + bx = n�,

y = � bax+ n�

a

FIGURE II.6.3a NF FIGURE II.6.3b NF

70

II.6.4Find a conformal map of the horizontal strip f�A < Im z < Ag ontothe right half-plane fRew > 0g. Hint. Recall the discussion of theexponential function, or refer to the preceding problem.

Solution (A. Kumjian)

FIGURE II.6:4az FIGURE II.6:4aw

It is clearly intended that A > 0 so we will assume this. We de�ne the maph : C! C by h (z) = e �z2A for all z 2 C. Then

h0 (z) =�

2Ae�z2A

for all z 2 C. Hence, h is analytic with a nowhere vanishing derivative andso h is conformal. Note that the exponential functions maps the horizontalstrip fz 2 C : ��=2 < Im z < �=2g onto the right half-plane

fw 2 C : Rew > 0g = fw 2 C : ��=2 < Argw < �=2g :Hence, h maps the horizontal strip fz 2 C : �A < Im z < Ag onto the righthalf-plane fw 2 C : Rew > 0g.

71

II.6.5Find a conformal map of the wedge f�B < arg z < Bg onto the righthalf-plane fRew > 0g. Assume 0 < B < �.

SolutionTry w = z�, multiply angels by �

2B, so eiB ! ei�=2, w = z�=(2B) does the trick

FIGURE II.6.5az FIGURE II.6.5aw

72

II.6.6Determine where the function f (z) = z+1=z is conformal and whereit is not conformal. Show that for each w, there are at most twovalues z for which f (z) = w. Show that if r > 1, f (z) maps the circlefjzj = rg onto an ellipse, and that f (z) maps the circle fjzj = 1=rgonto the same ellipse. Show that f (z) is one-to-one on the exteriordomainD = fjzj > 1g. Determine the image of D under f (z). Sketchthe images under f (z) of the circles fjzj = rg for r > 1, and sketchalso the images of the parts of the rays farg z = �g lying in D.

Solutionf (z) = z + 1=z, f 0 (z) = 1� 1=z2, f 0 (z) = 0 at z2 = 1, z = �1. Not de�nedat z = 0, not conformal at z = �1. If f (z) = �, then z =

���

p�2 � 4

�=2.

If z = �ei�, then w = u + iv = � cos � + i� sin � + cos �=� � i sin �=�, whichcan be written u2= (�+ 1=�)2 + v2= (�� 1=�)2 = 1, i.e. image is an ellipse,replace � by 1=� and get same ellipse. Since f (D) = f (D), and f (z) is atmost two�to�one, f (z) is one�to�one on D. Since f (z) maps @D onto theinterval [�2; 2], and f (z) maps onto C, the image of D is Cn [�2; 2].

73

II.6.7For the function f (z) = z + 1=z = u+ iv, sketch the families of levelcurves of u and v. Determine the images under f (z) of the tophalf of the unit disk, the bottom half of the unit disk, the partof the upper half-plane outside the unit disk, and the part of thelower half-plane outside the unit disk. Hint. Start by locatingthe images of the curves where u = 0, where v = 0, and wherev = 1. Note that the level curves are symmetric with respect to thereal and imaginary axis, and they are invariant under the inversionz 7! 1=z in the unit circle.

Solution

74

II.6.8 ( From Hints and Solutions)1 2 3 P L K

Consider f (z) = z + ei�=z, where 0 < � < �. Determine where f (z)is conformal and where it is not conformal and where it is notconformal. Sketch the images under f (z) of the unit circle fjzj = 1gand the intervals (�1;�1] and [+1;+1) on the real axis. Show thatw = f (z) maps fjzj > 1g conformally onto the complement of a slitplane in the w� plane. Sketch roughly the images of the segmentsof rays outside the unit circle farg z = �; jzj � 1g under f (z). Atwhat angels do they meet the slit, and at what angles do theyapproach 1?

Solutionf (z) = z + ei�=z , f 0 (z) = 1 � ei�=z2, f 0 (z) = 0 at z2 = ei�. Not de-�ned at z = 0, not conformal at z = �ei�=2. The expression f (z) =ei�=2

�e�i�=2z + 1=

�e�i�=2z

��, shows that f (z) is a composition of the ro-

tation by ��=2, the function of Exercises 6 and 7, and a rotation by �=2.Thus f (z) maps fjzj > 1g one�to�one onto the complement rotate of [�2; 2]by �=2.

75

II.6.9Let f = u+ iv be a continuously di¤erentiable complex-valued func-tion on a domain D such that the Jacobian matrix of f does notvanish at any point of D. Show that if f maps orthogonal curves toorthogonal curves, then either f or �f is analytic, with nonvanishingderivative.

SolutionSuppose f (0) = 0 and ux 6= 0 at 0. The tangents in the orthogonal directions(1; t) and (�t; 1) are mapped to the tangents in the directions (ux; vx) +t (uy; vy) and �t (ux; vx) + (uy; vy).The orthogonality of these directions for all t gives

h(ux; vx) + t (uy; vy) ;�t (ux; vx) + (uy; vy)i = 0;can be rewritten

h(ux; vx) ; (uy; vy)i+t (h(uy; vy) ; (uy; vy)i � h(ux; vx) ; (ux; vx)i)�t2 h(uy; vy) ; (ux; vx)i = 0:

The systems of equations

(1)(2)

�uxuy + vxvy = 0

u2y + v2y � (u2x + v2x) = 0

most hold to get orthogonality. Put (1) into (2) and get

u2x�u2y + v

2y � u2x � v2x

�= v2xv

2y + u

2xv2y � u4x � u2xv2x =

= v2x�v2y � u2x

�+ u2x

�v2y � u2x

�=�v2x + u

2x

� �v2y � u2x

�= 0:

If ux 6= 0 we are lead to ux = �vy and uy = �vx.Tangent to curve s! (s; ts) is tangent to image curve s+(u (s; ts) ; v (s; ts))(1; 0) ! (ux; vx), (0; 1) ! (uy; vy), orthogonal ) (uy; vy) = C (�vx; uy) forsome. A ????? gives C = �1, same sign holds in ?????.Either ux = vy, uy = �vx ) conformal in neighborhood of 0, or ux = �vy,uy = vx ) anti conformal in neighborhood of 0. ux = �vy and uy = �vx.So Cauchy Riemann is satis�ed for either f or �f .

76

II.7.1Compute explicitly the fractional linear transformations determinedby the following correspondences of triples(a) (1 + i; 2; 0) 7�! (0;1; i� 1) (e) (1; 2;1) 7�! (0; 1;1)(b) (0; 1;1) 7�! (1; 1 + i; 2) (f) (0;1; i) 7�! (0; 1;1)(c) (1; 1 + i; 2) 7�! (0; 1;1) (g) (0; 1;1) 7�! (0;1; i)(d) (�2; i; 2) 7�! (1� 2i; 0; 1 + 2i) (h) (1; i;�1) 7�! (1; 0;�1)

Solution

(w � w0) (w1 � w2)(w � w2) (w1 � w0)

=(z � z0) (z1 � z2)(z � z2) (z1 � z0)

(a)We have the points

z0 = 1 + i w0 = 0z1 = 2 w1 =1z2 = 0 w2 = i� 1

The mapping is

(w � w0) (1� w2=w1)(w � w2) (1� w0=w1)

=(z � z0) (z1 � z2)(z � z2) (z1 � z0)

;

take limit as w1 !1, to obtain

w � 0w � (i� 1) =

(z � (1 + i)) (2� 0)(z � 0) (2� (1 + i)) ) w =

2iz + (2� 2i)z � 2 :

(b)We have the points

z0 = 0 w0 = 1z1 = 1 w1 = 1 + iz2 =1 w2 = 2

The mapping is

(w � w0) (w1 � w2)(w � w2) (w1 � w0)

=(z � z0) (z1=z2 � 1)(z=z2 � 1) (z1 � z0)

;

take limit as z2 !1, to obtain

77

(w � 1) ((1 + i)� 2)(w � 2) ((1 + i)� 1) =

z � 01� 0 ) w =

2z � (1 + i)z � (1 + i) :

(c)We have the points

z0 =1 w0 = 0z1 = 1 + i w1 = 1z2 = 2 w2 =1

The mapping is

(w � w0) (w1=w2 � 1)(w=w2 � 1) (w1 � w0)

=(z=z0 � 1) (z1 � z2)(z � z2) (z1=z0 � 1)

;

take limit as z0; w2 !1, to obtain

w � 01� 0 =

(1 + i)� 2z � 2 ) w =

i� 1z � 2 :

(d)We have the points

z0 = �2 w0 = 1� 2iz1 = i w1 = 0z2 = 2 w2 = 1 + 2i

The mapping is

(w � w0) (w1 � w2)(w � w2) (w1 � w0)

=(z � z0) (z1 � z2)(z � z2) (z1 � z0)

;

obtain

(w � (1� 2i)) (0� (1 + 2i))(w � (1 + 2i)) (0� (1� 2i)) =

(z � (�2)) (i� 2)(z � 2) (i� (�2)) ) w = iz + 1:

(e)We have the points

z0 = 1 w0 = 0z1 = 2 w1 = 1z2 =1 w2 =1

78

The mapping is

(w � w0) (w1=w2 � 1)(w=w2 � 1) (w1 � w0)

=(z � z0) (z1=z2 � 1)(z=z2 � 1) (z1 � z0)

;

take limit as z2; w2 !1, to obtain

w � 01� 0 =

z � 12� 1 ) w = z � 1:

(f)We have the points

z0 = 0 w0 = 0z1 =1 w1 = 1z2 = i w2 =1

The mapping is

(w � w0) (w1=w2 � 1)(w=w2 � 1) (w1 � w0)

=(z � z0) (1� z2=z1)(z � z2) (1� z0=z1)

;

take limit as z1; w2 !1, to obtain

w � 01� 0 =

z � 0z � i ) w =

z

z � i(g)We have the points

z0 = 0 w0 = 0z1 = 1 w1 =1z2 =1 w2 = i

The mapping is

(w � w0) (1� w2=w1)(w � w2) (1� w0=w1)

=(z � z0) (z1=z2 � 1)(z=z2 � 1) (z1 � z0)

;

take limit as z2; w1 !1, to obtain

w � 0w � i =

z � 01� 0 ) w =

iz

z � 1 :

(h)

79

We have the points

z0 = 1 w0 = 1z1 = i w1 = 0z2 = �1 w2 = �1

The mapping is

(w � w0) (w1 � w2)(w � w2) (w1 � w0)

=(z � z0) (z1 � z2)(z � z2) (z1 � z0)

;

obtain

(w � 1) (0� (�1))(w � (�1)) (0� 1) =

(z � 1) (i� (�1))(z � (�1)) (i� 1) ) w =

iz + 1

z + i:

80

II.7.2Consider the fractional linear transformation in Exercise 1a above,which maps 1 + i to 0, 2 to 1, and 0 to i� 1. Without referring toan explicit formula, determine the image of the circle fjz � 1j = 1g,the image of the disk fjz � 1j < 1g, and the image of the real axis.

SolutionIt is obvious that the tripple (1 + i; 2; 0) lies on the circle jz � 1j = 1 in thez � plane. From exercise II.7.1a we know that points in the w � plane thetripple is mapped on, i.e,

z0 = 1 + i w0 = 0z1 = 2 w1 =1z2 = 0 w2 = i� 1

The circle jz � 1j = 1 is mapped to the straight line through 0 and i� 1, i.eImw = �Rew since tree points determines the "circle". By preservation oforientation the disk goes to the half-plane to lower left of the straight linethrough 0 and i� 1.Image of the real axis is "circle" orthogonal to image of jz � 1j = 1, that mustbe the straight line through i�1, with slope 1, i.e. the line Imw = Rew+2.

II.7.2 z­plane II.7.2 w­plane

81

II.7.3Consider the fractional linear transformation that maps 1 to i, 0to 1 + i, �1 to 1. Determine the image of the unit circle fjzj = 1g,the image of the open unit disk fjzj < 1g, and the image of theimaginary axis. Illustrate with a sketch.

SolutionWe have the points

z0 = 1 w0 = iz1 = 0 w1 = 1 + iz2 = �1 w2 = 1

The mapping is

(w � w0) (w1 � w2)(w � w2) (w1 � w0)

=(z � z0) (z1 � z2)(z � z2) (z1 � z0)

;

obtain

(w � i) ((1 + i)� 1)(w � 1) ((1 + i)� i) =

(z � 1) (0� (�1))(z � (�1)) (0� 1) ) w =

i� 1z + i

By our mapping w = (i� 1) = (z + i), we have that �i on the unitcircle inthe z � plane is mapped to1 in the w � plane. The circle jzj = 1 is mappedto the straight line through i and 1, i.e Imw = �Rew + 1 since tree pointsdetermines the "circle". By preservation of orientation the disk goes to thehalf-plane to upper left of the straight line through i and 1.The tripple (1; 0;�1) on the real axis is mapped to the points (i; 1 + i; 1) inthe w � plane, it must be so that the real axis in the z � plane is mappedonto the circle

��w � �12+ 1

2i��� = p

22in the w � plane.

Imaginary axis must be mapped to a circle through 1+i, that is orthogonal toimage of real line, this is the straight line through 1+i and 0, i.e Imw = Rew.

82

II.7.3 z­plane II.7.3 w­plane

83

II.7.4Consider the fractional linear transformation that maps �1 to �i,1 to 2i, and i to 0. Determine the image of the unit circle fjzj = 1g,the image of the open unit disk fjzj < 1g, and the image of theinterval [�1;+1] on the real axis. Illustrate with a sketch.

SolutionWe have the points

z0 = �1 w0 = �iz1 = 1 w1 = 2iz2 = i w2 = 0

The mapping is

(w � w0) (w1 � w2)(w � w2) (w1 � w0)

=(z � z0) (z1 � z2)(z � z2) (z1 � z0)

;

obtain

(w � (�i)) (2i� 0)(w � 0) (2i� (�i)) =

(z � (�1)) (1� i)(z � i) (1� (�1)) ) w =

(�6� 2i) z � 2 + 6i5z � 3 + 4i

It is obvious that the tripple (�1; 1; i) lies on the circle jzj = 1 in the z �plane. And we can see that thiese three points are mapped to three pointson the imaginary axis in the w � plane, i.e Rew = 0, since tree pointsdetermines the "circle". By preservation of orientation the disk goes to theright half-plane.The interval [�1; 1] must be mapped to a circle through �i and 2i that isorthogonal to image of the unit circle, this is an arc of the circle

��w � 12i�� = 3

2.

We have that w (0) = 65� 2

5i must be an point on the arc, thus the arc is the

circle��w � 1

2i�� = 3

2in the right half-plane from �i to 2i.

84

II.7.4 z­plane II.7.4 w­plane

85

II.7.5What is the image of the horizontal line through i under the frac-tional linear transformation that interchanges 0 and 1 and maps �1to 1 + i? Illustrate with a sketch.

SolutionWe have the points

z0 = 0 w0 = 1z1 = 1 w1 = 0z2 = �1 w2 = 1 + i

The mapping is

(w � w0) (w1 � w2)(w � w2) (w1 � w0)

=(z � z0) (z1 � z2)(z � z2) (z1 � z0)

;

obtain

(w � 1) (0� (1 + i))(w � (1 + i)) (0� 1) =

(z � 0) (1� (�1))(z � (�1)) (1� 0) ) w =

iz � iz � i

The tripple (0; 1;�1) on the real axis is mapped to the points (1; 0; 1 + i) inthe w � plane, it must be so that the real axis in the z � plane is mapped ontothe circle

��w � �12+ 1

2i��� = p

22in the w � plane, since tree points determines

the "circle". The real axis and the line through i have one common point atin�nity. The common point at in�nity for both real axis and the line throughi must be mapped to the same point in the w � plane, we have that the pointis w (1) = i. The two lines in the z � plane is parallell, thus the images ofthe lines must go through i in the w � plane and be parallell in that point,thus the mapping of the horizontal line through i must be the line throughi, which is tangent line to circle at i, i.e. Im z = Re z + 1. Remark that theimage can not be a circle because w (i) =1, so the image of the horizontalline must contain the point at in�nity.

86

II.7.5 z­plane II.7.5 w­plane

87

II.7.6Show that the image of a straight line under the inversion z 7! 1=zis a straight line or circle, depending on whether the line passesthrough the origin.

SolutionThe image of the straight line ax+by = c is (kontrollera x�iy i lösningshäftet)

w =1

z=1

z�z=

�z

jzj2= (x� iy) 1

r2

where u = x=r2 and v = �y=r2. Get

au� bv = ax+ by

r2=c

r2=

c

x2 + y2= c

�u2 + v2

�The image is the solution of c (u2 + v2)� au+ bv = 0.If c = 0 we have that

�au+ bv = 0;this is a straight line through 0.If c 6= 0 we have

c�u2 + v2

�� au+ bv = 0;

which can be rewritten as

�u� a

2c

�2+

�v +

b

2c

�2=

0@s� a2c

�2+

�b

2c

�21A2

;

it is a circle through 0.

Solution (K. Seip)Set f (z) = 1

z. Then 1 2 f (S), 0 2 S, which proves the claim.

88

II.7.7Show that the fractional linear transformation f (z) = (az + b) = (cz + d)is the identity mapping z if and only if b = c = 0 and a = d 6= 0.

SolutionWe have that

az + b

cz + d= z , az + b = cz2 + dz , cz2 + (d� a) z � b = 0:

Put in values on z for intance 0 and �1 the expession becomes

(1)(2)(3)

c� (d� a) = 0; z = �1:�b = 0 z = 0c+ (d� a) = 0; z = 1:

From (2) we have b = 0 and if we add (1) and (3) we have c = 0.Thus the fractional linear transformation f (z) is the identity mapping if andonly if b = c = 0, a = d 6= 0.

SolutionSet R� = R[ f1g. Suppose f (z) = az+b

cz+dmaps R� to R�. Then 0 7! b

d2 R�,

1 7! ac2 R�, 0 - � b

a2 R�, 1 - �d

c2 R�. At least one of theese rations

is di¤erent from 0 and 1, and since one of a; b; c; d can be chosen freely, theresult follows.

89

II.7.8Show that any fractional linear transformation can be representedin the form f (z) = (az + b) (cz + d), where ad� bc = 1. Is this repre-sentation unique?

SolutionIf

w = (�z + �) ( z + �)

divide each coe¢ cient by the square root of

�� � � to obtain representation with

ad� bc = 1:Representation is not unique, as can multiply all coe¢ cients by �1.

90

II.7.9Show that the fractional linear transformations that are real onthe real axis are precisely those that can be expressed in the form(az + b) = (cz + d), where a; b; c;and d are real.

Solution (K. Seip)Set R� = R[f1g. Suppose f (z) = az+b

cz+dmaps R� to R�. Then 0 7�! b

d2 R�,

1 7! ac2 R�, 0 - � b

a2 R�, 1 - �d

c2 R�. At least one of these is

di¤erent from 0 and 1, and since one of a; b; c; d can be chosen freely, theresult follows.

91

II.7.9Show that the fractional linear transformations that are real onthe real axis are precisely those that can be expressed in the form(az + b) = (cz + d), where a; b; c;and d are real.

Solution (A. Kumjian)Set R� = R[ f1g, note that a fractional linear transformation f : C� ! C�is real on the real axis i¤ f (R�) � R� (since f is continuous). For A =

�a bc d

�with detA 6= 0 we de�ne the fractional linear transformation fA : C� ! C�by the formula fA (z) = (az + b) = (cz + d). It remains to prove that given afractional linear transformation f , we have f (R�) � R� i¤ f = fA for some2� 2 matrix A with real entries.

Let f : C� ! C� be a fractional linear transformation. Suppose that f = fAwhere A is a 2 � 2 matrix with real entries a; b; c and d as above. Then forx 2 R� we have

f (x) = fA (x) =ax+ b

cx+ d2 R�:

This is clear for x 2 R but requires a little work for x =1. We have

f (1) = limz!1

az + b

cz + d=

�a=c if c 6= 0;1 if c = 0:

Hence, f (R�) � R�.Conversely, suppose that f (R�) � R�, we must show that f = fA for some 2�2 matrix with real entries. If A is a 2�2 matrix (with non-zero determinant),then fA�1 = (fA)

�1. Hence, f = fA for some 2�2 matrix A with real entriesi¤ its inverse f�1 has the same property.By the unlikeness part of the theorem on page 64, f is completely determinedby the three extended real numbers x0 = f (0), x1 = f (1) and x1 = f (1)(recall that we have assumed f (R�) � R�). So it su¢ ces to show thatthe fractional linear transformation g for which g (x0) = 0, g (x1) = 1 andg (x1) = 1 has the requisite properties, because we must have g = f�1

(again by the uniqueness part of the theorem).There are four cases to consider. The �rst case is when x0; x1; x1 6=1 andthe remaining three cases are x0 = 1; x1 = 1 and x1 = 1. The formulafor g (z) in the four cases is given as follows (in the �rst case k = x1�x1

x1�x0 2 R):

92

kx� x0x� x1

if x0; x1; x1 6=1;x1 � x1x� x1

if x0 =1;x� x0x� x1

if x1 =1;x� x0x1 � x0

if x1 =1:

In each case g (z) has the desired form. Hence, f = fA for some 2� 2 matrixA with real entries as required.

93

II.7.10Suppose the fractional linear transformation (az + b) = (cz + d) mapsR to R, and ad� bc = 1. Show that a; b; c; and d are real or they areall pure imaginary.

SolutionBecause R is mapped on R and the orientation is perserved, f (x) will beeither increasing or decreasing.

Case 1 : Suppose f (x) is increasing, then f 0 (x) = 1= (cx+ d)2 > 0 for allx) c; d are real. f (z) = (az + b) = (cz + d), then ax+ b is real for all x 2 R,so a; b are also real.

Case 2 : Suppose f (x) is decreasing, then f 0 (z) = 1= (cz + d)2 < 0 for allx) c; d are pure imaginary. f (z) = (az + b) = (cz + d), then ax+ b are pureimaginary for all x 2 R, so a; b are also pure imaginary.

94

II.7.11Two maps f and g are conjugate if there is h such that g = h�f �h�1.Here the conjugating map h is assumed to be one-to-one, withappropriate domain and range. We can think of f and g as the"same" map, after the change of variable w = h (z). A point z0 isa �xed point of f if f (z0) = z0. Show the following. (a) If f isconjugate to g, then g is conjugate to f .(b) If f1 is conjugate to f2 and f2 to f3, then f1 is conjugate to f3.(c) If f is conjugate to g, then f � f is conjugate to g � g, and moregenerally, the m�fold composition f � � � � � f (m times) is conjugateto g � � � � � g (m times).(d) If f and g are conjugate, then the conjugating function h mapsthe �xed points of f to �xed points of g. In particular, f and ghave the same number of �xed points.

Solution(a)If f is conjugate to g then

g = h � f � h�1 ) f = h�1 � g � hthus g is conjugate to f(b)If f1 = h � f2 � h�1 and f2 = g � f3 � g�1 we get

f1 = h � g � f3 � g�1 � h�1 = (h � g) � f3 � (h � g)�1 :(c)

g = h � f � h�1thenm timesz }| {

g � g � : : : � g == h � f � h�1 � h � f � h�1 � : : : � h � f � h�1 =

= h �m timesz }| {

f � f � : : : � f � h�1

(d)g = h � f � h�1, f (z0) = z0, and w0 = h (z0), then

95

g (w0) = h�f�h�1 (w0)

��= h (f (z0)) = h (z0) = w0:

We have that f and g have the same number of �xed points since every pointof g is mapped to a �xed point of f by the function h�1.

96

II.7.12Classify the conjugacy classes of fractional linear transformationsby establishing the following(a)A fractional linear transformation that is not the identity has either1 or 2 �xed points, that is, points satisfying f (z0) = z0.(b)If a fractional linear transformation f (z) has two �xed points, thenit is conjugate to the dilation z 7! az with a 6= 0, a 6= 1, that is, thereis a fractional linear transformation h (z) such that h (f (z)) = ah (z).Is a unique? Hint. Consider a fractional linear transformation thatmaps the �xed points to 0 and 1.(c)If a fractional linear transformation f (z) has exactly one �xedpoint, then it is conjugate to the translation � 7! � + 1. In otherwords, there is a fractional linear transformation h (z) such thath (f (h�1 (�))) = � + 1, or equivalently, such that h (f (z)) = h (z) + 1.Hint. Consider a fractional linear transformation that maps the�xed point to 1.

Solutiona) az+b

cz+d= z , az + b = cz2 + dz , cz2 + (d� a) z � b = 0. If this is = 0

for every z, we get w (z) = z. It is not identically equall to zero, so it haseither 1 or 2 �nite solutions. If c 6= 0, it has 2 solutions possibly one withmultiplicity and 1 is not a �xed point. If c = 0, it has one �nite solutionz1 = b= (d� a), and z2 =1 is a �xed point, so they are two.b) Make a change of variables h (z) = (�z + �) = ( z + �) that maps �xedpoints to 0 and 1 . Then h � f � h�1 is FLT with �xed points at 0 and 1.(h � f � h�1) (w) = aw for some a 6= 0, a 6= 1. f is conjugate to a dilationSuppose aw is conjugate to Aw, i.e. 9h, (h (ah�1)) (w) = Aw, h (az) =Ah (z). h must map �xed points to �xed points, so either h (z) = cz orh (z) = c=z. So either h (z) = cz ) h (az) = caz = Ah (z) = Acz ) A = aor h (z) = c=z ) h (az) = c= (az) = Ah (z) = Ac=z ) A = 1=a. Thus a isnot unique.c) Suppose h maps the �xed points to1, then h � f � h�1 has only one �xedpoint at1, so (h � f � h�1) (w) = aw+b. Since this has no �nite �xed points,aw+b = w has no solutions and a = 1, b 6= 0. Thus (h � f � h�1) (w) = w+b.

97

Now w + b is conjugate to w + 1, by a dilation g (w) = Aw. w ! w=A !w=A+ b! w + Ab, take A = 1=b. So

�g ��h � f � h�1

�� g�1

�(w) = w + 1 ()

()�(g � h) � f � (g � h)�1

�(w) = w + 1 ()

() ((g � h) � f) (z) = (g � h) (z) + 1:

98

III 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 1912345678

1

III.1.1Evaluate

R y2dx + x2dy along the following paths from (0; 0) to

(2; 4),(a) the arc of the parabola y = x2,(b) the horizontal interval from (0; 0) to (2; 0), followed by the ver-tical interval from (2; 0) to (2; 4), (c) the vertical interval from (0; 0)to (0; 4), followed by the horizontal interval from (0; 4) to (2; 4).

Solution(a)

Z

y2dx+x2dy =

�x = t dx = dt; 0 � t � 2y = t2 dy = 2t dt

�=

Z 2

0

�t2�2dt+

Z 2

0

t22t dt =72

5:

(b)

Z 1

y2dx+ x2dy =

�x = t dx = dt 0 � t � 2y = 0 dy = 0 dt

�=

Z 2

0

02dt+

Z 2

0

t2 0 dt = 0Z 2

y2dx+ x2dy =

�x = 2 dx = 0 dt 0 � t � 4y = t dy = dt

�=

Z 4

0

t2 0 dt+

Z 4

0

22dt = 16Z

y2dx+ x2dy =

Z 1

y2dx+ x2dy +

Z 2

y2dx+ x2dy = 16:

(c)

Z 1

y2dx+ x2dy =

�x = 0 dx = 0 dx 0 � t � 4y = t dy = dt

�=

Z 4

0

y2 0 dt+

Z 4

0

02dt = 0Z 2

y2dx+ x2dy =

�x = t dx = dt 0 � t � 2y = 4 dy = 0dt

�=

Z 2

0

42dt+

Z 2

0

t2 0 dt = 32Z

y2dx+ x2dy =

Z 1

y2dx+ x2dy +

Z 2

y2dx+ x2dy = 32:

2

III.1.2Evaluate

R xy dx both directly and using Green�s theorem, where

is the boundary of the square with vertices at (0; 0), (1; 0), (1; 1),and (0; 1).

Solution (A. Kumjian)

1

1

γ

γ

γ

γ

R

1234

VII.1.2

Denote the square by D and note that xy dx = P dx + Qdy where P = xyand Q = 0. Then P and Q are continuously di¤erentiable on �D and = @D,hence by Green�s Theorem we have,

Z

xy dx =

Z@D

P dx+Qdy =

ZZD

�@Q

@x� @P

dy

�dx dy =

Z 1

0

Z 1

0

�x dx dy =�12:

Observe that = 1 + 2 + 3 + 4 where 1 and 3 are the bottom andtop of the square while 2 and 4 are the last two sides taken in the orderindicated by the order of the vertices in the statement of the problem (so theboundary is oriented counter clockwise). Note that the path integrals on 2and 4 are zero because the edges are vertical.

Z

xy dx =

Z 1

xy dx+

Z 3

xy dx =

Z 1

0

x � 0 dx�Z 1

0

x � 1 dx = 0� 12= �1

2:

3

III.1.3Evaluate

R@Dx2dy both directly and using Green�s theorem, where

D is the quarter-disk in the �rst quadrant bounded by the unitcircle an the two coordinate axes.

SolutionEvaluate

R@Dx2dy directly, set @D = 1 + 2 + 3.

Z 1

x2dy =

�x = t dx = dt 0 � t � 1y = 0 dy = 0 dt

�=

Z 1

0

t2 0 dt = 0Z 2

x2dy =

�x = cos t dx = � sin t dt 0 � t � �=2y = sin t dy = cos t dt

�=

Z �=2

0

cos2t cos t dt =2

3Z 3

x2dy =

�x = 0 dx = 0 dt 0 � t � 1y = 1� t dy = �dt

�= �

Z 1

0

02dt = 0Z@D

x2dy =

Z 1

x2dy +

Z 2

x2dy +

Z 3

x2dy =2

3

Now we evaluateR@Dx2dy, this time using Green�s theorem. In this case,

P (x; y) = 0 and Q (x; y) = x2.

Z@D

x2dy =

ZZD

2x dxdy =

= 2

ZZD

x dxdy =

�x = r cos � dx dy = r dr d� 0 � r � 1y = r sin � 0 � � � �=2

�= 2

Z 1

0

Z �=2

0

r cos � rdrd� =

= 2

Z 1

0

r2dr �Z �=2

0

cos �d� =2

3:

4

III.1.4Evaluate

R y dx both directly and using Green�s theorem, where

is the semicircle in the upper half-plane from R to �R.

SolutionEvaluate

R y dx directly, set = 1 + 2.

Z 1

y dx =

�x = t dx = dt �R � t � Ry = 0 dy = 0 dt

�=

Z �R

�R0 dt = 0Z

2

y dx =

�x = R cos t dx = �R sin t dt 0 � t � �y = R sin t dy = R cos t dt

�= �

Z �

0

R sin t �R sin t dt = ��R2

2Z@D

x2dy =

Z 1

x2dy +

Z 2

x2dy = ��R2

2

Now we evaluateR y dx, this time using Green�s theorem. In this case,

P (x; y) = y and Q (x; y) = 0.

Z@D

y dx = �ZZ

D

dx dy =

= �ZZ

D

dx dy =

�x = r cos � dx dy = r dr d� 0 � r � Ry = r sin � 0 � � � �

�= �

Z R

0

Z �

0

r dr d� =

= �Z R

0

r dr �Z �

0

d� = ��R2

2:

5

III.1.5Show that

R@Dx dy is the area of D, while

R@Dy dx is minus the area

of D.

SolutionUsing in the two casesFor

R@Dx dy, we have P = 0 and Q = x, using Green�s theorem we haveZ

@D

x dx =

ZD

Zdx dy = AreaD:

ForR@Dy dx, we have P = y and Q = 0, using Green�s theorem we haveZ

@D

y dx = �ZD

Zdx dy = �AreaD:

6

III.1.6Show that if P and Q are continuous complex-valued functions ona curve , then R

Pdx

z � w+R

Qdy

z � w; (z = x+ iy)

is analytic for w 2 Cn . Express F 0 (w) as a line integral over .

SolutionDi¤erentiate by hand, use uniform convergence of

1

�w

�1

z � (w +�w) �1

z � w

�=

1

(z � w) (z � (w +�w)) !1

(z � w)2

as �w ! 0, uniformly for z 2 . Get

F 0 (w) =R

Pdx

(z � w)2+R

Qdx

(z � w)2; z = x+ iy

7

III.1.7Show that the formula in Green�s theorem is invariant under coor-dinate changes, in the sense that if the theorem holds for a boundeddomain U with piecewise smooth boundary, and if F (x; y) is asmooth function that maps U one-to-one onto another such domainV and that maps the boundary of U one-to-one smoothly onto theboundary of V , then Green�s theorem holds for V . Hint. First notethe change of variable formulae for line and area integrals, givenby

Z@V

P d� =

Z@U

(P � F )�@�

@xdx+

@�

@ydy

�;ZZ

V

Rd� d� =

ZZU

(R � F ) det JF dx dy;

where F (x; y) = (� (x; y) ; � (x; y)), and where JF is the Jacobian ma-trix of F . Use these formulae, with R = �@P=@�. The summandRQd� is treated similarly.

Solution

ZZV

�@P@�

d� d� =

ZZU

��@P

@�

�(� (x; y) ; � (x; y)) det (JF ) dx dy =

F (x; y) = (� (x; y) ; � (x; y))

�@F@x= @�

@x+ @�

@x@F@y= @�

@y+ @�

@y

det JF (x; y) =@�

@x

@�

@y� @�

@y

@�

@x=

= �ZZU

@P

@�(� (x; y) ; � (x; y))

�@�

@x

@�

@y� @�

@y

@�

@x

�dx dy

Using Green�s theorem

8

Z@V

Pd� =

=

Z@U

(P � F )�@�

@xdx+

@�

@ydy

�=

Z@U

P (� (x; y) ; � (x; y))

�@�

@xdx+

@�

@ydy

�=

=

ZZU

�� @

@y

�P@�

@x

�+

@

@x

�P@�

@y

��dx dy =

=

ZZU

�� @

@yP (� (x; y) ; � (x; y))

@�

@x� P

@2�

@y@x+

@

@xP (� (x; y) ; � (x; y))

@�

@y+ P

@2�

@x@y

�dx dy =

=

ZZU

���@P@�

@�

@y� @P

@�

@�

@y

�@�

@x+

�@P

@�

@�

@x+@P

@�

@�

@x

�@�

@y

�dx dy =

=

ZZU

��@P@�

@�

@y

@�

@x+@P

@�

@�

@x

@�

@y

�dx dy =

= �ZZU

@P

@�JF dx dy:

Similar argument as above hold forRQd�

Here we use that Z@V

Qd� =

Z@U

(Q � F )�@�

@xdx+

@�

@ydy

�and replace R = �@P=@� by R=@Q=@� to conclude thatZ

@V

Qd� =

ZZV

@Q

@�d�d�:

9

III.1.8Prove Green�s theorem for the rectangle de�ned by x0 < x < x1 andy0 < y < y1 (a) directly, and (b) using the result for triangles.

Solution

I

II

III

IV R

VII.1.8a

I

II

III

IV ST

VII.1.8b

(a)Make �gure and set the linesegment between (x0; y0) and (x1; y0) to 1 andproceed in this way in counterclockwise direction. We haveZ

1

Qdy =

Z 2

P dx =

Z 3

Qdy =

Z 4

P dx = 0:

Integrate around the rectangle

10

ZZR

�@Q

@x� @P

@y

�dx dy =

=

Z y1

y0

�Z x1

x0

@Q

@xdx

�dy �

Z x1

x0

�Z y1

y0

@P

@ydy

�dx =

=

Z y1

y0

[Q (x1; y)�Q (x0; y)] dy �Z x1

x0

[P (x; y1)� P (x; y0)] dx =

=

Z y1

y0

Q (x1; y) dy �Z y1

y0

Q (x0; y) dy�Z x1

x0

P (x; y1) dx+

Z x1

x0

P (x; y0) dx =

=

Z x1

x0

P (x; y0) dx+

Z y1

y0

Q (x1; y) dy �Z x1

x0

P (x; y1) dx�Z y1

y0

Q (x0; y) dy =

=

Z 1

P dx+

Z 2

Qdy +

Z 3

P dx+

Z 4

Qdy =

=

Z@R

(P dx+Qdy) :

(b)Write R = S [ T . The integrals on the diagonal are in di¤erent directionsand will cancel

Z@R

(P dx+Qdy) =

=

Z@S

(P dx+Qdy) +

Z@T

(P dx+Qdy) =

=

ZZS

�@Q

@x� @P

@y

�dx dy +

ZZT

�@Q

@x� @P

@y

�dx dy =

=

ZZR

�@Q

@x� @P

@y

�dx dy:

11

III.2.1Determine whether each of the following line integrals is indepen-dent of path. If it is, �nd a function h such that dh = P dx + Qdy.If it is not, �nd a closed path around which the integral is notzero. (a) x dx+ y dy, (b) x2dx+ y5dy, (c) y dx+ x dy, (d) y dx� x dy.

SolutionWe compute the partial derivatve because if h exist we have that dh =Pdx+Qdy where @P=@y = @Q=@x.(a)

P = x Q = y@P@y= 0 @Q

@x= 0R

Pdx = x2

2+ g (y)

RQdy = y2

2+ f (x)

) h = x2+y2

2

(b)

P = x2 Q = y5@P@y= 0 @Q

@x= 0R

Pdx = x3

3+ g (y)

RQdy = y6

6+ f (x)

) h = 2x3+y6

6

(c)

P = y Q = x@P@y= 1 @Q

@x= 1R

Pdx = xy + g (y)RQdy = xy + f (x)

) h = xy

(d)

P = y Q = �x@P@y= 1 @Q

@x= �1

The line integral is not independent of path, we use Greens�theoremZjzj=1

(y dx� x dy) =

ZZjzj<1

�2 dx dy = �2�:

12

III.2.2Show that the di¤erential

�ydx+ xdy

x2 + y2; (x; y) 6=(0; 0) ;

is closed. Show that it is not independent of path on any annuluscentered at 0.

Solution

P = �yx2+y2

Q = xx2+y2

@P@y=

�(x2+y2)+y(2y)(x2+y2)2

= y2�x2(x2+y2)2

@Q@x=(x2+y2)�x(2x)

(x2+y2)2= y2�x2

(x2+y2)2

Because @P=@y = @Q=@x;the di¤erential is closed and we can use Green�sTheorem

Ijzj=r

P dx+Qdy =

=

Ijzj=r

�yx2 + y2

dx+x

x2 + y2dy =

1

r2

Ijzj=r

�y dx+ x dy =

=1

r2

ZZjzj<r

(1 + 1) dx dy =1

r2� 2 � r2 � � = 2� 6= 0

Thus the line integral is not independent of path.

13

III.2.3Suppose P and Q are smooth functions on the annulus fa < jzj < bgthat satisfy @P=@y = @Q=dx. Show directly using Green�s theoremthat

Hjzj=r Pdx+Qdy is independent of the radius r, for a < r < b.

Solution (A. Kumjian)

γ

γ

1

2 D

III.2.3

Let r1; r2 be given so that a < r1 < r2 < b. We must prove thatIjzj=r1

Pdx+Qdy�=

Ijzj=r2

Pdx+Qdy:

Let D denote the annulus fz 2 C : r1 < jzj < r2g. Observe that D is abounded domain with piecewise smooth boundary @D = 1 [ 2 where 1denotes the circle fz 2 C : jzj = r1g with clockwise orientation while 2 de-notes the circle fz 2 C : jzj = r2g with counter clockwise orientation. More-over, both P and Q are continuously di¤erentiable on �D = D [ @D. Hence,Green�s Theorem applies and we obtainZ

@D

Pdx+Qdy =

ZZD

�@Q

@x� @P

@y

�dxdy = 0;

and since, as noted above, @D consists of the two circles with opposite ori-entation, Z

@D

Pdx+Qdy =

Ijzj=r2

Pdx+Qdy �Ijzj=r1

Pdx+Qdy:

Hence, equation (�) holds andHjzj=r Pdx+Qdy is independent of the radius

r, for a < r < b.

14

III.2.4Let P and Q be smooth functions on D satisfying @P=@y = @Q=@x.Let 0 and 1 be two closed paths in D such that the straight linesegment from 0 (t) to 1 (t) lies in D for every parameter valuet. Then

R 0P dx + Qdy =

R 1P dx + Qdy. Use this to give another

solution to the preceding exercise.

SolutionUse the theorem on page 81. Use the straight lines to deform 0 to 1. De�ne s (t) = s 1 (t) + (1� s) 0 (t), 0 6 s 6 1, a1 6 t 6 b1. The theorem on page81 applies. Obtain the result for the annulus above by parameterizing thecircles jzj = r0 and jzj = r1 by 0 (t) = r0e

2�it, 1 (t) = r1e2�it, 0 6 t 6 1. The

straight line segments joining 0 (t) to 1 (t) are radical are in the annulus,so the �rst part applies.

15

III.2.5Let 0 (t) and 1 (t), 0 � t � 1, be paths in the slit annulus fa < jzj < bg n (�b;�a)from A to B. Write down explicitly a family of paths s (t) from Ato B in the slit annulus that deforms continuously to 1.Suggestion. Deform separately the modulus and the principal valueof the argument.

SolutionWrite 0 (t) = r0e

i�0(t), 0 6 t 6 1 , when �� < �0 (t) < �, r0 (t), �0 (t)continuous. (Use the fact that � (t) = Arg 0 (t) is continuous on the slitannulus) Also 1 (t) = r1e

i�1(t).Then consider

s (t) = [sr1 (t) + (1� s) r0 (t)] ei[s�1(t)+(1�s)�0(t)]; 0 � s � 1:

This does the trick, it deform 0 to 1 continuously in D, and each s is apath in D from A to B.We see that s (t) belongs to the slit annulus for 0 � s � 1 and 0 � t � 1since for every 0 � 5 � 1 we have

a < min (r0 (t) ; r1 (t)) � sr1 (t) + (1� s) r0 (t) � max (r0 (t) ; r1 (t)) < b

and

�� < min (�0 (t) ; �1 (t)) � s�1 (t) + (1� s) �0 (t) � max (�0 (t) ; �1 (t)) < �

16

III.2.6Show that any closed path (t), 0 � t � 1, in the annulus fa < jzj < bgcan be deformed continuously to the circular path � (t) = (0) e2�imt,0 � t � 1, for some integer m. Hint. Reduce to the case wherej (t)j � j (0)j is constant. Then start by �nding a subdivision0 = t0 < t1 < � � � < tn = 1 such that arg (t) has a continuous deter-mination on each interval tj�1 � t � tj.

SolutionWe �rst note that there are two cases. For the closed path (t) enclosingthe origin we can deform (t) to a curve with constant modulus. Supposej (t)j = r for 0 6 t 6 1 and (0) = (1) = 1 where a < r < b. Followhint, write (t) = rei�j(t), tj�1 6 t 6 tj. Note �j (tj)� �j+1 (tj) is a integralmultiple of 2�, since (t) is continuous. Add multiples of 2� to the �j �s,obtain � (t) continuous for 0 6 t 6 1 such that � (0) = 0 and (t) = rei�(t).Note � (1) = 2�m for some integer m. Deform by s (t) = rei[(1�s)�(t)+2�sm].The second case is when (t) does not enclose the origin. For this case (t)the counter of a region that lies entirely inside the annulus. Therefrom thecurve (t) may be deformed to any point of this region. In particular itcan be continuously deformed to the point (0) = � (t) = (0) e2�imt whenm = 0.

17

III.2.7Show that if 0 and 1 lie in di¤erent connected components of thecomplement C�nD of D in the extended complex plane, then thereis a closed path in D such that

R d� 6= 0. Hint. The hypothesis

means that there are � > 0and a bounded subset E of CnD suchthat 0 2 E; and every point of E has distance at least 5� from everypoint of CnD not in E. Lay down a grid of squares in the planewith side length �;and let F be the union of the closed squares inthe grid that meet E or that border on a square meeting E. Showthat @F is a �nite union of a closed paths in D, and that

R@Fd� = 2�.

SolutionProceed as in the hint. Assume that 0 is the centre of one of the squaresS0 in the grid. Then

R@S0

d� = 2�, whileR@S

d� = 0 for any other square inthe grid. Thus

PR@Sj

d� = 2�, where we sum over squares. S0; S1; : : : ; Snin F . If two squares in F are ?????, the corresponding integrals along theedge cancel. Then this is the sum over the integrals over the edges of theSj ´s that have F on one side and CnF on the other, that is the edges thatform @F . We have thus

R@Fd� = 2�. Note that by construction, @F � D.

Now note how edges @F can meet. Either a vertex in @F have one edgecoming in and one out, or it have two coming in and two out. By starting onfollowing edges, and making a left turn at vertices where 4 edges meet, wemust eventually end where we started, with a closed path. Three paths mustbe disjoint, except for vertices with four edges. Call the paths 1; : : : ; m.Since

PR jd� = 2�, we must have

R d� 6= 0 for one of the j �s. (In fact,

we will haveR jd� = 2� on the j ´s we get by moving from 0 to @F .)

18

III.3.1For each of the following harmonic functions u, �nd du, �nd dv, and�nd v, the conjugate harmonic functions of u.(a) u (x; y) = x� y (c) u (x; y) = sinhx cos y(b) u (x; y) = x3 � 3xy2 (d) u (x; y) = y

x2+y2

Solution(a)Use Cauchy-Riemanns equations page 83.

u (x; y) = x� ydu = dx� dydv = dx+ dyv (x; y) = x+ y

(b)Use Cauchy-Riemanns equations page 83.

u (x; y) = x3 � 3xy2du = (3x2 � 3y2) dx� 6xy dydv = 6xy dx+ (3x2 � 3y2) dyv (x; y) = 3x2y � y3

(c)Use Cauchy-Riemanns equations page 83.

u (x; y) = sinhx cos ydu = cosh x cos y dx � sinh x sin y dydv = sinhx sin y dx+ coshx cos y dyv (x; y) = coshx sin y

(d)Use Cauchy-Riemanns equations page 83.

u (x; y) = yx2+y2

du = �2xy(x2+y2)2

dx+ x2+y2�y(2y)(x2+y2)2

dy = �2xy(x2+y2)2

dx+ x2�y2(x2+y2)2

dy

dv = y2�x2(x2+y2)2

dx� 2xy

(x2+y2)2dy

v (x; y) = xx2+y2

19

III.3.2Show that a complex-valued function h (z) on a star-shaped domainD is harmonic if and only if h (z) = f (z)+ g (z), where f (z) and g (z)are analytic on D.

SolutionWrite h = u + iw where u, w are harmonic. By the Theorem on page 83both u and w have harmonic conjugate in D, (since D is star-shaped), sothus there are ', analytic such that u = Re', w = Re .

u = ('+ ') =2; w =� +

�=2;

gives

h = 12

�'+ '+ i + i

�= 1

2['+ i ] + 1

2

�'+ i

�:

Take

f = 12['+ i ] ; g = 1

2['� i ] ;

then h (z) = f (z) + g (z).

To show the oposite direction. Assume that h (z) = f (z) + g (z) where f (z)and g (z) are analytic on D. Then h = Re f + Re g + i (Im f + Im g) andfrom Cauchy-Riemanns equations follows that

@2 (Re f)

@x2+@2 (Re g)

@x2+ i

�@2 (Im f)

dx2� @2 (Im g)

@x2

�+

+@2 (Re f)

@y2+@2 (Re g)

@y2+ i

�@2 (Im f)

@y2� @2 (Im g)

@y2

�=

=@2 Im f

@x@y+@2 Im g

@x@y+ i

��@

2Re f

@x@y+@2Re g

@x@y

�� @2 Im f

@y@x� @2 Im g

@y@x+

+ i

�@2Re f

@y@x� @2Re g

@y@x

�= 0

so h (z) is harmonic.

20

III.3.3Let D = fa < jzj < bg n (�b;�a), an annulus slit along the negativereal axis. Show that any harmonic function on D has a harmonicconjugate on D.Suggestion. Fix c between a and b, and de�ne v (z) explicitly as a lineintegral along the path consisting of the straight line from c to jzjfollowed by the circular arc from jzj to z. Or map the slit annulusto a rectangle by w = Log z.

SolutionIntegrate the C�R equations we have

v (z) =

Z z

z0

�1r

@u

@�dr + r

@u

@rd� = �

Z@u

@ydx+

Z@u

@xdy

integrate along r � interval, then � � interval. Perception is same as gettingfrom z0 to z1, then from z1 to little disk. Integral is well-de�ned, continuous,harmonic and is harmonic conjugate for u.

21

III.3.4 (From Hints and Solutions)Let u (z) be harmonic on the annulus fa < jzj < bg. Show that thereis a constant C such that u (z) � C log jzj has a harmonic conjugateon the annulus. Show that C is given by

C =r

2�

Z 2�

0

@u

@r

�rei�

�d�;

where r is any �xed radius, a < r < b.

Solutionu has harmonic conjugate v1 on the annulus slit along (�b;�a), and also aharmonic conjugate v2 on the annulus slit (a; b). Since v1 � v2 is constantabove the slit (�b;�a), and also constant below the slit, v1 jumps by aconstant across the slit. Arg z also jumps by a constant across the slit. Byappropriate choice of C, v1 � C Arg z is continuous across the slit (�b;�a),and u�C log jzj has a harmonic conjugate v1�C log jzj on the annulus. Forthe identity, use the polar form of the Cauchy-Riemann equations to convertthe r�derivative of u to a ��derivative of v.

22

III.3.5The �ux of a function u across a curve is de�ned to beZ

@u

@nds =

Z

ru � n ds;

where n is the unit normal vector to and ds is arc length. Showthat if a harmonic function u on a domain D has a conjugate har-monic function v on D, then the integral giving the �ux is inde-pendent of path in D. Further, the �ux across a path in D fromA to B is v (B)� v (A).

Solution�!t =

�dsdx; dsdy

�, �!n =

�dyds;�dx

ds

�,R

ru�!n ds =

R

�@u@x

dyds� @u

@ydxds

�ds =

R �@u@ydx+ @u

@xdy =|{z}

�R @v@xdx+ @v

@ydy =

R dv

(�) C�R equations. If goes from P to Q, this is v (Q)� v (P ), which showsthat the integral is independent of path in D.

23

III.4.1Let f (z) be a continuous function on a domain D. Show that iff (z) has the mean value property with respect to circles, as de�nedabove, then f (z) has the mean value property with respect to disks,that is if z0 2 D and D0 is a disk centered at z0 with area A andcontained in D, then f (z0) =

1A

RRD0f (z) dx dy.

Note: This MVP for area is not exactly what is de�ned as the MVP fordisks. What is then is that if h (z0) is the average of h (z) for any circlefjz � z0j = rg, 0 < r < 1. Then h (z) is the average of h (z) over any diskfjz � z0j = r0g, 0 < r < r0. Express dx dy in polar coordinates at z0 andintegrate �rst with respect to �.

SolutionSuppose that f satis�es the mean value property,

f (z0) =1

2�

Z 2�

0

f�z0 + rei�

�d�

for all z0 2 D and r > 0 such thatBr (z0) �D. LetD0 = fz 2 C : jz � z0j � Rg �D. We have, A = �R2 and parametrizing in polar coordinates, and use themean value property with respect to circles,

1

A

ZZD0

f (z) dx dy =1

�R2

ZZD0

f (z) dx dy =

�x = x0 + r cos � dx dy = r dr d� 0 � r � Ry = y0 + r sin � 0 � � � 2�

�=

=1

�R2

Z 2�

0

Z R

0

f�z0 + rei�

�r dr d� =

1

�R2

Z R

0

�Z 2�

0

f�z0 + rei�

�d�

�r dr =

1

�R2

Z R

0

2�f (z0) r dr =

=2f (z0)

R2

Z R

0

r dr =1

R2

�r2

2

�R0

f (z0) = f (z0) :

Another note: It�s better to use something other than 1=A as A was alreadyused as the circle average.

24

III.4.2Derive (4.2) from the polar form of the Cauchy-Riemann equations(Exercise II.3.8).

SolutionThe polar form of Cauchy-Riemann�s equations (from Exercise II.3.8)

@u

@r=1

r

@v

@�;

@u

@�= �r@v

@r:

Now we have

r

Z 2�

0

@u

@r

�z0 + rei�

�d� =

Z 2�

0

@v

@�

�z0 + rei�

�d� = 0

25

III.4.3A function f (t) on an interval I = (a; b) has the mean value propertyif

f

�s+ t

2

�=f (s) + f (t)

2; s; t 2 I:

Show that any a¢ ne function f (t) = At + B has the mean valueproperty. Show that any continuous function on I with the meanvalue property is a¢ ne.

Solution (A. Kumjian)Let I := (a; b) and let f : I ! R be given. Suppose �rst that f is an a¢ nefunction, that is, there are A;B 2 R such that f (t) = At + B for all t 2 I.Now let s; t 2 I be given, then

f

�s+ t

2

�= A

�s+ t

2

�+B =

1

2(As+B) +

1

2(At+B) =

f (s) + f (t)

2:

Hence, f has the mean value property.Next suppose that f is continuous and that it has the mean value property.We must prove that f is a¢ ne. We claim that for any c; d 2 I, with c < d,and t 2 [0; 1] we have

f (tc+ (1� t) d)�= tf (c) + (1� t) f (d) :

So let c; d 2 I be given with c < d. We will �rst prove that equation (�)holds for t = k=2m where k;m are nonnegative integers with k � 2m. Wedo this by induction on m. For m = 0, we have t = k = 0; 1 and theassertion is obvious. Now suppose that the assertion holds for some integerm � 0 (that is, equation (�) holds for t = k=2m for all k = 0; : : : ; 2m).This inductive step entails showing that equation (�) holds for t = k=2m+1

for any nonnegative integer k with k � 2m+1. By the inductive hypothesiswe may assume k = 2j + 1 for some integer j = 0; : : : ; 2m � 1 (for if kwere even we could rewrite t as j=2m for some integer j with 0 � j � 2m).Then setting t0 = j=2m and t1 = (j + 1) =2m, we have t = (t0 + t1) =2 and1� t = ((1� t0) + (1� t1)) =2, thus,

26

tc+ (1� t) d =1

2((t0c+ (1� t0) d+ (t1c+ (1� t1) d))

hence,

f (tc+ (1� t) d) =

= f

�1

2((t0c+ (1� t0) d+ (t1c+ (1� t1)d)

�=

=1

2(f (t0c+ (1� t0) d) + f (t1c+ (1� t1) d)) = by the mean value property

=1

2(t0f (c) + (1� t0) f (d) + t1f (c) + (1� t1) f (d)) = by inductive hypothesis

= tf (c) + (1� t) f (d) :

So we have proved that equation (�) holds for t = k=2m where k;m arenonnegative integers with k � 2m. Since such numbers are dense in [0; 1]the claim follows by the continuity of f . It is now straightforward to verifythat for any c; d 2 I, with c < d, there are unique constants A and B suchthat f (x) = Ax + B for all x 2 [c; d] (take A = (f (c)� f (d)) = (c� d) andB = f (c)�Ac ). The desired result now follows by observing that A and Bdo not depend on c; d.

27

III.4.4Formulate the mean value property for a function on a domainin R3, and show that any harmonic function has the mean valueproperty.Hint. For A 2 R3 and r > 0, let Br be the ball of radius r centeredat A, with volume element d� , and let @Br be its boundary sphere,with area element d� and unit outward normal vector n. Apply theGauss divergence theoremZZ

@Br

F � n d� =ZZZ

Br

r � F d�

to F = ru.

SolutionProve MV theorem for harmonic functions in Rn , as follows u (R�!x ) �u��!0�=R R0

@u@r(r�!x ) dr. Let d� = area measure, �!x = unit vector, getR

@u@r(R�!x ) d� �

R@u@r(r0�!x ) d� =

R R0

RS@2u@r2(r�!x ) drd� (�!x ). Get directional

derivative of Strokes formula. Better: Apply stroke�s theorem to shell f�0 < � < �1g,getR�

: : :R �!V � �!n d surface =

RVolume

: : :Rr�!V volume. Apply to ru = �!v ,

getR�

: : :Rru�!n d surface =

R: : :Rr2�!n d volume = 0,R

j�!x�x0j=�1: : :R

@u@�d� �

Rj�!x�x0j=�0

: : :R

@u@�d�

@u@�

Rj�!x�x0j=�

: : :R_ud� is constant.

28

III.5.1Let D be a bounded domain, and let u be a real-valued harmonicfunction on D that extends continuously to the boundary @D. Showthat if a � u � b, then a � u � b on D.

SolutionSince D [ @ is a compact set we have that u attains both maximum andminimum values. Assume that u (z0) = c > b fore some z0 2 D. Then itfollows by the maximum principle that u � c onD, but this is a contradictionto the asumption that u is continuously extended to @D. Thereby u � b onD.For �u we have that �b � �u � �a on @D. Assume that �u (z0) = c0 > �afor seme z0 2 D. Then by the maximum principle �u � c0 on D, but thiscontradicts that �u is continuously extended to @D with �u � �a on @D.So �u � �a on D, equivalently a � u � b on D.This show that a � u � b on D.

29

III.5.2Fix n � 1, r > 0 and � = �ei'. What is the maximum modulus ofzn+� over the disk fjzj � rg? Where does zn+� attain its maximummodulus over the disk?

Solution (A. Kumjian)We claim that the maximummodulus of zn+� over the disk fz 2 C : jzj � rgis rn + � and this is attained at � = rei� where � = (2k� + ')=n for k =0; 1; : : : ; n � 1. By the maximum modulus principle the maximum modulusoccurs on the boundary fz 2 C : jzj = rg. For z 2 C with jzj = r we haveby the triangle inequality

jzn + �j � jznj+ j�j = rn + � =��(rn + �) ei'

�� = j�n + �j :so the claim is proven.

30

III.5.3Use the maximum principle to prove that the fundamental theoremof algebra, that any polynomial p (z) of degree n � 1 has a zero, byapplying the maximum principle to 1=p (z) on a disk of large radius.

SolutionIt is su¢ cient to show that any p (z) has one root, for by division we canthen write p (z) = (z � z0) g (z), with g (z) of lower degree.Note that if

p (z) = anzn + an�1z

n�1 + � � �+ a0;

then as jzj ! 1, jp (z)j ! 1. This follows as

p (z) = zn���an + an�1

z+ � � �+ a0

zn

��� :Assume p (z) is non-zero everywhere. Then 1

p(z)is bounded when jzj � R.

Also, p (z) 6= 0, so 1p(z)

is bounded for jzj � R by continuity. Thus, 1p(z)

is abounded entire function, which must be constant. Thus, p (z) is constant, acontradiction which implies p (z) must have a zero.

Solution (K. Seip)If p (z) has no zeros, then 1=p (z) is an entire function. Also, if we denote bym (R) the maximum of 1=p (z) on the circle jzj = R, then m (R) ! 0 whenR ! 1, unless p (z) is a constant. By the maximum principle j1=p (z)j �m (R) when jzj � R, which means 1=p (z) = 0. This is impossible, and so1=p (z) is not an entire function.

31

III.5.4Let f (z) be an analytic function on a domain D that has no zeroson D. (a) Show that if jf (z)j attains its minimum on D, then f (z) isconstant. (b) Show that if D is bounded, and if f (z) extends con-tinuously to the boundary @D of D, then f (z) attains its minimumon @D.

Solution(a)If jf (z)j attains its minimum on D, then j1=f (z)j attains its maximum onD, which can only happens if f (z) is a constant, by the maximum principle.(b)Since D[@D is compact and f (z) is continuous it follows that jf (z)j attainsboth maximum and minimum value on D [ @D. Assume that jf (z)j doesnot attain its minimum on @D. Then jf (z)j attains its minimum on D andit follows by (a) that f is constant so jf (z)j attains its minimum on @D.

32

III.5.5Let f (z) be a bounded analytic function on the right half-plane.Suppose that f (z) extends continuously to the imaginary axis andsatis�es jf (iy)j � M for all points iy on the imaginary axis. Showthat jf (z)j �M for all z in the right half-plane.Hint. For " > 0 small, consider (z + 1)�" f (z) on a large semidisk.

SolutionNow we consider the function (z + 1)�" f (z) in the right half-plane, becausef (z) is bounded here, we assume that jf (z)j � C in this domain. On asemidisk of radius r in the right half-plane we have��(z + 1)�" f (z)�� � C

jr � 1j"

For r = R su¢ cient large, we have��(z + 1)�" f (z)�� � C

(R� 1)" �M

for all jzj � R, when Re z � 0.On the imaginary axis we have that��(z + 1)�" f (z)�� �M

because jf (iy)j �M for all points iy on the imaginary axis.By the maximum principle, we have��(z + 1)�" f (z)�� �M

for jzj � R, when Re z � 0.Now let "! 0, we have

jf (z)j �M

for all z in the right half-plane.

Solution (K. Seip)We assume that jf (z)j � C for Re z > 0 and jf (iy)j �M . For " > 0 set

F" (z) = (1 + z)�" f (z) :

33

Then

jF" (iy)j �M

and ��F" �Rei���� � C �R�" �M

for su¢ ciently large R.Thus for arbitary z, Re z > 0 we have

jF" (z)j �M;

or

jf (z)j �M (1 + jzj)" :This holds for every " > 0, thus jf (z)j �M .

34

III.5.6Let f (z) be a bounded analytic function on the right half-plane.Suppose that lim sup jf (z)j � M as z approaches any point of theimaginary axis. Show that jf (z)j �M for all z in the half-plane.Remark . This is a technical improvement on the preceding exercisefor students who can deal with lim sup (see Section V.1).

SolutionSet

g (z) = (z + 1)�" f (z) :

Then

jg (z)j 6 jf (z)jbecauce jz + 1j > 1 for all z in the right half-plane.Take R large so that jg (z)j 6M for jzj > R. The lim sup condition impliesthat there is � > 0 such that

jg (z)j 6M + "; jzj < R; 0 < Re z 6 �:

Apply the maximum principles to the domain fzjjzj < R;Re z > �g, to ob-tain jg (z)j 6M + ". Then g (z)! f (z) as "! 0.

35

III.5.7Let f (z) be a bounded analytic function on the open unit disk D.Suppose there are a �nite number of points on the boundary suchthat f (z) extends continuously to the arcs of @D separating thepoints and satis�es

��f �ei���� �M there. Show that jf (z)j �M on D.Hint. In the case that there is only one exceptional point z = 1,consider the function (1� z)" f (z).

SolutionLet the points be a1; : : : ; an. Then (z � aj)

", where 1 � j � n is analytic(take analytic branch). We have that jz � ajj" ! 0 as z ! aj for 1 � j � n.Then [

Q(z � aj)

"] f (z) = f" (z) satis�es f" (z) is continuous in D [ @D.jf" (z)j 6 M on @D. jf" (z)j 6 M on D. Let " ! 0, obtain jf (z)j 6 M onD.

SolutionLet C = supz2D jf (z)j. By de�nition jf (z)j � C. Let z1; : : : ; zn 2 @D be allthe points for which f (z) does not extend continuously.Now suppose that jf (z)j �M for all z 2 @D except fz1; : : : ; zng and to thecontrary M < C, then there is a z0 2 D such that jf (z0)j = M + � with0 < � � C �M .Let

g (z) = (z � z1)" � : : : � (z � zn)

" f (z)

where " is chosen so that 8z 2 D [ @D.We have that

M + �=2

M + �� j(z � z1)

" � : : : � (z � zn)"j and j(z � z1)

" � : : : � (z � zn)"j � M + �=3

M

Note that g (z) extends continuously to @D and that g (z) is analytic in D.By the maximum modulus principle, and since

jg (z0)j = j(z0 � z1)" � : : : � (z0 � zn)

"j f (z0) � (M + �)M + �=2

M + �

we have that

36

M + �=2

M + ��(M + �) � jg (z0)j � sup

z2@Dj(z � z1)

" � : : : � (z � zn)"j �jf (z)j � M + �=3

M�M

which is a contradiction.

37

III.5.8Let f (z) be a bounded analytic function on a horizontal strip inthe complex plane. Suppose that f (z) extends continuously to theboundary lines of the strip and satis�es jf (z)j � M there. Showthat jf (z)j � M for all z in the strip. Hint. Find a conformal mapof the strip onto D and apply Exercise 7.

SolutionAssume strip is

���2< Im z < �

2

�, so it is mapped to right half-plane by

w = �i log z, z = eiw. Then g (w) = f (eiw) hold on right half�plane, 6 Mat all point of iR except at w = 0. Modify proof idea of Exercise III.5.5 toget jgj 6M on the half�plane.

38

III.5.9Let D be an unbounded domain, D 6= C, and let u (z) be a harmonicfunction on D that extends continuously to the boundary @D. Sup-pose that u (z) is bounded below on D, and that u (z) � 0 on @D.Show that u (z) � 0 on D. Hint. Suppose 0 2 @D, and considerfunctions of the form u (z) + � log jzj on D \ fjzj > "g.

Solution (From Hints and Solutions)Let " > 0. Take � > 0 such that u (z) > �" for z 2 D, 0 < jzj < �. Let� > 0. Take R > 0 so large that u (z) + � log jzj > 0 for jzj > R. By themaximum principle, u (z) + � log jzj > �" + � log � for z 2 D, � < jzj < R,hence for all z 2 D such that jzj > �. Let � ! 0, then let " ! 0, to obtainu (z) > 0 on D.

Solution (K. Seip)We may assume 0 2 @D. By continuity of u at 0, we can �nd � and � so that(�; � > 0)

u (z) + � log jzj � �"; z 2 D \ fjzj = �gfor arbitary " > 0. Since u is bounded below, there is also R0 > 0 such thatfor R > R0

u (z) + � log jzj � 0; z 2 D \ fjzj = Rg ;Applying the maximum principle to �u (z)� � log jzj in D \ f� < jzj < Rg,we get u (z) � "� � log jzj. Since " and � can be chosen arbitarily small, theresult follows.

39

III.5.10Let D be a bounded domain, and let z0 2 @D. Let u (z) be a har-monic function on D that extends continuously to each boundarypoint of D except possibly z0. Suppose u (z) is bounded below onD, and that u (z) � 0 for all z 2 @D, z 6= z0. Show that u (z) � 0 onD.

SolutionD hold, z0 2 @D, u > �M on D, u > 0 (@D) n fz0g. The map ' (z) =1= (z � z0) maps D onto an unbounded domain, to which the result of Ex-ercise 9 applies. Or: consider w (z) = u (z) + � log (1= jz � z0j). Assumejz � z0j 6 R on D, then take " > 0 small, assume R > 1. Then w (z) >" log (1=R) at all points of (@D) n fz0g, w (z) ! +1 as z ! z0. By themaximum principle, w (z) > " log (1=R) on D. Let " ! 0, get w (z) > 0 onD.

40

III.5.11Let E be a bounded set of integer lattice points in the complexplane. A point m + ni of E is an interior point of E if its fourimmediate neighbors m� 1 + ni, m+ ni� i belong to E. Otherwisem + ni is a boundary point of E. A function E is harmonic if itsvalue at any interior point of E is the average of its values at thefour immediate neighbors. Show that a harmonic function on abounded set of lattice points attains its maximum modulus on theboundary of the set.

SolutionSuppose u (z) attains its maximum at m+ ni, call it M . Then u (m+ ni) isthe average of its values at its four neighbors, so all their values must coincidewith M. Consider fm+ ni : u (m+ ni) =Mg. This is a �nite set. It have apoint with largest m. This point have a neighbor in the boundary of E. uattains value M at that point. u = c at some point of boundary of E.For complex�valued case, follow same approach as in book. Remark: Shouldjust consider real-valued u, and show it attains its maximum and minimumat boundary points of E.

41

III.6.1Consider the �uid �ow with constant velocity V = (2; 1). Find thevelocity potential � (z), the stream function (z), and the complexvelocity potential f (z) of �ow. Sketch the streamlines of the �ow.Determine the �ux of the �ow across the interval [0; 1] on the realaxis and across the interval [0; i] on the imaginary axis.

SolutionV = (2; 1), V = r� , � = 2x+ y, @�

@x= @

@y, @�@y= �@

@x) = 2y � x

f (z) = �+ i = 2z � iz = (z � i) zStreamlines are straight line with slope 1=2, �ux across [0; 1] = (1)� (0) =�1, �ux across [0; i] = (i)� (0) = 2.

42

III.6.2Fix real numbers � and �, and consider the vector �eld given inpolar coordinates by

V (r; �) =�

rur +

�

ru�;

where ur and u� are the unit vectors in r and � directions, respec-tively.(a)Show that V (r; �) is the velocity vector �eld of a �uid �ow, and �ndthe velocity potential � (z) of the �ow.(b)Find the stream function (z) and the complex velocity potentialf (z) of the �ow.(c)Determine the �ux of the �ow emanating from the origin. Whenis 0 a source and when is 0 a sink?(d)Sketch the streamlines of the �ow in the case � = �1 and � = 1.

Solutiona)V (r; �) = �

rur +

�ru�

r� = @�@rur +

1r@�@�u� =

�rur +

�ru�

Get @�@r= �

r, � = � log r + h (�)

1r@�@�= h0 (�) = �

r, h0 (�) = �, h (�) = �� ) V = r�

� = � log r + �� is harmonic locally. is the velocity vector �eld of a �owb)� = Re (� log z + � arg z) stream function = = Im (� log z + � arg z) =� arg z � � log jzjcomplex velocity of a point f (z) = (�� i�) log zc)Flux = of around a circle centred at 2��, sinu : � > 0, sinu : � > 0.d)� = �1, � = 1. Note: � and are only locally de�ned.

43

III.6.3Consider the �uid �ow with velocity V = r�, where � (r; �) =(cos �) =r. Show that the streamlines of the �ow are circles andsketch them. Determine the �ux of the �ow emanating from theorigin.

Solution� = cos �

r= r cos �

r2= Re

�1z

�= Re

�zjzj2

� = Im

�1z

�= � sin �

r

Since is single-valued, �ux =Rjzj=" d = 0, and there is no source or sink

at 0. Set w = 1=z, z = 1=w, these streamlines are the curves Im (1=z) =constant, which correspond to the curves Imw = constant get modi�ed totangent to real line at 0 = z (1), and real axis. r� = @�

@rur +

1r@�@�u� =

� cos �r2ur � sin �

r2u�.

44

III.6.4Consider the �uid �ow with velocity V = r�, where

� (z) = log

����z � 1z + 1

���� :Show that the streamlines of the �ow are arcs of circles and sketchthem. Determine the �ux of the �ow emanating from each of thesingularities at �1.

Solution� is the composition of an analytic function w = (z � 1) = (z + 1) and the har-monic function log jwj , so � is harmonic, and � = Re log ((z � 1) = (z + 1)).The stream function is = Im (log (z � 1) = (z + 1)) = arg ((z � 1) = (z + 1)), then is not single-valued.

Hjz�1j="d =

Hjz�1j="d arg (z � 1) = 2� = �ux

emanating from +1.Hjz�1j="d =

Hjz�1j="d arg (z + 1) = �2� = �ux ema-

nating from �1. z = 1 is a source, z = �1 is a sink.Set � = (z � 1) = (z + 1), w = log �. The line Imw = constant in the w�plane correspond to rays issuing from 0 in the �� plane. (since � = ew ), andthis correspond to arcs of circles from 1 to �1 in the z�plane, since FLT´smaps circles to circles. Hence streamlines are arcs of circle from 1 to �1,including the circle through 1.

45

III.6.5Consider the �uid �ow in the horizontal strip f0 < Im z < �g witha sink at 0 and equal sources at �1. Find the stream function (z) and the velocity vector �eld V (z) of the �ow. Sketch thestreamlines of the �ow. Hint. Map the strip to a half-plane by� = ez and solve a Dirichlet problem with constant boundary valueson the three intervals in the boundary separating sinks and sources.

SolutionConsider �uid �ow in a horizontal strip. f0 < Im z < �g with a sink 0, butno source at �1 . Find the stream function (z). Sketch the �ow lines.Find the complex vector �eld V (z). Hint : Map the half-plane by.Map to half-plane � = ez. A arg � + B arg (� � 1) + C, arg � = arg ez = y,Ay+C+B arg (ez � 1), c3 = 0, requires �A+�B+C = 0) C = �� (A+B),A (y � �) + B [arg (ez � 1)� �]. C1 = �C2 require �A� + B� � B� =� [�A� �B�], A [� (y � �) + 2 [arg (ez � 1)� �]] = A [� (� + y) + 2 arg (ez � 1)]

46

III.6.6For a �uid �ow with velocity potential � (z), we de�ne the conju-gate �ow to be the �ow whose velocity potential is the conjugateharmonic function (z) of � (z). What is the stream function ofthe conjugate �ow? What is the complex velocity potential of theconjugate �ow?

SolutionStream function of conjugate �ow in direction r is ��. Complex velocitypotential is g (z) = � i� = �if (z) = �i (�+ i )

47

III.6.7Find the stream function and the complex velocity potential of theconjugate �ow associated with the �uid �ow with velocity vectorur=r. Sketch the streamlines of the conjugate �ow. Do the particlesnear the origin travel faster or slower than particles on the unitcircle?

SolutionFlow ur=r, � = log jzj, = arg z. Complex velocity potential of conjugate�ow is �i log z. Streamlines of the conjugate �ow are �� = � log jzj =constant , r = jzj = constant, which are circles centered in origin. Speed= jV (z)j = jf 0 (z)j = 1= jzj. Particles travel faster near z = 0.

48

III.6.8Find the stream function of the conjugate �ow of

V (r; �) =1

r(�ur + u�) :

Sketch the streamlines of both the �ow and the conjugate �ow onthe same axis. (See Exercise 2d.)

SolutionThis is the vector �eld from 2d. Hence V = r (� log r + �). The velocitypotential is � = � log r + �. The conjugate �ow has velocity potential =�� � log r = Re ((i� 1) log z). The stream function of the conjugate �owis then Im ((i� 1) log z) = �� + log r = ��. (The steam function of theconjugate is always the negative of the velocity potential of the �ow. SeeEx.6) Streamline of conjugate �ow are orthogonal to the stream of the �ow(Exercise 2d) and velocity vector �eld. Conjugate �ow is the rotation by 90�

of the velocity vector �eld of the �ow, so the origin is a sink of both �ows inthis case.

49

III.7.1Find the steady-state heat distribution u (x; y) in a laminar platecorresponding to the half disk fx2 + y2 < 1; y > 0g, where the semi-circular top edge is held at temperature T1 and the lower edge(�1; 1) is held at temperature T2. Find and sketch the isothermalcurves for the heat distribution. Hint. Consider the steady-stateheat distribution for the full unit disk with the top held at temper-ature T1 and the bottom temperature T3, where T2 = (T1 + T3) =2.

Solution� (z) = 2

�(Arg (1 + z)� Arg (1� z)) has temp +1 on top, �1 on bottom

half and is 0 on the interval (�1; 1). Take u = A� + B,�T1 = A+BT2 = B

)�A = /T 1 � T2B = T2

, get u = (T1 � T2)� (z) + T2

u = (T1 � T2)2�[Arg (1 + z)� Arg (1� z)] + T2

The isothermal curves are the curves where u = konstant. these are thecurves

Arg

�1 + z

1� z

�= konsant

then

(1 + z) (1� �z) = 1 + 2i Im z � jzj2

and

Im z

1� jzj2= konstant.

50

III.7.2Find the potential function � (x; y) for the electric �eld for a con-ducting laminar plate corresponding to the quater-disk fx2 + y2 < 1; x > 0; y > 0g,where the two edges on the coordinate axis are grounded (that is,� = 0 on the edges), and the semicircular edge is held at constantpotential V1. Find and sketch the equipotential lines and the linesof force for the electric �eld. Hint. Use the conformal map � = z2

and the solution to the preceding exercise.

Solution� (z) = 2

�(Arg (1 + w (z))� Arg (1� w (z))), �1 = V1

2�(Arg (1 + z2)� Arg (1� z2))

51

III.7.3Find the potential function � (x; y) for the electric �eld for a con-ducting laminar plate corresponding to the unit disk where theboundary quater-circles in each quadrant are held at a constantvoltages V1; V2; V3 and V4. Hint. Map the disk to the upper half-plane by w = w (z) and consider potential functions of the formArg (w � a).

SolutionLet w = i (1� z) = (1 + z), w (1) = 0, w (�1) = 1, w (i) = 1, w (�i) = �1.z ! w (z) maps unit disk to upper half�plane. Su¤ers to �nd a har-monic function u (w) in upper half-plane with boundary values constanton each interval, as above. Any such function is a linear combination ofu�1 (w) =

1�Arg (w + 1), u0 (w) = 1

�Argw, u1 (w) = 1

�Arg (w � 1), and if

� = A�1u�1 + A0u0 + A1u1 + C, we get a system we can back solve.8>><>>:A�1 + A0 + A1 + C = V3

A0 + A1 + C = V3A1 + C = V1

C = V2

)

8>><>>:C = V2A1 = V1 � V2A0 = V4 � V1A�1 = V3 � V4

That does it-just plug in.Example: V2 = 1, V1 = V3 = V4 = 0, this solution is �2 (z) = 1� 1

�Arg

�i1�z1+z

�,

and we can get other elementary solutions, �1 (z), �3 (z), �4 (z), and then� (z) =

Pvj�j (z).

52

III.7.4Find the steady-state heat distribution in a laminar plate corre-sponding to the vertical half-strip fjxj < �=2; y > 0g, where the ver-tical sides at x = ��=2 are held at temperature T0 = 0 and thebottom edge (��=2; �=2) on the real axis is held at temperatureT1 = 100. Make a rough sketch of the isothermal curves and thelined of heat �ow. Hint. Use w = sin z to map the strip to theupper half-plane, and make use of harmonic functions of the formArg (w � a).

Solution

Try � = aArg (w + 1) + bArg (w � 1) + c, want

8<:a+ b+ c = 0a� b+ c = 100�a� b+ c = 0

)8<:a = 50b = �50c = 0

� (z) = 50Arg (sin z + 1)� 50Arg (sin z � 1).

53

III.7.5Find the steady-state heat distribution in a laminar plate corre-sponding to the vertical half-strip fjxj < �=2; y > 0g, where the sidex = ��=2 is held at constant temperature T0, the side x = �=2 isheld at constant temperature T1, and the bottom edge (��=2; �=2)on the real axis is insulated, that is, no heat passes through thebottom edge, so the gradient ru of the solution u (x; y) is parallelthere to the x� axis. Hint. Try linear functions plus constants.

Solutionu (x; y) = ax+ b is harmonic. @u

@y= 0�

��2a+ b = T0

�2a+ b = T1

)�a = T1�T0

�

b = T0+T12

u = 1�(T1 � T0)x+

T0+T12.

54

III.7.6Find the steady-state heat distribution in a laminar plate corre-sponding to the upper half-plane fy > 0g, where the interval (�1; 1)is insulated, the interval (�1;�1) is held at temperature T0, andthe interval (1;1) is held at temperature T1. Make a rough sketchof the isothermal curves and the lines of heat �ow. Hint. Use thesolution to Exercise 5 and the conformal map from Exercise 4.

SolutionLet z = z (w) map the vertical half-strip to the upper half-plane withz (��=2) = �1 and z (�=2) = 1, w = w (z) : upper half�plane ! verticalsemi strip. Take z = sinw, this has desired mapping property, w = sin�1 (z).Then solution is composition of sine function with 2

�(T1 � T0) Rew +

T0+T12.

Get u = 2�(T1 � T0) Re

�sin�1 (z)

�+ T0+T1

2, where we take branch sin�1 z

with values in the vertical semi-strip. Note : This conformal map was usedin Exercise 4.

55

III.7.7The gravitational �eld near the surface of the earth is approxi-mately constant, of the form F = ck, where k is the unit vector inthe z� direction in (x; y; z)� space and the surface of the earth isrepresented by the plane where z = 0 (the �at earth theory). Showthat F is conservative, and �nd a potential function � for F .

Solution� (x; y; z) = cz, r� = c

�!k

Since

r� F =

�@

@x;@

@y;@

@z

�� (0; 0; c) = (0; 0; 0) ;

we have that F is conservative.

56

III.7.8Show that the inverse square force �eld F = ur=r

2 on R3 is con-servative. Find the potential function � for F , and show that � isharmonic.

Solution�!r = �!u r

r2= r�, � (x; y; z) = �1

r

57

III.7.9For n � 3, show that the function 1=rn�2 is harmonic on Rnn f0g.Find the vector �eld F that has this function as its potential.

Solution1

rn�2 =1

(x21+:::+x2n)n�22= u = r2�n = (x21 + : : :+ x2n)

1�n2

@u@xj

=�1� n

2

�(x21 + : : :+ x2n)

�n2 � 2xj = (2� n)xjr

�n

@2u@x2j

= (2� n) r�n + (2� n)xj��n2

�(x21 + : : :+ x2n)

�n2�1 � 2xj =

2�nrn+

(2�n)( �n)x2jrn+2

�u = n2�nrn� (2�n)n

rn+2

Px2j = 0 this shows that u is harmonic on Rnn f0g

F = �u =P

@u@xjej =

(2�n)rn

Pxjej =

2�nrn�1 �

�xjrej =

(2�n)~urrn�1

F = (2r) rrn= c�!u

rn�1RF � d

P= c

rn�1Area (s) = (n� 2)Area(S)rn�1

58

IV 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 1912345678

1

IV.1.11 2 3 P L K F

KKK FFFLet be the boundary of the triangle f0 < y < 1� x; 0 < x < 1g, with theusual counterclockwise orientation. Evaluate the following integrals(a)

R Re z dz (b)

R Im z dz (c)

R z dz

Solutiona)We begin do split the contur into three parts so that = 1 + 2 + 3.And integrate along each side in the triangle

= 1 + 2 + 3

We parametrizize 1 = [0; 1], thus 1 (t) = t, for 0 � t � 1, then Re (t) = tand dz = dt Z

1

Re z dz =

Z 1

Re (t) dz =

Z 1

0

t dt =1

2:

We parametrizize 2 = [1; i], thus 2 (t) = 1 � t + it, for 0 � t � 1, thenRe (1� t+ it) = 1� t and dz = (�1 + i) dt, thenZ

2

Re z dz =

Z 1

0

(1� t) (�1 + i) dt = �12+1

2i:

We parametrizize 3 = [i; 0], thus 3 (t) = i (1� t), for 0 � t � 1, then,Re (i (1� t)) = 0 and dz = �i dt, thenZ

3

Re z dz =

Z 1

0

(0) (�i) dt = 0:

2

And now we take the parts in the integral togetherZ

Re z dz =

Z 1

Re z dz +

Z 2

Re z dz +

Z 3

Re z dz =i

2

b) For details see a)

Z 1

Im z dz =

Z 1

0

0 dt = 0;Z 2

Im z dz =

Z 1

0

t (�1 + i) dt = �12+1

2i;Z

3

Im z dz =

Z 1

0

(1� t) (�i) dt = �12i:

thus Z

Im z dz = �12:

c) For details see a)

Z 1

z dz =

Z 1

0

t dt =1

2;Z

2

z dz =

Z 1

0

(1� t+ it) (�1 + i) dt = �1Z 3

z dz =

Z 1

0

(i (1� t)) (�i) dt = 1

2:

thus Z

z dz = 0:

3

IV.1.21 2 3 P L K111 LLL

Absolutbeloppet i c) n�agot �ar fel

Let be the unit circle fjzj = 1g, with the usual counterclockwiseorientation. Evaluate the following integrals, for m = 0;�1;�2; : : :.(a)

R zm dz (b)

R �zm dz (c)

R zm jdzj

SolutionSet z = ei� , where 0 � � � 2�, and thus dz = iei�d�.(a)

Z

zmdz =

Z 2�

0

eim�iei�d� = i

Z 2�

0

ei(m+1)�d� =

�ei(m+1)�

m+ 1

�2�0

=

=1

m+ 1

�ei2�(m+1) � 1

�=

�0; m 6= �12�i m = �1:

(b)

Z

�zmdz =

Z 2�

0

e�im�iei�d� = i

Z 2�

0

ei(1�m)�d� =

=

�ei(1�m)�

1�m

�2�0

=1

1�m�ei2�(m�1) � 1

�=

�0; m 6= 12�i m = 1:

(c)

Z

zm jdzj =Z 2�

0

eim�d� =

�eim�

im

�2�0

=1

im

�ei2�m � 1

�=

�0; m 6= 02� m = �1:

4

IV.1.31 2 3 P L K111Let be the circle fjzj = Rg, with the usual counterclockwise ori-entation. Evaluate the following integrals, for m = 0;�1;�2; : : :.(a)

R jzmj dz (b)

R jzmj jdzj (c)

R �zm dz

SolutionSet z = Rei� , where 0 � � � 2�, and thus dz = iRei�d� and jzj = R.(a)

Z

jzmj dz =Z 2�

0

��Rmeim��� iRei�d� = iRm+1 Z 2�

0

ei�d� = Rm+1�ei��2�0= Rm+1

�ei2� � 1

�= 0:

(b)

Z

jzmj jdzj =Z 2�

0

��Rmeim��� ��iRei��� d� = Rm+1 Z 2�

0

d� = Rm+1 [�]2�0 = Rm+1 (2� � 0) = 2�Rm+1:

(c)

Z

�zmdz =

Z 2�

0

Rme�im�iRei�d� = iRm+1

Z 2�

0

ei(1�m)�d� = Rm+1�ei(1�m)�

1�m

�2�0

=

=Rm+1

1�m�ei2�(1�m) � 1

�=

�0; m 6= 1;2�iRm+1; m = 1;

=

�0; m 6= 1;2�iR2; m = 1:

5

IV.1.41 2 3 P L K

LLLShow that if D is a bounded domain with smooth boundary, thenZ

@D

�z dz = 2i Area (D) :

SolutionSubstitute �z = x� iy, dz = dx+ idy, and apply Green�s theorem.

6

IV.1.51 2 3 P L K 138111 KKKShow that ����I

jz�1j=1

ez

z + 1dz

���� � 2�e2:SolutionWe begin with parametrizize the circle jz � 1j = 1, as z = 1+eit = 1+cos t+i sin t where 0 � t � 2�.For the nominator in the integrand ez= (z + 1) we have

jezj =��eRe z+i Im z�� = ��eRe zei Im z�� = ��eRe z�� � e2;

since Re z � 2 on the circle.And for the denominator in the integrad we have

j1 + zj = j2 + cos t+ i sin tj � Re (2 + cos t+ i sin t) = 2 + cos t � 1;

Thus

M =

���� ez1 + z

���� = jezjjz + 1j �

e2

1= e2:

Becauce the integrationcontour is a circle with radius 1 we have that L = 2�.ML� estimate gives ����I

jz�1j=1

ez

z + 1dz

���� � 2�e2:

7

IV.1.61 2 3 P L K111 KKKShow that there is a strict inequality����I

jzj=R

Log z

z2dz

���� � 2p2� logRR ; R > e�:

SolutionFor the integrand Log z=z2, with the nominator Log z = logR + i� with theprincipalargument �� � � � � we have

����LogRR2���� =

plog2R + �2

R26plog2R + �2

R26plog2R + log2R

R2=

p2 logR

R2:

there the last inequalitys follows by given fact R > e� that gives �2 < log2R.Becauce the integrationcontour is a circle with radius R we have that L =2�R.ML� estimate gives����Hjzj=RLog zz2 dz

���� 6 p2 logR

R22�R = 2

p2�logR

R; R > e�:

8

IV.1.71 2 3 P L K

Show that there is a strict inequality����Ijzj=R

zn

zm � 1dz���� � 2�Rn+1

Rm � 1 ; R > 1;m � 1; n � 1:

Solution

If jf (z)j 6 m0 < m on an arc 0 of the circle, = 0[ 1, then�����R f (z) dz

����� =�����R 0 f (z) dz +R 1

f (z) dz

����� 6m0length ( 0)+mlength ( 1) = (m0 �m) length ( 0)+

mlength ( ) < mlength ( )

Thus if���R f (z) dz��� = ML, then jf (z)j = M on . In this case, jzmj = Rn,�� 1

zm�1

�� 6 1Rm�1 , with strict inequality unless z = R � e

2�ik=m| {z }th root of w

. ) M = Rn

Rm�1 ,

L = 2�R, get���Rjzj=R zn

zm�1dz��� < Rn

Rm�1 � 2�R =2�Rn+1

Rm�1 , R > 1, m > 1, n > 0.

9

IV.1.81 2 3 P L K

LLLSuppose the continuous function f

�ei��on the unit circle satis�es��f �ei���� � M and

���Rjzj=1 f (z) dz��� = 2�M . Show that f (z) = c�z forsome constant c with modulus jcj =M .

SolutionMultiply f by a unimodular constant, assume jf (z)j 6M ,

Hjzj=1f (z) dz =R 2�

0f�ei��d� = 2�M . Take real parts, have

R 2�0Re�f�ei��iei��d� = 2�M ,

Re�f�ei��iei��d� 6

��f �ei���� 6 M . If have strict inequality somewhere,then

R 2�0< 2�M . We conclude thatRe

�f�ei��iei��d� �M . Since

���f �ei�� iei���� 6M , we have Im

�f�ei��iei��d� � 0. ) f

�ei��iei� � M , f

�ei��� �ie�i�M ,

f (z) � �iM �z. ) f (z) = c�z for a constant, jcj =M .

10

IV.1.91 2 3 P L K

Suppose h (z) is a continuous function on a curve . Show that

H (w) =

Z

h (z)

z � wdz; w 2 Cn ;

is analytic on the complement of , and �nd H 0 (w).

SolutionFor the analyticity, di¤erentiate by hand. (See Exercise III.1.6). The Deriv-

ative is H 0 (w) = limw!1

1�w

R

hh(z)

z�(w+�w) �h(z)z�w

idz =

limw!1

1�w

R

h(z)�w(z�(w+�w))(z�w)dz =

R h(z)dz

(z�w)2

11

IV.2.11 2 3 P L K K111 LLL KKK 246Evaluate the following integrals, for a path that travels from ��ito �i in the right half-plane, and also for a path from ��i to �iin the left half-plane.(a)

R z4dz (b)

R ezdz (c)

R cos z dz (d)

R sinh z dz

SolutionThe integrals are all independent of path and can be evaluated by �nding aprimitive.(a) Z

z4dz =

�z5

5

��i��i

=(�i)5

5� (��i)

5

5=2�5i

5:

(b) Z

ezdz = [ez]�i��i = e�i � e��i = �1� (�1) = 0:

(c)

Z

cos z dz = [sin z]�i��i =

= sin (�i)� sin (��i) = ei(�i) � e�i(�i)2i

� ei(��i) � e�i(��i)

2i=

=2 (e�� � e�)

2i= i

�e� � e��

�:

(d)

Z

sinh z dz = [cosh z]�i��i = cosh (�i)� cosh (��i) =

=e�i + e��i

2� e

��i + e�(��i)

2= 0:

12

IV.2.21 2 3 P L K K111 LLL 246Using an appropriate primitive, evaluate

R 1=z dz for a path that

travels from ��i to �i in the right half-plane, and also for a path from ��i to �i in the left half-plane. For each path give a precisede�nition of the primitive used to evaluate the integral.

SolutionIn right half�plane use f (z) = Log z = log jzj + i�, where �� < � < � as aprimitive, getZ

1

zdz = [Log z]�i��i =

�log j�ij+ i�

2

���log j�i�j � i�

2

�= �i:

In left half�plane use f (z) = log0 z = log jzj + i�, where 0 < � < 2� as aprimitive, getZ

1

zdz = [log0z]

�i��i =

�log j�ij+ i�

2

���log j�i�j+ i3�

2

�= ��i:

13

IV.2.31 2 3 P L K111 LLL KKKShow that if m 6= �1, then zm has a primitive on Cn f0g.

SolutionThe primitive is z

m+1

m+1, which is analytic for z 6= 0 and m 2 Z, m 6= �1.

14

IV.2.41 2 3 P L K

LetD = Cn (�1; 1], and consider the branch ofpz2 � 1 onD that is positive

on the interval (1;1).(a)Show that z +

pz2 � 1 omits the negative real axis, that is, the range of the

function on D does not include any values in the interval (�1; 0] on the realaxis.(b)Show that Log

�z +

pz2 � 1

�is a primitive for 1=

pz2 � 1 on D.

(c)Evaluate Z

dzpz2 � 1

;

where is the path from �2i to +2i in D counterclockwise around the circlejzj = 2.(d)Evaluate the integral above in the case is the entire circle jzj = 2, orientedcounterclockwise. (Note that the primitive is discontinuous at z = �2.)

SolutionD = Cn (�1; 1],

pz2 � 1

a)z +

pz2 � 1 = �t (t > 0) )

pz2 � 1 = �t � z, z2 � 1 = t2 � 2z + z2,

2z = t2 + 1, z = (t2 + 1) =2.But values of the branch are > 0 on (1;1) = D \ R. ) It assumes nonegative values.b)Log

�z +

pz2 � 1

�is the composition of z+

pz2 � 1 and Logw onCn (�1; 0],

so it�s analytic. By the chain role, ddzLog

�z +

pz2 � 1

�= 1

z+pz2�1

�1 + 1

2(z2 � 1)�1=2

�2z =

1z+pz2�1

�1 + zp

z2�1

�= 1p

z2�1 , where we use everywhere the constant branch

ofpz2 � 1, by ????? it on (1;1) or using d

dzza = aza

z

c)

15

R

dzpz2�1 =

�Log

�z +

pz2 � 1

��2i�2i = Log

�2i+ i

p5��Log

��2i� i

p5�= �i

Use fact thatpz2 � 1 is positive imaginary on positive imaginary axis, and

also Log�2i+ i

p5�� Log

��2i� i

p5�?????. d)R

dzpz2�1 =

�Log

�z +

pz2 � 1

���2+i0�2+i0

At �2 + 0t , haveLog�z +

pz2 � 1

�!

log���2 +p3�� + iArg �z +pz2 � 1� at z = �2. As z ! �2 + i0, have

argpz2 � 1! +�, jz2 � 1j ! 3. ) z+

pz2 � 1! �2+(�1)

p3 = �2�

p3,

arg�z +

pz2 � 1

�! �. As z ! �2+i0, have arg

pz2 � 1! ��, jz2 � 1j !

3. ) z +pz2 � 1 ! �2 + (�1)

p3 = �2 �

p3, arg

�z +

pz2 � 1

�! ��.

Values coincide )R

dzpz2�1 =

�Log

�z +

pz2 � 1

���2+i0�2+i0 = log jj � log jj+ �i� (��i) = 2�i

16

IV.2.51 2 3 P L K

Show that an analytic function f (z) has a primitive in D if andonly if

R f (z) dx = 0 for every closed path in D.

Solution (A. Kumjian)Let f be an analytic function de�ned on a domain D. Suppose �rst that fhas a primitive in D. Then by the Fundamental Theorem of Calculus forAnalytic Functions (Part I) we have

R f (z) dx = 0 for every closed path

2 D (since for any such path A = B).Conversely, suppose

R f (z) dx = 0 for every closed path 2 D. By formula

(1.1) on p. 102, we haveZ

f (z) dx+ if (z) dy =

Z

f (z) dx = 0

for every close path 2 D. It follows that the integral on the left is pathindependent for paths which are not necessarily closed and hence, by theLemma on page 77, there is a continuously di¤erentiable function F on Dsuch that dF = fdx+ ifdy, that is, @

@xF = f and @

@yF = if . It follows that

the real and imaginary parts of F satisfy the Cauchy-Riemann equations:

@

@xReF = Re

@F

@x= Re f = Im if = Im

@F

@y=@

@yImF;

@

@yReF = Re

@F

@y= Re if = � Im f = � Im @F

@x= � @

@xImF:

Moreover, F 0 (z) = f (z) for all z 2 D. Hence, f has a primitive inD, namelyF .

17

IV.3.11 2 3 P L K

LLLBy integrating e�z2=2 around a rectangle with vertices �R; it�R;andsending R to 1, show that

1p2�

Z 1

�1e�x

2=2e�itxdx = e�t2=2; �1 < t <1:

Use the known value of the integral for t = 0. Remark. Thisshows that e�x2=2 is an eigenfunction of the Fourier transform witheigenvalue 1. For more see the next exercise.

SolutionR@De�z

2=2dz =R R�R e

�x2=2dx�R t0e�(�R+iy)

2=2idy+R t0e�(R+iy)

2=2idy+R R�R e

�(x+it)2=2dx =0���e�(�R+iy)2=2���, 0, uniformly for 0 6 y 6 1 asR!1.

p2� =

R1�1 e

�x2=2dx =R1�1 e

�(x+it)2=2dx =R1�1 e

�(x2+2ixt�t2)=2dx =

et2=2R1�1 e

�x2=2e�ixtdx

) 1p2�

R1�1 e

�x2=2dx = e�t2=2

18

IV.3.21 2 3 P L K

We de�ne the Hermite polynomials Hn (x) and Hermite orthogonal functions�n (x) for n � 0 by

Hn (x) = (�1)n ex2 dn

dxn

�e�x

2�; �n (x) = e

�x2=2Hn (x) :

(a)Show that Hn (x) = 2nxn+ � � � is a polynomial of degree n such that is evenwhen n is even and odd when n is odd.(b)By integrating the function

e(z�it)2=2 d

n

dzn

�e�z

2�

around a rectangle with vertices �R; it�R and sending R to 1, show that

1p2�

Z 1

�1�n (x) e

�itxdx = (�i)n �n (t) ; �1 < t <1:

Hint. Use the identity from Exercise 1, and also justify and use the identity

dn

dxne�(x+it)

2

=1

indn

dtne�(x+it)

2

:

(c)Show that �

00

n � x�n + (2n+ 1)�n = 0.(d)Using

R�00

n�m dx =R�n�

00

n dx and (c), show thatZ 1

�1�n (x)�m (x) dx = 0; n 6= m:

Remark. This shows that the �n�s form an orthogonal system of eigenfuctionsfor the (normalized) Fourier transform operator F with eigenvalues �1 and�i. Thus F extends to a unitary operator on square-integrable functions.Further, F4is the identity operator, and the inverse Fourier transform is givenby (F�1f) (x) = (Ff) (�x).

Solution

19

Part A0 =

R@DR

e(z�it)2=2 dn

dzne�z

2dz

!R1�1 e

(x�it)2=2 dndxne�x

2dx =

R1�1 e

�t2=2e�itxex2=2 dn

dxne�x

2dx =

e�t2=2R1�1 e

�itx (�1)n �n (x) dx = e�t2=2 (�1)n �̂n (t)R

(x+it)\@DR !R1�1 e

x2=2 dn

dxne�(x+it)

2

dx =R1�1 e

x2=2 dn

d(it)ne�(x+it)

2

dx =

(�i) dndtn

R1�1 e

x2=2e�(x+it)2

dx = (�i) dndtn

R1�1 e

x2=2e�2itxet2dx =

(�i) dndtne�t

2

Z 1

�1e�x

2=2e�2itxdx| {z }p2�e�(2t)

2=2

=p2� (�i)n dn

dtne�t

2=p2�e�t

2=2in�n (t)

If we set these equal, we get(�i)n e�t2=2�̂n (t) =

p2�e�t

2=2in�n (t),1p2��̂n (t) = (�i)

n �n (t)Part B0 =

R@DR

ez2=2 dn

dzne�z

2dzR1

�1 ex2=2 d

dxe�z

2dx =

R1�1 e

x2=2 dn

dxne�(x+it)

2

dx = (�i)nR1�1 e

z2=2 dn

dtne�(x+it)

2

dxR1�1 e

(x�it)2=2 ddzne�x

2dx =

R1�1 e

x2=2 ddzne�(x+it)

2

dx =

e�t2=2R1�1 e

�itx ex2=2 d

dxne�x

2| {z }(�1)n�n(x)

dx+ (�i)nR1�1 e

x2=2 ddxne�(x+it)

2

dx

e�t2=2�̂n (t) = i

n dn

dtne�t

2+ (�i)n d

dxn

�et2 R1

�1 e�x2=2e�2ixtdx

��̂n (t) = (�i)

n � (t) + (�i)np2� d

dxn

�et2e�2t

2�= (�i)n

p2� d

dxnet2

�n (x) = e�x2=2Hn (x), Hn (x) = (�1)n ex

2 � dndxn

�e�x

2=

H0 (x) = 1, H1 (x) = e�x2 ddx

�e�x

2�= �e�x2 (�2x) e�x2 = 2x

H2 (x) = e�x2 d2

dx2

�e�x

2�= ex

2�4x2e�x

2 � 2e�x2�= 4x2 � 2

Hn (x) = 2nxn+ lower order polynomial.

R1�1 �n (x) e

�itxdx =R1�1 e

�x2=2 (�1)n ex2 dndxn

�e�x

2�e�ixtdx =

(�1)nR1�1D

nx

�ex

2�ex

2=2�ixtdx =R1�1 e

�x2=2Dnx

�ex

2=2�ixt�dx =

1R�1

e�x2Dnx

�e(x�it)

2=2�et2=2dx = (i)n

R1�1 e

�x2Dnt

�e(x�it)

2=2�et2=2dx =

et2=2inDn

t

R1�1 e

�x2e(x�it)2=2dx = et

2=2inDnt

R1�1 e

�x2=2e�ixte�t2=2dx =p

2�et2=2inDn

t e�t2=2e�t

2=2

20

IV.3.31 2 3 P L K

Let f (z) = c0 + c1z + � � �+ cnzn be a polynomial.(a)If the ck�s are real, show thatZ 1

�1f (x)2 dx � �

Z 2�

0

��f �ei���� d�2�= �

nXk=0

c2k:

Hint. For the �rst inequality, apply Cauchy�s theorem th the function f (z)2

separately on the top half an the bottom half of the unit disk.(b)If the ck�s are complex, show thatZ 1

�1jf (x)j2 dx � �

Z 2�

0

��f �ei���� d�2�= �

nXk=0

jckj2 :

(c)Establish the following variant of Hilbert�s inequality, that�����

nXj;k=0

cjckj + k + 1

����� � �nXk=0

jckj2 ;

with strict inequality unless the complex numbers c0; : : : ; cn are all zero. Hint.Start by evaluating

R 10f (x)2 dx.

Solution(a)1R�1f (x)2 dx = �

R� 1

f (z)2 dz 6�R0

��f �ei����2 d�R 1�1 f (x)

2 dx = �R� 1

f (z)2 dz 6R 2��

��f �ei����2 d�Add, get 2

R 1�1 f (x)

2 dx =6R 2��

��f �ei����2 d�R 2�0

��f �ei����2 d� = 2� nPk=0

c2k , by from��f �ei����2 = � nP

k=0

ckeik�

��nPk=0

cme�im�

�(b)

��f (x)2�� = ��P akxk + i

Pbkx

k��2 = �P akx

k�2+�P

bkxk�2R 1

�1

��f (x)2�� dx = R 1�1 �P akxk�2dx+

R 1�1�P

bkxk�2dx = �

�Pakx

k +Pbkx

k�=

�Pckx

k

21

�hR 1�1

��P akeik��� d�2�+R 1�1

��P bkeik��� d�2�

i(c)R 10(Pcjx

j)2dx =

R 10

nPj;k=0

cjckxj+kdx =

"nP

j;k=0

cjckxj+k+1

j+k+1

#10

=nP

j;k=0

cjckj+k+1

jj 6R 10jf (x)j2 dx 6

R 10jf (x)j2 dx 6 �

Pjckj2, for equality, must haveR 1

0jf (x)j2 =

R 10jf (x)j2, so f (x) = 0, for �1 6 x 6 0, and then f (x) � 0

Suppose P (z) = a0 + a1z + ::: + anzn is a polynomial. By integrating

P (z)2 log z, for an appropriate branch of log z, around a keyhole contour,

show that

����� nPj;k=0

ajakj+k+1

����� =���� 1R0

P (x)2 dx

���� 6 �nPk=0

jakj2.R@D

P (z)2 log z dz = 0,

givesR 10P (z)2 log x dx�

R 10P (z)2 (log x+ 2�i) dx+

R 2�0P�ei��2�iei� d� = 0.

Get 2�iR 10P (x)2 dx = i

R 2�0P�ei��ei��d�. Note that

RP�ei��d� , so latter

integral is iR 2�0P�ei��2ei� (� � �) d�. Thus

R 10P (x)2 dx =

R 2�0P�ei��ei� (� � �) d�

2�

)���R 10 P (x)2 dx��� 6 � R 2�0 ��P �ei����2 d�

2�= �

nPk=0

jakj2

22

IV.3.41 2 3 P L K

Prove that a polynomial in z without zeros is constant (the fundamentaltheorem of algebra) using Cauchy�s theorem, along the following lines. IfP (z) is a polynomial that is not a constant, write P (z) = P (0) + zQ (z),divide by zP (z), and integrate around a large circle. this will lead to acontradiction if P (z) has no zeros.

SolutionP (z) is a polynomial with no zeros. Assume P (0) = 1 . Write p (z) =1 + zQ (z).Hjzj=R

1zdz =

Hjzj=R

P (z)zP (z)

dz =H

jzj=R

�1

zP (z)+ Q(z)

P (z)

�dz =

Hjzj=R

1zP (z)

dz2 = �i

! 0 by ML-estimate if degP > 1.M = max

z=R

��� 1zP (z)

��� � 1Rn+1

, L � R, when n = detPML � 1

Rn! 0, if R!1, if n > 1.

23

IV.3.51 2 3 P L K

Suppose that D is a bounded domain with piecewise smooth boundary, andthat f (z) is analytic on D [ @D. Show that

supz2@D

j�z � f (z)j � 2 Area (D)Length (D)

:

Show that this estimate is sharp, and that in fact there exist D and f (z) forwhich equality holds. Hint. Consider

R@Dj�z � f (z)j dz, and use Exercise 4

in section 1.

SolutionR@D

[�z � f (z)] dz =R@D

�zdz = 2iArea (D)��R@D[�z � f (z)] dz

�� 6 supz2@D

j�z � f (z)jLength (@D)

) supz2@D

j�z � f (z)j > 2 Area(D)Length(@D)

To see its sharp, take D = D = unit disk, get supz2@D

j�z � f (z)j = 1

2 Area(D)Length(@D) = 2 �

�2�= 1

24

IV.3.61 2 3 P L K

Suppose f (z) is continuous in the closed disk fjzj � Rg and analyticon the open disk fjzj < Rg. Show that

Hjzj=R f (z) dz = 0. Hint.

Approximate f (z) uniformly by fr (z) = f (rz).

Solution (A. Kumjian)To prove

Hjzj=R f (z) dz = 0 it su¢ ces to show that

���Hjzj=R f (z) dz��� � ", forevery " > 0. So let " > 0 be given. Since f is continuous in the closed diskfjzj � Rg, it must be uniformly continuous there (since the close disk is bothclosed and bounded). Hence, there is a � > 0 such that for all z1; z2 in theclosed disk, we have

jf (z1)� f (z2)j <"

2�Rif jz1 � z2j < �:

For r 2 (0; 1), de�ne fr on the disk fz : jzj < R=rg by fr (z) = f (rz) for allz such that jzj < R=r. Then since fr is analytic on a domain containing theclosed disk fjzj � Rg, we have

Hjzj=R f (z) dz = 0 by Cauchy�s theorem. Now

let r be chosen such that 1 � �=R < r < 1. Then for all z with jzj = R wehave jz � zrj = R (1� r) < � and so by the choice of � we have

jf (z)� fr (z)j = jf (z)� f (rz)j <"

2�R:

Hence we have

����Ijzj=R

f (z) dz

���� = ����Ijzj=R

(f (z)� fr (z)) dz���� � I

jzj=Rj(f (z)� fr (z))j jdzj �

"

2�R2�R = ":

by the ML-estimate theorem on p. 105.

25

IV.4.11 2 3 P L K111 LLLEvaluate the following integrals, using the Cauchy integral formula:(a)

Hjzj=2

zn

z�1dz; n � 0 (e)Hjzj=1

ez

zmdz; �1 < m <1

(b)Hjzj=1

zn

z�2dz; n � 0 (f)Hjz�1�ij=5=4

Log z

(z�1)2dz

(c)Hjzj=1

sin zzdz (g)

Hjzj=1

dzz2(z2�4)ez

(d)Hjzj=1

cosh zz3dz (h)

Hjz�1j=2

dzz2(z2�4)ez

SolutionWe rearragnde the Cauchy integral formula. Let D be bounded domain withpiecewise smooth boundary @D and let f be an analytic function de�ned ina domain which contains �D. Then for any z0 2 D and positive integer m wehave, Z

@D

f (z)

(z � z0)mdz =

2�i

(m� 1)!

�dm

dzmf (z)

�z=z0

:

(a)The nominator has one zero at z = 1 inside jzj = 2, thus by by Cauchy´stheorem, H

jzj=2zn

z � 1dz =2�i

0![zn]z=1 = 2�i:

(b)The nominator has no zeros inside jzj = 1, thus by by Cauchy´s theorem,H

jzj=1zn

z � 2dz = 0:

(c)The nominator has one zero at z = 0 inside jzj = 1, thus by by Cauchy´stheorem, H

jzj=1sin z

zdz =

2�i

0![sin z]z=0 = 0:

(d)The nominator has one tripel zero at z = 0 inside jzj = 1, thus by byCauchy´s theorem,

26

Hjzj=1

cosh z

z3dz =

2�i

2!

�d2

dz2cosh z

�z=0

= �i:

(e)The nominator has one tripel zero at z = 0 inside jzj = 1, thus by byCauchy´s theorem.If m � 0, then the integrand z�mez de�nes an analytic function on C, hence,by Cauchy�s Theorem the integral must be zero.If m � 1, the nominator has a zero of order m at z = 0 inside jzj = 1, henceby by Cauchy´s theorem.I

jzj=1

ez

zmdz =

2�i

(m� 1)!

�dm�1

dzm�1ez�z=0

=2�i

(m� 1)! :

Thus Ijzj=1

ez

zmdz =

�0 if m � 0;2�i

(m�1)! if m � 1:

(f)Let f be the principal branch of the logarithm de�ned on the slit plane, thatis, f (z) = Log z for all z 6= 0 with Arg z 6= ��. Then f is analytic on adomain jz � 1� ij � 5=4. The nominator has one double zero at z = 1 insidejz � 1� ij < 5=4, thus by by Cauchy´s theorem.

Hjz�1�ij=5=4

Log z

(z � 1)2dz =

2�i

1!

�d

dz(Log z)

�z=1

= 2�i:

(g)The nominator has one double zero at z = 0 inside jzj = 1, thus by byCauchy´s theorem,

Hjzj=1

dz

z2 (z2 � 4) ez =2�i

1!

�d

dz

�e�z

z2 � 4

��z=0

=�i

2:

(h)The nominator has one zero at z = 0, and one zero at z = 2 inside jz � 1j = 2,thus by by Cauchy´s theorem,

Ijz�1j=2

dz

z (z2 � 4) ez =2�i

0!

�1

(z2 � 4) ez

�z=0

+2�i

0!

�1

z (z + 2) ez

�z=2

= ��i2+�i

4e2:

27

IV.4.21 2 3 P L K

LLLShow that a harmonic function is C1, that is, a harmonic functionhas partial derivatives of all orders.

SolutionWrite u = Re f , and note that @

@xu = Re f 0, @

@yu = @u

@y= � @v

@x= Re (f 0).

v = Im f = Re (�if), @v@x= Re (�if). Now take successive derivatives, get

@mu@xp@yq

a multiple of Re f 0 is Re (if 0).

28

IV.4.31 2 3 P L K

LLLUse the Cauchy integral formula to derive the mean value propertyof harmonic functions, that

u (z0) =

Z 2�

0

u�z0 + �e

i�� d�2�; z0 2 D;

whenever u (z) is harmonic in a domain D and the closed diskjz � z0j � � is contained in D.

SolutionWrite f (z0) =

Hjz�z0j=�

f(z)z�z0dz =

H 2�0f�z0 + �e

i��d�2�, which is the MVP.

29

IV.4.41 2 3 P L K

Let D be a bounded domain with smooth boundary @D, and let z0 2 D.Using the Cauchy integral formula, show that there is a constant C suchthat

jf (z0)j � D sup fjf (z)j : z 2 @Dgfor any function f (z) analytic on D [ @D. By applying this estimate tof (z)n, taking nth roots, and letting n ! 1, show that the estimate holdswith C = 1. Remark. This provides an alternative proof of the maximumprinciple for analytic function.

SolutionWrite f (z0) = 1

2�i

R@D

f(z)z�z0dz . Letkfk = maxz2@D

jf (z)j, 1jz�z0j 6

1dist(z0;@D)

= 1d.

By ML�estimate, L = length @D, jf (z0)j 6 12�dkfkL. Apply to fn, get

jf (z0)jn 6 12�dkfnkL. Use kfnk = kfkn, take nth roots, get jf (z0)j 6

kfk�L2�d

�1=n. Let n!1, get jf (z0)j 6 kfk.

30

IV.5.11 2 3 P L K111 LLL

Denna beh�over Nog fixas till Lite

Show that if u is a harmonic function on C that is bounded above,then u is constant.Hint. Express u as the real part of an analytic function, and expo-nentiate.

SolutionLet u be the real part of the analytic function f = u + iv, where we knowthat u � C, for some constant C. Then set g = eu+iv where g is an entirefunction. Now we have that

jgj =��eu+iv�� = ��eueiv�� = jeuj ��eiv�� = eu:

We can se that becauce u is bounded above, so is g bounded above. Nowwe now that g is both analytic and bounded, by Liouville�s theorem g isconstant. Thus f is constant.

31

IV.5.21 2 3 P L K111 LLLShow that if f (z) is an entire function, and there is a nonemptydisk such that f (z) does not attain any values in the disk, thenf (z) is constant.

SolutionIf f (z) does not attain values in the disk jw � cj < ", then 1= (f (z)� c) isa bounded entire function, hence constant by Liouville�s theorem, and f (z)is constant.Let w0 be the center of the disk that f omits. Then 1= (f (z)� w0)

32

IV.5.21 2 3 P L K

LLLShow that if f (z) is an entire function, and there is a nonemptydisk such that f (z) does not attain any values in the disk, thenf (z) is constant.

Solution (A. Kumjian)Let f be such an entire function. Then there is a disc D = Dr (z0) for somez0 2 C and r > 0 such that f (z) 62 D for all z 2 C. That is, jf (z)� z0j � rfor all z 2 C. We de�ne an entire function g by the formula

g (z) =1

f (z)� z0for all z 2 C:

It follows that jg (z)j � 1=r for all z 2 C. Hence, g is a bounded entirefunction an so by Liouville�s Theorem g must be a constant function. Thedesired result now follows immediately.

33

If f does not attain values in the disk jw � cj < ", then 1= (f � c) is bounded,hence constant by Liouville�s theorem, and f is constant.

34

IV.5.31 2 3 P L K

LLLA function f (z) on the complex plane is double periodic if thereare two periods w0 and w1 of f (z) that do not lie on the same linethought the origin (that is, w0 and w1 are linearly independent overthe reals, and f (z + w0) = f (z + w1) = f (z) for all complex numbersz). Prove that the only entire functions that are doubly periodicare the constants.

SolutionThe "lattice" of pointsmw0+nw1, m;n 2 Z are the corners of parallelogramsthat cover C, each parallelogram being a translate F +mw0 + nw1 of "thefundamental region" F as shown in the �gure:

Figure IV.5.3

Thus each � 2 C can be written � = z +mw0 + nw1 with z 2 F . Since weassume f (�) = f (z), it follows that f i bounded, thus a constant.

35

If jf (z)j 2 M for jzj 6 R. Suppose jw0j, jw0j 6 R. Let P = parallelogramwith vertices 0; w0; w1; w0 + w1, and suppose for z 2 P . Any z 2 C haveform � + mw0 + nw1 where � 2 P . Thus jf (z)j = jf (�)j 6 M , so f (z) isbounded and entire.

36

IV.5.41 2 3 P L K

Suppose that f (z) is an entire function such that f (z) =zn is boundedfor jzj � R. Show that f (z) is a polynomial of degree at most n.What can be said if f (z) =zn is bounded on the entire complexplane?

SolutionApply the Cauchy estimates for f (m+1) (z) to disk jz � z0j < R, and letR ! 1, to obtain f (m+1) (z0) = 0. If f (z) is a polynomial of degree 6 msuch that f (z) =zm is bounded near 0, then f (z) = czm.

37

IV.5.51 2 3 P L K

Show that if V (z) is the velocity vector �eld for a �uid �ow in the entirecomplex plane, and if the speed jV (z)j is bounded, then V (z) is a constant�ow.

SolutionUse V (z) = f 0 (z) (section III.6 p92), when f is entire. Get f 0 constant, byLiouville�s theorem, i.e. V (z) is constant.

38

IV.6.11 2 3 P L K

LLLLet L be a line in the complex plane. Suppose f (z) is a continuouscomplex-valued function on a domain D that is analytic on DnL.Show that f (z) is analytic on D.

SolutionRotate and apply result in text.

39

IV.6.21 2 3 P L K

Let h (t) be a continuous function on the interval [a; b]. Show that the Fouriertransform

H (z) =

Z b

a

h (t) e�itzdt

is an entire function that satis�es

jH (z)j � CeAjyj; z = x+ iy 2 C;for some constants A;C > 0. Remark. An entire function satisfying such agrowth restriction is called an entire function of �nite type.

SolutionThe analyticity of H (z) follows from the theorem in the text. If jhj 6 M ,take C = M (b� a) and A = max (jaj ; jbj). We have je�itzj = ety. jH (z)j 6M (b� a) max

a�t�bety 6M (b� a) eAjyj

40

IV.6.31 2 3 P L K

Let h (t) be a continuous function on a subinterval [a; b] of [0;1). Showthat the Fourier transform H (z), de�ned as above, is bounded in the lowerhalf-plane.

Solution

41

IV.6.41 2 3 P L K

Let be a smooth curve in the plane R2, let D be a domain in the complexplane, and let P (x; y; �) and Q (x; y; �) be continuous complex-valued func-tions de�ned for (x; y) on and � 2 D. Suppose that the functions dependanalytically on � for each �xed (x; y) on .Show that

G (�) =

Z

P (x; y; �) dz +Q (x; y; �) dy

is analytic on D.

SolutionReduce to Riemann integral and apply theorem in the text.

42

IV.7.11 2 3 P L K

Find an application for Goursat�s theorem in which it is not patently clearby other means that the function in question is analytic.

SolutionAnswer to this is not known (by me).

43

IV.8.11 2 3 P L K

LLLShow from the de�nition that@@zz = 1; @

@�zz = 0; @

@z�z = 0; @

@�z�z = 1:

SolutionWe use the �rst-order di¤erential operators (page 124)

@

@z=1

2

�@

@x� i @@y

�and

@

@�z=1

2

�@

@x+ i

@

@y

�:

We have that

@

@zz =

1

2

�@

@x� i @@y

�(x+ iy) =

1

2(1 + 1) = 1

@

@�zz =

1

2

�@

@x+ i

@

@y

�(x+ iy) =

1

2(1� 1) = 0

@

@z�z =

1

2

�@

@x� i @@y

�(x� iy) = 1

2(1� 1) = 0

@

@�z�z =

1

2

�@

@x+ i

@

@y

�(x� iy) = 1

2(1 + 1) = 1

44

IV.8.21 2 3 P L K

LLLCompute @

@�z(az2 + bz�z + c�z2). Use the result to determine where

az2 + bz�z + c�z2 is complex-di¤erentiable and where it is analytic.(See Problem II.2.3)

SolutionBy the Leibnitz rule and the results of Exercise 1, the �z �derivative is bz+2c�z.The function is complex di¤erentiable at z if and only if bz + 2c�z = 0. Ifb = c = 0, the function is entire. Otherwise this locus is either a point f0gor a straight line through the origin, and there is no open set on which thefunction is analytic.

45

IV.8.31 2 3 P L K

LLLShow that the Jacobian of a smooth function f is given by

det Jf =

����@f@z����2 � ����@f@�z

����2 :Solution

Jf = det

�@u@x

@v@x

@u@y

@v@y

�= @u

@x@v@y� @u

@y@v@x��@f

@z

��2 � ��@f@�z

��2 = 14

���@(u+iv)@x� i@(u+iv)

@y

���2 � 14

���@(u+iv)@x+ i@(u+iv)

@y

���2= 1

4

��@u@x+ @v

@y

�2+�@v@x+ @u

@y

�2�� 1

4

��@u@x� @v

@y

�2+�@v@x+ @u

@y

�2�= 1

4

h2@u@x

@v@y� 2 @v

@x@u@y+ 2@u

@x@v@y� 2 @v

@x@u@y

i= @u

@x@v@y� @v

@x@u@y= det Jf

46

IV.8.41 2 3 P L K

LLLShow that

@2

@x2+@2

@y2= 4

@2

@z@�z:

Deduce the following, for a smooth complex-valued function h.(a) h is harmonic if and only if @2h=@z@�z = 0.(b) h is harmonic if and only if @h=@z is conjugate-analytic.(c) h is harmonic if and only if @h=@�z = 0.(d) If h is harmonic, then any mth order partial derivative of h isa linear combination of @mh=@zm and @mh=@�zm.

Solution4 @2

@z@�z=�@@x� i @

@y

��@@x+ i @

@y

�= @2

@x2� i @2

@x@y� i @2

@y@x+ @2

@y2= @2

@x2+ @2

@y2

(a) clear(b) h harmonic , @2h

@�z@z= 0, @h

@zis analytic

(c) h harmonic , @2h@z@�z

= 0 , @@z

�@h@�z

�= 0. Since @h

@z= @�h

@�z, this ????? @h

@�zis

analytic , @h@�zis conjugate analytic.

(d) If h is analytic, then @k+lh@zk@�zl

= 0 unless l = 0. Only non-vanishing deriva-tive is @

mh@zm

= hm (z). If h is conjugate analytic, ????? @k+lh@�zk@zl

= 0 unless l = 0.If h is harmonic, then h = analytic + conjugate analytic, so all derivativesof h vanish except for @

kh@zk

= @lh@�zk.

47

(f)Rjz�1�ij=5=4

Log z

(z�1)2dz =2�i1!

ddzLog zjz=1 = 2�i 1z

��z=1

= 2�i

(g)Rjzj=1

dzz2(z2�4)ez = 2�i d

dze�z

z2�4

���z=0

= 2�i(�e�z)(z2�4)�e�z(2z)

(z2�4)2

����z=0

= 2�i�416

= �i2

(h)Rjz�1j=2

dzz(z2�4)ez = 2�i 1

(z2�4)ez

���z=0

+ 2�i 1(z+2)ez

���z=2

=

2�i��14+ 1

8e2

�= �i

��12+ 1

4e2

�

48

IV.8.51 2 3 P L K

LLLWith d�z = dx� idy, show that for a smooth function f (z) that

df =@f

@zdz +

@f

@�z:

Solutiondf = @f

@xdx+ @f

@ydy

@f@zdz + @f

@�zd�z = 1

2

�@f@x� i@f

@y

�(dx+ idy) + 1

2

�@f@x+ i@f

@y

�(dx� idy) =

12@f@xdx+ 1

2@f@ydy + 1

2@f@xdx+ 1

2@f@ydy + i

2

h@f@xdy � @f

@ydx� @f

@xdy + @f

@ydxi= df

49

IV.8.61 2 3 P L K

Show that if D is a domain with smooth boundary, and if f (z) and g (z) areanalytic on D [ @D, thenZ

@D

f (z) g (z) dz = 2i

ZZD

f (z) g0 (z) dx dy:

Compare this formula with Exercise 1.4.

SolutionBy (8.4),

R@Df (z) g (z) dz = 2i

RRD

@@E(f �g) dz = 2i

RRDfgdz

Using the Leibnitz rule, obtain @@�z(f �g) = f @�g

@�z+ �g @f

@�z= f @g

@z= fg0

To get Ex1.4, take f = 1, g (z) = z, getR@D�z dz = 2i

RRDdxdt = 2iArea

50

IV.8.71 2 3 P L K

LLLShow that the Taylor series expansion at z0 = 0 of a smooth functionf (z), through the quadratic terms, is given by

f (z) = f (0)+@f

@z(0) z+

@f

@�z(0) �z+

1

2

�@2f

@z2(0) z2 + 2

@2f

@z@�z(0) jzj2

�+o�jzj3�:

SolutionSince a smooth function f is a linear combination of the functions 1; z; �z; z2; �z2; jzj2and a reminder term O

�jzj3�, it su¢ ces to check the formula for these six

functions. The formula holds fro each of these. For instance, for jzj2 = z�zare harmonic. @f

@z(0) = 0, @f

@�z(0) = 0, @

2f@�z2(0) = 0, f (0) = 0, @2f

@z@�z(0) = 1.

The term O�jzj3�has Taylor series 0 +O

�jzj3�. That ????? it.

51

IV.8.81 2 3 P L K

LLLEstablish the following version of the chain rule for smooth complex-valued functions w = w (z) and h = h (w).

@

@�z(h � w) =

@h

@w

@w

@�z+@h

@ �w

@ �w

@�z;

@

@z(h � w) =

@h

@w

@w

@z+@h

@ �w

@ �w

@z:

SolutionSuppose w (0) = 0, h (0) = 0, write h (w) = aw + b �w + O

�jwj2

�, a = @h

@w(0),

b = @h@ �w(0) and w (z) = �w + � �w + O

�jzj2�, � = @w

@z(0), b = @w

@�z(0) then

h (w (z)) = a��z + ��z +O

�jzj2��+ b

����z + ��z +O

�jzj2��+O

�jwj2

�=

a�z + a��z + b���z + b��z +O�jzj2�=�a�+ b��

�z + (a� + b��) �z +O

�jzj2�

This gives,@h�w@�z

(0) = a� + b�� = @h@w

@w@�z+ @h

@ �w@w@z @ �w=@�z

= @h@w

@w@�z+ @h

@ �w@ �w@�z

@h�w@z

(0) = a�+ b�� = @h@w

@w@z+ @h

@ �w

@w

@�z|{z}@ �w=@z

= @h@w

@w@z+ @h

@ �w@ �w@z

52

IV.8.91 2 3 P L K

LLLShow with the aid of the preceding exercise that if both h (w) andw (z) are analytic, then (h � w) (z) and (h � w)0 (z) = h0 (w (z))w0 (z).

SolutionIf the functions h and w are analytic, then @h�w

@�z= 0 + 0 = 0 so h � w is

analytic. (h � w)0 (z) = @@zh � w = @h

@w@w@z+ 0 = h0 (w (z))w0 (z)

53

IV.8.101 2 3 P L K

Let g (z) be a continuously di¤erentiable function on the complex plane thatis zero outside of som compact set. Show that

g (w) = � 1�

ZZC

@g

@�z

1

z � wdx dy; w 2 C:

Remark. If we integrate this formally by parts, we obtain

g (w) =1

�

ZZCg (z)

@

@�z

�1

z � w

�dx dy:

Thus the "distribution derivative" of 1= (� (z � w)) with respect to z is thepoint mass w ("Dirac delta-function"), in the sense that it is equal to 0 awayfrom w, and it is in�nite at w in such a way that its integral (total mass) isequal to 1.

SolutionApply Pompeeiu�s formula to a large disk fjzj < Rg.

54

V 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 1912345678

1

V.1.11 2 3 P L K

(Harmonic Series) Show that

nXk=1

1

k� log n:

Deduce that the seriesP1

k=11kdoes not converge. Hint. Use the

estimate

1

k�Z k+1

k

1

xdx:

SolutionLet k be a nonnegative integer. Then since the function x 7! 1=x is decreasingon the positive reals, we have for all x 2 [k; k + 1] that 1=x � 1=k . Hencelocking at the oversum in the interval [k; k + 1]

1

k((k + 1)� k) = 1

k�Z k+1

k

1

xdx:

Therefore, for n � 1nXk=1

1

k�

nXk=1

Z k+1

k

1

xdx =

Z n+1

1

1

xdx = log (n+ 1) � log n:

Where the last inequality follows from the fact that x 7�! log x is an increas-ing function on the positive reals. And thus is the sum

Pnk=1

1khave not limit

as n!1 and does not converge.

2

V.1.21 2 3 P L K

Show that if p < 1, then the seriesP1

k=1 1=kp diverges. Hint. Use

Exercise 1 and the comparison test.

SolutionIf k > 1 then, kp < k for 0 < p < 1, then

1

kp>1

k:

Thus we have

1Xk=1

1

kp>

1Xk=1

1

k:

By the comparison test, we have that the since the sumPn

k=1 1=k diverge asn!1 by Exercise 1, also the sum

Pnk=1 1=k

p diverge as n!1, which wasto be shown.

3

V.1.31 2 3 P L K

Show that if p > 1, then the seriesP1

k=1 1=kp converges to S, where�����S �

nXk=1

1

kp

����� < 1

(p� 1)np�1 :

Hint. Use the estimate 1kp<R kk�1

dxxp.

SolutionLet k be a nonnegative integer. Then since the function x 7! 1=x is decreasingon the positive reals, we have for all x 2 [k � 1; k] that 1=k � 1=x. Becaucep > 1, it follows that 1=kp � 1=xp. Hence locking att the undersum in theinterval [k � 1; k]

1

kp(k � (k � 1)) = 1

kp�Z k

k�1

1

xpdx

Therefore

nXk=1

1

kp�

nXk=1

Z k

k�1

dx

xp= 1+

Z n

1

dx

xp= 1+

�x�p+1

�p+ 1

�n1

= 1+1

p� 1�n�p+1

p� 1 � 1+1

p� 1 :

Hence the serie converge to S, where

S � 1 + 1

p� 1 =p

p� 1 ;

since its terms are > 0.Now we have that

NXk=n+1

1

kp6Z N

n

dx

xp= 1 +

�x�p+1

�p+ 1

�Nn

=n�p+1 �N�p+1

p� 1 ! n�p+1

p� 1

as N !1.Thus we have

4

NXk=n+1

1

kp= S �

nXk=1

1

kp6 n�p+1

p� 1 =1

(p� 1)np�1

thus �����S �NX

k=n+1

1

kp

����� � 1

(p� 1)np�1 :

5

V.1.41 2 3 P L K

Show that the series

1Xk=1

(�1)k+1

k= 1� 1

2+1

3� 14+ � � �

converges. Hint. Show that the partial sums of the series satisfyS2 < S4 < S6 < � � � < S5 < S3 < S1.

SolutionWe de�ne the sum Sn by

Sn =nXk=1

(�1)k+1

k= 1� 1

2+1

3� 14+ : : : :

This is the �alternating series test�, which is standard signs alternate, andterms ! 0, jtermsj # , so series converge.

6

V.1.51 2 3 P L K

Show that the series

1 +1

3� 12+1

5+1

7� 14+1

9+1

11� 16+ � � �

converges to 3S=2, where S is the sum of the series in Exercise 4.(It turns out that S = log 2.) Hint. Organize the terms in the seriesin Exercise 4 in groups of four, and relate it to the groups of threein the above series.

SolutionWe have the series S from exercise V.1.5 and arrange the terms in groups offour

S = 1� 12+1

3� 14+1

5� 16+1

7� 18+1

9� 1

10+1

11� 1

12+ : : :

And now we arrange the terms in groups of four

S =

�1� 1

2+1

3� 14

�+

�1

5� 16+1

7� 18

�+

�1

9� 1

10+1

11� 1

12

�+ : : :

Now we rearrange the terms in groups of two, and divide the terms by 2

S

2=

�1

2� 14

�+

�1

6� 18

�+

�1

10� 1

12

�+ : : :

Now add one group of four from the sum S with a group of two from thesum S=2, and so on

3

2S =

�1 +

1

3� 2 � 1

4

�+

�1

5+1

7� 2 � 1

8

�+

�1

9+1

11� 2 � 1

12

�+ : : :+ =

=

�1 +

1

3� 12

�+

�1

5+1

7� 14

�+

�1

9+1

11� 16

�+ : : :

Since series S in Exercise 4 converge, thus also 3S=2 converge by ......

7

V.1.61 2 3 P L K

Show thatZ R

e

1

k log k= [log (log k)]Re = log (logR)� log e (log e)!1

diverges whileP1

k=11

k(log k)2converges.

Solution 1 (Comparison Test)For the �rst sum

2n+1Xk=2n+1

1

k log k

there are 2n terms between k = 2n+1 and k = 2n+1, each of the terms greaterthan 1

2n+1 log 2n+1, so these 2n terms have sum greater than 1

2(n+1) log 2. Thus

series diverges, by comparison with the harmonic series. The other series istreated similarly, using an upper estimate.

For the other sum

2n+1Xk=2n+1

1

k (log k)2

there are 2n terms between k = 2n + 1 and k = 2n+1, each of the terms lessthan 1

2n+1 log 2n+1, so these 2n terms have sum less than 1

2(n+1) log 2. Thus series

converges, by comparison with the harmonic series.

Solution 2 (Integral Test)Note that x log x and x (log x)2 are increasing for x > 1, so the terms of bothseries are positive and decreasing. Vi can use Cauchy�s integralkriterium andlook att them corresponding integrals. Becauce of di¢ cult with the integrandwe split the integral into two partsZ R

e

1

x log x= [log (log x)]Re = log (logR)� log (log e)!1;

8

as R!1. Z R

e

1

x (log x)2=

�� 1

log x

�Re

=1

log e� 1

logR! 1;

as R!1.We have that the series

1Xk=1

1

k log k

diverges, while the series

1Xk=1

1

k (log k)2

konverges.

9

V.1.71 2 3 P L K

Show that the seriesPak converges if and only if

Pk=nk=m

1k(log k)2

converges.

SolutionIf Sn is the partial-sum

nPk=1

ak, then Sn�Sm�1 =nP

k=m

ak. This follows imme-

diately fromnP

k=m

ak = Sn � Sm�1, when Sn =nPk=1

ak is the nth partial-sum of

the series.V.2.1 d

dxxk

k+x2k=

kxk�1(k+x2k)�xk(2k)x2k�1

(1+x2k)2 = 0 at k2xk�1 + kx3k�1 = 2kx3k�1,

k2xk�1 = k3k�1, k = x2k,pk = xk. ) Function " for 0 6 x 6 k1=2k, # for

x > k1=2k, ! 0 as x ! 1. xk

k+x2k6

pk

k+k= 1

2pk= worst-case estimation. )

Converge to 0 uniformly.

10

V.2.11 2 3 P L K

Show that fk (x) = xk=�k + xk

�converges uniformly to 0 on [0;1).

Hint. Determine the worst-case estimator "k by calculus.SolutionSuppose fk (x) converge pointwise to a function f (x) as k !1.We have

fk (x) =xk

k + x2k;

thus

f0

k (x) =kxk�1

�k + x2k

�� xk (2k)x2k�1

(k + x2k)2;

there f0k (x) = 0 for x1 = 0 and x2 = k

1=(2k).We have

x 0 k1=(2k)

f0k (x) 0 + 0 �fk (x) % &

and the function is increasing for 0 � x � k1=(2k), and decreasing for x �for x � k1=2k, and fk (x) ! 0 as x ! 1. We have that fk (x) attains itsmaximum for x = k1(2k).We determine f (x) applying the works case estimator "k to the functionfk (x),

f (x) = limk!1

xk

k + x2k� lim

k!1

pk

k + k= lim

k!1

1

2pk= 0:

The series fk (x) = xk=�k + xk

�converge uniformly to 0 on [0;1).

11

V.2.21 2 3 P L K

Show that gk = xk=�1 + xk

�converges pointwise on [0;1) but not

uniformly. What is the limit function? On which subsets of [0;1)does the sequence converge uniformly?

SolutionSuppose gk (x) converge pointwise to a function g (x) as k !1, we start toshow that is is an increasing function

g0

k (x) =kxk�1 � xkkxk�1

(1 + xk)2=

kxk�1

(1 + xk)2> 0

also is the function gk (x) increasing, now we take the limit in three intervals.If 0 � x < 1, in this interval we have that xk ! 0 as k !1, and we have

g (x) = limk!1

xk

1 + xk=

0

1 + 0= 0:

If x = 1, then

g (1) = limk!1

1k

1 + 1k=

1

1 + 1=1

2:

If x > 1, in this interval we have that xk ! 0 as k !1, and we have

g (x) = limk!1

xk

1 + xk= lim

k!1

1

1=xk + 1=

1

0 + 1= 1:

We also have that gk (x)! g (x), where

g (x) =

8<:0 for 0 � x < 1;12for x = 1;

1 for x > 1

Convergence is uniform on [0; 1� "] for any " > 0. Not uniform on [0; 1],because limit function is not continuous at 1. Convergence is uniform on[1 + ";+1] for any " > 0.

12

V.2.31 2 3 P L K

Show that fk (z) = zk=k converges uniformly for jzj < 1. Show thatf0k (z) does not converge uniformly for jzj < 1. What can be saidabout uniform convergence of f 0k (z)?

SolutionSuppose fk (z) converge pointwise to a function f (z) as k !1 in the intervaljzj � 1, in this domain we have that

��zk=k�� � 1=k thusf (z) = lim

k!1

����zkk���� � lim

k!1

1

k= 0:

Thus fk (z) converges uniformly for jzj < 1.Now suppose f

0k (z) converge pointwise to a function f

0(z) as k !1 in the

interval jzj � 1, now we must split the interval when we take the limit

f0(z) = lim

k!1

��zk�1�� = � 0 for jzj < 1;1 for jzj = 1:

If jzj = 1, then��f 0k (z)�� = 1 so the series can not converge uniformly for

jzj < 1. But for any " > 0 the series f 0k (z) = zk�1 converges uniformly forjzj � 1� ".

13

V.2.41 2 3 P L K

Show that

1Xk=1

1

k2xk

1 + x2k

converges uniformly for �1 < x < +1.

SolutionWe apply the Weierstrass M�test (p. 135.) with Mk =

1k2, where the sumP1

k=0Mk converges. First we show that��xk�� � ��1 + x2k�� for all �1 < x <

+1.We have ��xk�� � 1 � 1 + x2k = ��1 + x2k�� for jxj � 1��xk�� � x2k < 1 + x2k = ��1 + x2k�� for jxj > 1This establishes the fact that ���� xk

1 + x2k

���� � 1for all �1 < x < +1.Hence, for each k � 1 and �1 < x < +1,���� 1k2 xk

1 + x2k

���� � 1

k2=Mk:

Thus convergence is uniform on �1 < x < +1 by the WeierstrassM�test.

14

V.2.51 2 3 P L K

For which real numbers x doesP

1k

xk

1+x2kconverge?

SolutionSuppose gk (x) converge pointwise to a function g (x) as k ! 1, and takelimit in three intervalsIf x < 1, in this interval we have that xk ! 0 as k !1, and we have

1Xk=1

1

k

xk

1 + x2k<

1Xk=1

1

1=xk + xk<

1Xk=1

1

1=xk=

1Xk=1

xk

which is a convergent series.If x = 1, then xk = x2k = 1 then

1Xk=1

1

k

xk

1 + x2k=

1Xk=1

1

2k

which is a divergent series.If x > 1, in this interval we have that xk !1 as k !1, and we have

1Xk=1

1

k

xk

1 + x2k=

1Xk=1

1

k

1

1=xk + xk<

1Xk=1

1

xk

which is a convergent series.

15

V.2.61 2 3 P L K

Show that for each " > 0, the seriesP

1k

xk

1+x2kconverges uniformly

for x � 1 + ".

SolutionWe have

fk (x) =xk

1 + x2k;

thus

f0

k (x) =kxk�1

�1 + x2k

�� xk (2k)x2k�1

(1 + x2k)2=xk�1 � x3k�1

(1 + x2k)2

there f0k (x) = 0 for x1 = 0 and x2 = 1.

We have

x 0 1f0k (x) 0 + 0 �fk (x) % &

and the function is increasing for 0 � x � 1, and decreasing for x � 1, andfk (x)! 0 as x!1. We have that fk (x) attains its maximum for x = 1.If x = 1, then xk = x2k = 1 then

1Xk=1

1

k

xk

1 + x2k=

1Xk=1

1

2k

which is a divergent series.If x > 1, in this interval we have that xk !1 as k !1, and we have

1Xk=1

1

k

xk

1 + x2k=

1Xk=1

1

k

(1 + ")k

1 + (1 + ")2k=

1Xk=1

1

k

1

1= (1 + ")k + (1 + ")k�

�1Xk=1

1

k

1

(1 + ")k

16

V.2.71 2 3 P L K

LLLLet an be a bounded sequence of complex numbers. Show that for each " > 0,the series

P1n=1 ann

�z converges uniformly for Re z � 1 + ". Here we choosethe principal branch of n�z.

Solution1Pn=1

ann�z, janj 6 C, Re z > 1 + ". Apply Weierstrass M-test, jann�zj 6

Cn�Re z 6 Cn1+"

= Mn,PMn < 1 )

Pann

�z converges uniformly forRe z > 1 + ".

17

V.2.81 2 3 P L K

LLLShow that

Pzk

k2converges uniformly for jzj < 1.

SolutionApply the Weierstrass M � test with M = 1=k2 in jzj � 1, we �nd that

1Xk=1

����zkk2���� = 1X

k=1

jzjk

k2<

1Xk=1

1

k2;

which is a convergent series by Weierstass M � test. Thus the series con-verges uniformly for jzj < 1.

18

V.2.91 2 3 P L K

LLLShow that

Pzk

kdoes not converges uniformly for jzj < 1.

SolutionSuppose that we have uniform convergence for jzj < 1, then the series mustconverge for jzj � 1Apply the Weierstrass M � test with M = 1=k in jzj � 1, we �nd that

1Xk=1

����zkk���� = 1X

k=1

jzjk

k2<

1Xk=1

1

k;

which is a divergent series by Weierstass M � test. Thus the series do notconverge uniformly for jzj < 1.

19

V.2.101 2 3 P L K

Shoe that if a sequence of functions ffk (x)g converges uniformly on Ej for1 � j � n, then the sequence converges uniformly on the union E = E1 [E2 [ � � � [ En.

SolutionUse Cauchy criterion. Let " > 0. Choose Nj such that jfn (x)� fk (x)j < kform; k > Nj and x 2 Ej. LetN = max fN1; : : : ; Nng. Then jfm (x)� fk (x)j <" for m; k > N and x 2 E1 [ : : : [ En. ) ffmg converges uniformly.

20

V.2.111 2 3 P L K

Suppose that E is a bounded subset of a domainD � C at a positive distancefrom the boundary of D, that is � > 0 such that jz � wj � � for all z 2 Eand w 2 CnD. Show that E can be covered by a �nite number of closeddisks contained in D. Hint. Consider all closed disks with centers at points(m+ ni) �=10 and radius �=10 that meet E.

SolutionFollow the hint. There are only �nitely many of the disks in ????? (becauseE is bounded); and they cover E, because the collection of all disks centred at(m+ ni) �=10 of radius �=10 covers the complex plan, and they are containedin D, because dist ("; @D) > �.

21

V.2.121 2 3 P L K

Let f (z) be analytic on a domain D, and suppose jf (z)j �M for all z 2 D.Show that for each � > 0 and m � 1,

��f (m) (z)�� � m!M=�m for all z 2 Dwhose distance from @D is at least �. Use this to show that if ffk (z)g is asequence of analytic functions on D that converges to f (z) on D, then foreach m the derivatives f (m)k (z) converge uniformly to f (m) (z) on each subsetof D at a positive distance from �D.

SolutionUse the Cauchy estimates for f (m) (w) on a disk centred at w of radius � <d (w; @D), get

��f (m) (w)�� 6 m!m�m.

22

V.3.11 2 3 P L K

LLLFind the radius of convergence of the following power series:

(a)1Pk=0

2kzk (d)1Pk=0

3kzk

4k+5k(g)

1Pk=1

kk

1+2kkkzk

(b)1Pk=0

k6kzk (e)

1Pk=1

2kz2k

k2+k(h)

1Pk=3

(log k)k=2 zk

(c)1Pk=1

k2zk (f)1Pk=1

z2k

4kkk(i)

1Pk=1

k!zk

kk

Solution(a)We apply the ratio test, thus

R = limk!1

���� akak+1���� = lim

k!1

���� 2k2k+1���� = lim

k!1

1

2=1

2:

(b)We apply the ratio test, thus

R = limk!1

���� akak+1���� = lim

k!1

���� k6k 6k+1k + 1

���� = limk!1

���� 6kk + 1���� = lim

k!1

6

1 + 1=k= 6:

(c)We apply the ratio test, thus

R = limk!1

���� akak+1���� = lim

k!1

���� k2

(k + 1)2

���� = limk!1

k2

k2 + 2k + 1= lim

k!1

1

1 + 2=k + 1=k2= 1:

(d)We apply the ratio test, thus

R = limk!1

���� akak+1���� = lim

k!1

���� 3k

4k + 5k4k+1 + 5k+1

3k+1

���� == lim

k!1

������3k � 5k+1

��45

�k+1+ 1�

3k+1 � 5k��

45

�k+ 1������� = lim

k!1

������5��

45

�k+1+ 1�

3��

45

�k+ 1������� = 5

3:

23

(e)We apply the ratio test, thus

R = limk!1

���� akak+1���� = lim

k!1

����� 2kz2kk2 + k

(k + 1)2 + k + 1

2k+1z2(k+1)

����� = limk!1

�k + 2

k

�1

2 jzj2=

= limk!1

�1 + 2=k

1

�1

2 jzj2=

�0; 2 jzj2 < 11; 2 jzj2 > 1 :

The convergence radius R = 1=p2.

(f)We apply the ratio test, thus

R = limk!1

���� akak+1���� = lim

k!1

����� z2k4kkk4k+1 (k + 1)k+1

z2(k+1)

����� = limk!1

�����4 (k + 1)z2(k + 1)k

kk

����� == lim

k!1

4 (k + 1)

jzj2�1 + 1=k

1

�k=1:

(g)We apply the ratio test, thus

R = limk!1

���� akak+1���� = lim

k!1

����� kk

1 + 2kkk1 + 2k+1 (k + 1)k+1

(k + 1)k+1

����� = limk!1

����� kk

(k + 1)k1 + 2k+1 (k + 1)k+1

(1 + 2kkk) (k + 1)

����� == lim

k!1

������

1=k

1 + 1=k

�k 1

(1 + 2kkk) (k + 1)+2k+1 (k + 1)k

1 + 2kkk

!����� == lim

k!1

������

1=k

1 + 1=k

�k 1

(1 + 2kkk) (k + 1)+ 2

�k + 1

k

�k1

1= (2kkk) + 1

!����� = 2:(h)We apply the root test, thus

24

R =1

limk!1

kpjakj

=1

limk!1

k

q(log k)k=2

=1

limk!1

log k= 0:

(i)We apply the ratio test, thus

R = limk!1

���� akak+1���� = lim

k!1

����� k!kk (k + 1)k+1(k + 1)!

����� = limk!1

�k + 1

k

�k= lim

k!1

�1 +

1

k

�k= e:

25

V.3.21 2 3 P L K

LLLDetermine for which z the following series converge.

(a)1Pk=1

(z � 1)k (c)1Pm=0

2m (z � 2)m (e)1Pn=1

nn (z � 3)n

(b)1Pk=0

(z�i)kk!

(d)1Pm=0

(z+1)m

m2 (f)1Pn=3

2n

n2(z � 2� i)n

Solution(a)We apply the ratio test for ak = 1, thus

R = limk!1

���� akak+1���� = lim

k!1

1

1= 1:

hence the radius of convergence is R = 1. Therefore, the series converges forall z satisfying jz � 1j < 1 and diverges for all z satisfying jz � 1j > 1. Nowconsider z such that jz � 1j = 1 then

jz � 1j = 1:Since

P1k=1 1 diverges, the power series diverges for such z also. Hence, the

given series converges for all z satisfying jz � 1j < 1.

(b)We apply the ratio test for ak = 1=k!, thus

R = limk!1

���� akak+1���� = lim

k!1

1

k!

(k + 1)!

1= lim

k!1(k + 1) =1:

Therefore, the series converges for all z in C.

(c)We apply the ratio test for am = 2m, thus

R = limm!1

���� amam+1���� = lim

m!1

2m

2m+1= lim

m!1

1

2=1

2:

hence the radius of convergence is R = 1=2. Therefore, the series convergesfor all z satisfying jz � 2j < 1=2 and diverges for all z satisfying jz � 2j > 1=2.Now consider z such that jz � 2j = 1=2 then

26

j2m (z � 2)mj = 2m�1

2

�m= 1:

SinceP1

m=1 1 converges, the power series diverges for such z also. Hence,the given series converges for all z satisfying jz � 2j < 1=2.

(d)We apply the ratio test for am = 1=m2, thus

R = limm!1

���� amam+1���� = lim

m!1

1

m2

(m+ 1)2

1= lim

m!1

m2 + 2m+ 1

m2= lim

m!1

1 + 2=m+ 1=m2

1= 1:

hence the radius of convergence is R = 1. Therefore, the series converges forall z satisfying jz + ij < 1 and diverges for all z satisfying jz + ij > 1. Nowconsider z such that jz + ij = 1 then����(z + i)mm2

���� = 1m

m2=

1

m2:

SinceP1

m=1 1=m2 converges, the power series converges for such z also.

Hence, the given series converges for all z satisfying jz + ij � 1.

(e)We apply the ratio test for an = nn, thus

R = limn!1

���� anan+1���� = lim

n!1

nn

(n+ 1)n+1= lim

n!1

nn

(n+ 1)n (n+ 1)= lim

n!1

�n

n+ 1

�n1

n+ 1= 0:

Therefore, the series diverges for all z satisfying jz � 3j > 0. Now consider zsuch that jz � 3j = 0 then

jnn (z � 3)nj = nn � 0n = 0:Since

P1n=1 0 converges, the power series converges for z = 3. Hence, the

given series only converges for z = 3.

(f)We apply the ratio test for ak = 2k=k2, thus

27

R = limk!1

���� akak+1���� = lim

k!1

2k

k2(k + 1)2

2k+1= lim

k!1

k2 + 2k + 1

2k2= lim

k!1

1 + 2=k + 1=k2

2=1

2:

hence the radius of convergence is R = 1=2. Therefore, the series con-verges for all z satisfying jz � 2� ij < 1=2 and diverges for all z satisfyingjz � 2� ij > 1=2. Now consider z such that jz � 2� ij = 1=2 then����2nn2 (z � 2� i)n

���� = 2n

n2

�1

2

�n=1

n2:

SinceP1

k=1 1=n2 converges, the power series converges for such z also. Hence,

the given series converges for all z satisfying jz � 2� ij � 1=2.

28

V.3.31 2 3 P L K

LLLFind the radius of convergence of the following series.

(a)1Pn=0

z3n= z + z3 + z9 + z27 + z81 + � � �

(b)P

p primezp = z2 + z3 + z5 + z7 + z11 + � � �

Solutions(a)(b)

R =1

lim sup kpjakj

=1

1= 1:

Neither series converges at z = 1, so R = 1.

29

V.3.41 2 3 P L K

Show that the function de�ned by f (z) =Pzn! is analytic on the open unit

disk fjzj < 1g. Show that jf (r�)j ! +1 as r ! 1 whenever � is a root ofunity. Remark. Thus f (z) does not extend analytically to any larger openset than the open unit disk.

SolutionCauchy-Hadamard formula gives R = 1 (radius of convergence) ) f (z) isanalytic for jzj < 1. Suppose � is an N th root of unity, �N = 1. Then

(r�)n! = rn!��N�n!=N

= rn! for n > N . Now1Prn! " 1 as r " 1, because it is

increasing, and the �nite partial sums ! 1 as r ! 1. ) jf (r�)j ! +1 asr ! 1.

30

V.3.51 2 3 P L K

LLLWhat functions are represented by the following power series?

(a)1Pk=1

kzk (b)1Pk=1

k2zk

Solution(a)Di¤erentiate the geometric series

P1k=0 z

k, obtain

1Xk=0

zk =1

1� z )1Xk=1

kzk�1 =1

(1� z)2:

Multiply by z obtain

1Xk=1

kzk =z

(1� z)2:

(b)Di¤erentiate the geometric series

P1k=0 z

k twice, obtain

1Xk=0

zk =1

1� z )1Xk=1

kzk�1 =1

(1� z)2)

1Xk=2

k (k � 1) zk�2 = 2

(1� z)3:

Multiply by z2 obtain

1Xk=2

�k2 � k

�z2 =

2z2

(1� z)3:

We get

1Xk=2

k2zk =

1Xk=2

kzk +2z2

(1� z)3=

z

(1� z)2� z + 2z2

(1� z)3

31

V.3.61 2 3 P L K

Show that the seriesPakz

k, the di¤erentiated seriesPkakz

k�1, and theintegrated series

P akk+1zk+1 all have the same radius of convergence.

SolutionCan use Cauchy-Hadamard formula, and k

pk = 1. Then lim sup

k!1

kpjakj =

lim supk!1

kpk jakj = lim sup

k!1

k

qjakjk+1, so all series have the same radius of conver-

gence. Can also use the characterization of R as the radius of the largestdisk to which the function extends and ?????. This largest disk is clearly thesame for f , f 0, and ????.

32

V.3.71 2 3 P L K

Consider the series

1Xk=0

�2 + (�1)k

�kzk:

Use the Cauchy-Hadamard formula to �nd the radius of conver-gence of the series. What happens when the ratio test is applied?Evaluate explicitly the sum of the series.

SolutionWe apply the Cauchy-Hadamard formula where ak =

�2 + (�1)k

�k. First

observe that

kpjakj = k

r�2 + (�1)k

�k= 2 + (�1)k

Note that

2 + (�1)k =�3 if k is even,1 if k is odd.

Hence

kpjakj =

�3 if k is even,1 if k is odd.

Hence, since the lim sup of this sequence is 3 we have by the Cauchy-Hadamardformula

R =1

lim sup kpjakj

=1

3:

Applying the ratio test we have

���� akak+1���� =

�2 + (�1)k

�k�2 + (�1)k+1

�k+1 = � 3k if k is even,1

3k+1if k is odd.

����33

Since the sequence does not converge, the ratio test is inconclusive, in thatlimk!1 jakj = jak+1j does not exist. However since jakj = jak+1j � 3, it doesconverge for jzj < 1=3.Then we have with jzj < 1=3

1Xk=0

�2 + (�1)k

�kzk =

=

1Xk=0

�2 + (�1)2n

�2nz2n +

1Xk=0

�2 + (�1)2n+1

�2n+1z2n+1 =

=1Xn=0

32nz2n+1Xn=0

12n+1z2n+1 =1Xn=0

�9z2�n+ z

1Xn=0

�z2�n=

=1

1� 9z2 +z

1� z2 :

Note: The series is convergent because j9z2j ; jz2j < 1 since jzj < 1=3, andthat the series have singularities at z = �1=3 and z = �1.

34

V.3.7Let L = lim sup k

pjakj. The de�nition of n ln supn implies that given " > 0,

9 > 0, 9N3 for k > N we have kpjakj 6 L+", but k

pjakj > L�" for in�nitely

many k0s. If k > N , then jakj < (L+ ")k. If jzj < 1= (L+ "), thenPakz

k

converges. (L+ ") jzj < 1,���kzk�� 6 (L+ ")k jzjk = ((L+ ") jzj)k converges

by comparison with geometric series. ) R > 1= (L+ ") ) jakj > (L� ")k.If jzj > L� ", then

��akzk�� > (L� ")k � 1

(L�")k for in�nitely many k0s. ), andP

akzk diverge. ) R 6 1

L�" . Let "! 0, get R 6 1L.

35

V.3.81 2 3 P L K

Write out a proof of the Cauchy-Hadamard formula (3.4).

Solution

36

V.4.11 2 3 P L K

LLLFind the radius of convergence of the power series for the followingfunctions, expanded about the indicated point.(a) 1

z�1 about z = i; (d) Log z about z = 1 + 2i;(b) 1

cos zabout z = 0; (e) z3=2 about z = 3;

(c) 1cosh z

about z = 0; (f) z�iz3�z about z = 2i:

Solution(d)Since the values of Log z are unbounded in any neighborhood of the origin itmay not be extended to an analytic function on any open disk containing theorigin. But Log z is de�ned and analytic on the disk of radius j1 + 2ij =

p5

centered at 1 + 2i. Hence, the desired radius of convergence isp5 by the

second Corollary on p. 146.

(a) R =p2 (b) R = �=2 (c) R = �=2 (d) R =

p5 (e) R = 3 (f) R = 2

37

V.4.21 2 3 P L K

LLLShow that the radius of convergence of the power series expansion of (z2 � 1) = (z3 � 1)about z = 2 is

p7.

SolutionRewrite f (z) as z+1

(z�e2�i=3)(z+e2�i=3). Singularities of f (z) are at �e2�i=3, and

distance from 2 to nearest singularity isp7.

38

V.4.31 2 3 P L K

LLLFind the power series expansion of Log z about the point z = i � 2. Showthat the radius of convergence of the series is R =

p5. Explain why this

does not contradict the discontinuity of Log z at z = �2.

Solution

(From Hints and Solutions)Log z extends to be analytic for jz � (i� 2)j <

p5, through the extension

does not coincide with Log z in the part of the disk in the lower half�plane.

39

V.4.41 2 3 P L K

LLLSuppose f (z) is analytic at z = 0 and satis�es f (z) = z + f (z)2. What isthe radius of convergence of the power series expansion of f (z) about z = 0?

Solution(From Hints and Solutions)Near 0 the function coincides with one of the branches of

�1�

p1� 4z

�=2.

The radius of convergence of the power series of either branch is 1=4, whichis the distance to the singularity at 1=4.

40

V.4.51 2 3 P L K

Deduce the identity eiz = cos z + i sin z from the power series expansions.

Solutioneiz =

1Pn=0

inzn

n!=Pn even

+Pn odd

=1Pm=0

(�1)mz2m(2m)!

+ i1Pm=0

(�1)mz2m+1(2m+1)!

, where n = 2m

in the �rst and n = 2m+ 1 in the second series.

41

V.4.61 2 3 P L K

LLLFind the power series expansions of cosh z and sinh z about z = 0. What arethe radii of convergence of the series?

Solution(a) cosh z = cos iz =

1Pn=0

z2n

(2n)!= 1 + z2

2!+ z4

4!+ : : :

(b) sinh z = �i sin (iz) = (�i)�(iz)� (iz)3

3!+ (iz)5

5!� : : :

�= z + z3

3!+ z5

5!+ : : :

=1Pn=0

z2n+1

(2n+1)!

42

V.4.71 2 3 P L K

Find the power series expansion of the principal branch Tan�1 (z) ofthe inverse tangent function about z = 0. What is the radius of con-vergence of the series? Hint. Find it by integrating its derivative(a geometric series) term by term.

SolutionWe have

d

dzTan�1 z =

1

1 + z2=

1Xk=0

��z2

�k;

thus

Tan�1 z =1Xk=0

(�1)k z2k+1

2k + 1:

And we have R = 1.

43

V.4.81 2 3 P L K

Expand Log (1 + iz) and Log (1� iz) in power series about z = 0. Bycomparing power series expansions (see the preceding exercise),establish the identity

Tan�1 z =1

2iLog

�1 + iz

1� iz

�:

(See Exercise 5 in section I.8)

Solution

Log (1 + iz) = �1Xk=0

(�i) k+1 zk+1

k + 1; Log (1� iz) = �

1Xk=0

ik+1zk+1

k + 1:

Thus

1

2iLog

1 + iz

1� iz =2i

2i

1Xj=0

i2jz2j+1

2j + 1=

1Xj=0

(�1)j z2j+1

2j + 1:

44

V.4.91 2 3 P L K

Let a be real, and consider the branch of za that is real and positive on(0;1). Expand za in a power series about z = 1. What is the radius ofconvergence of the series? Write down the series explicitly.

Solutionf (z) = za = eaLog z, Re z > 0. Assume a 6= 0; 1; 2; : : :. f (m) (z) = a (a� 1) : : : (a�m+ 1) za�m, f (m) (1) = a (a� 1) : : : (a�m+ 1)f (z) =

1Pm=0

am (z � 1)m, am = f (m)(1)m!

= a(a�1):::(a�m+1)m!

=�am

�, m > 1, and

a0 = a1 =�a0

�. f (z) =

1Pm=0

�am

�(z � 1)m. R = 1 since

���am+1am

��� = ��a�mm+1

�� ! 1 as

m!1. If a is an integer > 0 get R =1.

45

V.4.10 (From Hints and Solutions)1 2 3 P L K

Recall that for a complex number �, the binomial coe¢ cient "�choose n" isde�ned by

��0

�= 1; and

��n

�=� (�� 1) � � � (�� n+ 1)

n!; n � 1:

Find the radius of convergence of the binomial series

1Xn=0

��n

�zn:

Show that the binomial series represents the principal branch of the function(1 + z)�. For which � does the binomial series reduce to a polynomial?

SolutionUse f (n) = � (�� 1) : : : (�� n+ 1) (1 + z)��n and the formula for the coe¢ -cient of zn. The series reduces to a polynomial for � = 0; 1; 2; : : :. Otherwiseradius of convergence is 1, which is distance to the singularity at �1. Wecan obtain the radius of convergence for the series also from the ratio test,anan+1

= �(��1):::(��n+1)�(��1):::(��n+1)(��n)

(n+1)!n!

= n+1��n ! �1 as n!1.

46

V.4.111 2 3 P L K

For �xed n � 0, de�ne the function Jn (z) by the power series

Jn (z) =

1Xk=0

(�1)k zn+2kk! (n+ k)!2n+2k

:

Show that Jn (z) is an entire function. Show that w = Jn (z) satis�es thedi¤erential equation

w00 +1

zw0 +

�1� n

2

z2

�w = 0:

Remark. This is Bessel�s di¤erential equation, and Jn (z) is Bessel�s functionof order n.

SolutionJn (z) =

1Pk=0

(�1)kzn+2kk!(n+k)!2n+2k

= zn

n!2n+O (zn+2)

J 0n (z) =1Pk=0

( �1)k(n+2k)zn+2k�1k!(n+k)!2n+2k

= zn�1

(n�1)!2n +O (zn+2)

J 00n (z) =1Pk=0

( �1)k(n+2k)(n+2k�1)zn+2k�2k!(n+k)!2n+2k

= zn�2

(n�2)!2n +O (zn)

Term multiplying zn+2k in z2J 00n + zJ0n + (z

2 � n2) Jn is ( �1)k(n+2k)(n+2k�1)k!(n+k)!2n+2k

+( �1)k(n+2k)k!(n+k)!2n+2k

� ( �1)kn2k!(n+k)!2n+2k

�( �1)k

(k�1)!(n+k�1)!2n+2k�2 =

(�1)kk!(n+k)!2n+2k

�(n+ 2k) (n+ 2k � 1) + (n+ 2k)� n2 � 4k (n+ k)

=0

�Should cheek separately to the sum terms of J0 + J1, when we divide by 0above, ???? work out. (constant term ????? J0, z term for J1 ). It works,because the case k = 0 we replace 1= (k � 1)! by k=k! = 0. Ratio test givesradius of convergence =1. (It�s not clear).

1

47

V.4.121 2 3 P L K

LLLSuppose that the analytic function f (z) has power series expansionPanz

n. Show that if f (z) is an even function, then an = 0 for nodd. Show that if f (z) is an odd function, then an = 0 for n even.

SolutionIf f (z) is analytic in Dr (0) then f (�z) is also analytic in the same regionand f (�z) =

P1n=0 (�1)

n anzn. If f is even then f (z)� f (�z) = 0, and soP1

n=0 2a2n+1z2n+1 = 0. A power series is 0 i¤ all of its contents are 0 (since

an = f(n) (0) =n! which shows an = 0 for n odd. The result for f (z) odd is

analogous.

48

f (z) even, f (�z) = f (z) . Chain rule: f 0 (�z) = �f 0 (z), f 00 (�z) = f 00 (z),etc. f (k) (�z) = (�1)k f (k) (z). f (k) (0) = (�1)k f (k) (0) ) f (k) (0) = 0 for kodd. ) ak = f (k) (0) =k! for k odd. If f (z) is odd, f (�z) = �f (z), sameargument gives ak = 0 for k even. Alternatively, apply ever result to zf (z)which is even.

49

V.4.131 2 3 P L K

Prove the following version of L�Hospitals�s rule. If f (z) and g (z) are ana-lytic, f (z0) = g (z0) = 0, and g (z) is not identically zero, then

limz!z0

f (z)

g (z)= lim

z!z0

f 0 (z)

g0 (z);

in the sense that either both limits are �nite and equal, or both limits arein�nite.

SolutionAssume z0 = 0 , f (z) =

1Pk=1

akzk = f 0 (0) z+O (z2), g (z) = g0 (0) z+O (z2) =

g0 (0) z+aNzn+O

�zN+1

�. f(z)g(z)

! f 0(0)g0(0) if g

0 (0) 6= 0!

8<:1 if g (0) = 0 and f (0) 6= 0 or k < NaNbN

if g (0) = 0 , f (0) 6= 0, k = N0 if g (0) = 0 , f (0) 6= 0, k > N

f 0(z)g0(z) =

f 0(0)+kakzk�1+O(zk)g0(0)+NbNzN�1+O(zN )

, f0(z)g0(z) !

f 0(0)g0(0) if g

0 (0) 6= 0

!

8<:1 g (0) = 0 and either f (0) 6= 0 or k < NNaNNbN

= aNbN

g (0) = 0 , f (0) 6= 0, k = N0 g (0) = 0 , f (0) 6= 0, k > N

Then the limits are equal in all cases.

50

V.4.141 2 3 P L K

Let f be continuous function on the unit circle T = fjzj = 1g. Show thatf can be approximated uniformly on T by a sequence of polynomials in zif and only if f has an extension F that is continuous on the closed diskfjzj � 1g and analytic on the interior fjzj < 1g. Hint. To approximate suchan F , consider dilates Fr (z) = F (rz).

SolutionIf f has an analytic extension F , then Fr (z) = F (rz) has power series

F (z) =Panz

k, jzj < 1. Fr (z) =1Pn=j

anrnzk, jzj < 1

r. Fr converge uniformly

for jzj 6 1, to F (rz) and F (rz)! f (z) uniformly for jzj = 1 as v " 1. For" > 0, take r < 1 with jFr (z)� F (z)j < " for jzj = 1, then take N such that���� NPn=0

anrnzn � Fv (z)

���� 6 "2for jzj = 1. Then

���� NPn=0

anrnzn � F (z)

���� 6 "2+ "

2= "

for jzj = 1. Conversely if pn are polynomials, pn ! f uniformly for jzj = 1,then jpn � pmj ! 0 uniformly for jzj = 1 as n;m!1, so by the maximumprinciple, jpn � pmj ! 0 uniformly for jzj 6 1. ) fpng converges uniformlyto same function F (z) for jzj 6 1 . Since pn (z) ! f (z) for jzj = 1,F (z) = f (z) for jzj = 1. Since pn is analytic for jzj < 1, also F (z) isanalytic for jzj < 1.

51

V.5.11 2 3 P L K

LLLExpand the following functions in power series about 1(a) 1

z2+1(b) z2

z3�1 (c) e1=z2(d) z sinh (1=z)

Solution(b)Denote the above function by f and de�ne g by

g (w) = f (1=w) =(1=w)2

(1=w)3 � 1=

w

1� w3 for w 6= 0

and g (0) = 0. Note that g is analytic on the disk D1 (0) with power seriesrepresentation

g (w) =w

1� w3 = w1Xk=0

�w3�n=

1Xk=0

w3n+1

for all w 2 D1 (0) (note that jw3j � 1 if jwj � 1). Hence, we have

f (z) = g (1=z) =1Xk=0

(1=z)3n+1 =1Xk=0

1

z3n+1

for all z with jzj > 1.

(a) 1z2+1

= 1� z2 + z4 � : : : =1Pn=0

(�1)n z2n 1z2

11+1=z2

= 1z2(�1)nz2n

=

1Pn=0

(�1)nz2n+2

(b) z2

z3�1 =1z

11�1=z3 =

1z

1Pn=0

1z3n=

1Pn=0

1z3n+1

(c) e1=z2=

1Pn=0

1n!

1z2n

(d) z sinh 1z= z

�1z+ 1

3!z3+ 1

5!z5+ : : :

�=

1Pn=0

1(2n+1)!

1z2n

52

V.5.21 2 3 P L K

Suppose f (z) is analytic at1, with series expansion (5.1). With the notationf (1) = b0 and f 0 (1) = b1, show that

f 0 (1) = limz!1

z jf (z)� f (1)j :

Solutionf (z) =

1Pk=0

bkzk= f (1)+ f 0(1)

z+

1Pk=2

bkzk. z [f (z)� f (1)] = f 0 (1)+

1Pk=1

bk+1zk!

f 0 (1) as z !1, since1Pk=1

bk+1wk ! 0 as w ! 0.

53

V.5.31 2 3 P L K

Suppose f (z) is analytic at 1, with series expansion (5.1). Let� � 0 be the smallest number such that f (z) extends to be analyticfor jzj > �. Show that the series (5.1) converges absolutely forjzj > � and diverges for jzj < �.

Solution (A. Kumjian)Recall that f is analytic at 1 i¤ there is a function at 0 such that f (z) =g (w) with w = 1=z wherever f (z) makes sense. We may suppose that f (z)is de�ned for jzj > � and that f is analytic at these points. Then settingR = 1=� if � > 0 and R = 1 if � = 0, we have that g is analytic onDR (z0) and cannot be extended to an analytic function on any larger opendisk centered at z0. Hence, g has a power series representation

g (w) =1Xk=0

bkwk for jwj < R;

with radius of convergence R. Thus, the series converges absolutely for jwj <R and diverges for jwj > R. It follows that by substituting w = 1=z, we have

f (z) =1Xk=0

bkzk

for jzj > �;

where the series converges absolutely for jzj > � and diverges for jzj < �.

54

V.5.4 (From Hints and Solutions)1 2 3 P L K

Let E be a bounded subset of the complex plane C over which area integralscan be de�ned, and set

f (w) =

ZZE

dx dy

w � z ; w 2 CnE;

where z = x + iy. Show that f (w) is analytic at 1, and �nd a formula forthe coe¢ cients of the power series of f (w) at 1 in descending powers of w.Hint. Use a geometric series expansion.

SolutionIf jzj 6 M for z 2 E, and R > M , then 1= (w � z) =

Pzn=wn+1 converges

uniformly for z 2 E and jwj > R . Integrate term by term, obtain f (w) =1PN=0

bnwn+1

, jwj > R, where bn =RRE

zndxdy.

55

V.5.5 (From Hints and Solutions)1 2 3 P L K

Determine explicitly the function f (w) de�ned in Exercise 4, in the case thatE = fjwj � 1g is the unit disk. Hint. There are two formulae for f (w), onevalid for jwj � 1 and the other for jwj � 1. Be sure thy agree for jwj = 1.

SolutionTo �nd the formula for jwj < 1 , break the integral into two pieces corre-sponding to jzj > jwj and to jzj < jwj, and use geometric series. f (w) =RRD

dxdyw�z =

ZZjzj<jwj

dxdy

w � z| {z }(1)

+

ZZjzj>jwj

dxdy

w � z| {z }(2)

= (1)+(2). If jwj < 1, getRRzn�1dxdy =

RRrndrd�ei(n�1)� = 0 , f (w) =

RR1

w�zdxdy =�w, jwj > 1.

If jwj < 1, get (1) 1w

RRjzj<jwj

dxdy1�z=w =

1w

P RRjzj6jwj

zn

wndxdy = 1

w

RRjzj6jwj

dxdy = 1w� jwj2 =

� �w

(2)RR

1>jzj>jwj

dxdyw�z = �

RR1zdxdy1�w=z = �

1Pn=0

wnRR

dxdyzn+1

= 0

f (w) =

��=w; jwj > 1� �w; jwj < 1

56

V.6.11 2 3 P L K

LLLCalculate the terms through order seven of the power series expansion aboutz = 0 of the function 1= cos z.

Solution1

cos z= 1 +

�z2

2!� z4

4!� z6

6!+O (z8)

�+ (: : :)2 + (: : :)3 +O (z)

1 + z2

2!+ a4z

4 + a6z6 +O (z8)

z4 : � 14!+�12!

�2= 1

4� 1

24= 5

24, z6 : � 2

2!4!+�12!

�3= 1

8� 1

24= 1

12

sin zcos z

=�z � z3

3!+ z5

5!� z7

7!+ : : :

��1 + z2

2+ 5

24z4 � 1

12z6 + : : :

�z : 1z3 : 1

2� 1

3!= 1

2� 1

6= 1

3

z5 : 15!� 1

2� 13!+ 5

24= 1

120� 1

12+ 5

24= 16

120= 4

30= 2

15

z7 : 112� 5

24�3! +12�5! �

17!= 7�6�5�2�52�7+3�7�1

7!=

420�175+21�17!

= 2657!= 53

7�6�4�3�2

57

V.6.21 2 3 P L K

LLLCalculate the terms through order �ve of the power series expansion aboutz = 0 of the function z= sin z.

Solutionzsin z

= zz�z3=3!+z5=5!�::: =

11�z2=3!+z4=5!�:::

= 1 +�z2

3!� z4

5!+ z6

7!� : : :

�+ (: : :)2 + (: : :)3 +O (z2) =

1 + z2=6 + a4z4 + a6z

6 +O (z7)a4 = � 1

5!+ 1

(3!)2= 1

36� 1

120= 10�3

360= 7

360

a6 =17!= � 2

3!5!+ 1

(3!)3= 6(1�14)+4�5�7

3!7!= 140�78

3!7!= 62

3!7!= 3

2

zsin z

= 1 + z2

6+ 7z4

360+O (z6)

58

V.6.31 2 3 P L K

LLLShow that

ez

1 + z= 1 +

1

2z2 � 1

3z3 +

3

8z4 � 11

30z5 + � � �

Show that the general term of the power series is given by

an = (�1)n�1

2!� 1

3!+ � � �+ (�1)

n

n!

�; n � 2:

What is the radius of convergence of the series?

SolutionWe have

1

1 + z=

1Xk=0

(�1)k zk for all jzj < 1 and1Xk=0

zk

k!for all z 2 C:

Since the given function can be expressed as product of these two we have

ez

1 + z=

1

1 + zez =

1Xn=0

cnzn for all jzj < 1

where

an =(�1)n

0!+(�1)n�1

1!+(�1)n�2

2!+ � � �+ (�1)

0

n!

by formula (6.1). Hence, c0 = 1, c1 = 0 and for n � 2 we have, since0! = 1! = 1,

an = (�1)n�1 (�1 + 1)+(�1)n�2

2!+� � �+(�1)

0

n!= (�1)n

�1

2!� 1

3!+ � � �+ (�1)

n

n!

�:

The radius of convergence is 1.

59

Solve ez= (1 + z) = a0 + a1z + a2z2 + : : : by multiplying by (1 + z) and com-

paring with the coe¢ cients of ez.

a0 = 1a0 + a1 = 1a1 + a2 = 1=2an�1 + an = 1=n!

9>>=>>;)Recursivelya0 = 1an = 1=n!� a�n

So an = 1n!� 1

(n�1)! +1

(n�2)! � : : :+ (�1)n 10!(the two last terms cancel.) The

radius of convergence is 1 because the function has a pole at �1 . (In fact,(�1)n an ! e�1 when n!1 .)

60

V.6.41 2 3 P L K

De�ne the Bernoulli numbers Bn by

z

2cot (z=2) = 1�B1

z2

2!�B2

z4

4!�B3

z6

6!� � � � :

Explain why there are no odd terms in this series. What is the radius ofconvergence of the series? Find the �rst three Bernoulli numbers.

Solutionz2cot (z=2) = 1�B1 z

2

2!�B2 z

4

4!�B3 z

6

6!� : : :,

cot (z=2) = cos(z=2)sin(z=2)

is odd, so z2cot (z=2) is even, so only even terms appear in

the series. The function has singularities at z=2 = ��, i.e., at �2�. Distancefrom 0 to nearest singularity = 2� = radius of convergence.

Find the two BernoullinumbersB1+B2. w cotw = w coswsinw

=1�w2

2!+w4

4!�w6

6!+:::

1�w2

3!+w4

5!�w6

7!+:::

=h1� w2

2!+ w4

4!+O (w6)

i �1 + w2

6� w4

120+�w2

6

�2+O (w6)

��1� w2

2+ w4

24

��1 + w2

6+�136� 1

120

�w4�+O (w6) =

1 +�16� 1

2

�w2 +

�7360� 1

12+ 1

24

�w4 +O (w6) =

1� 13w2 � 1

45w4 +O (w6)

cotw = 1�B1w2

2!�B2 2

4w4

4!� : : :

) 22B12!= 1

3, B1 = 1

6, 2

4

4!B2 =

145, B2 = 4!

24�45 , B2 =130

For B3 , must keep all powers ????? to and ????? w8. Better to just ?????for B1 and B2.

61

V.6.51 2 3 P L K

De�ne the Euler numbers En by

1

cosh z=

1Xn=0

Enn!zn:

What is the radius of convergence of the series? Show that En = 0 for n odd.Find the �rst four nonzero Euler numbers.

Solution1

cosh z=

1Pn=0

Enn!zn, cosh z has zeros at ��

2i;�3�

2i; : : :. Distance from 0 to

nearest singularity is �=2 = R . Since cosh z is even, En = 0 for n odd.1

cosh z= 1

(1+z2=2!+z4=4!+:::)= 1 + E2

z2

2!+ E4

z4

4!+ E6

z6

6!=

1��z2

2!+ z4

4!+ : : :

�+�z2

2!+ z4

4!+ z6

6!: : :�2� (: : :)3, E0 = 1

E22!= � 1

2!, E2 = �1, E44! = �

14!+�12!

�2= 1

4� 1

24= 5

24, E4 = 5

E66!= � 1

6!+ 2 1

2!4!� 1

(2!)3, E6 = �61

E6 = �1 + 6 � 5� 6!33= 29� 6 � 5 � 3 = 29� 90 = �61

62

V.6.61 2 3 P L K

Show that the coe¢ cients of a power series "depend continuously" on thefunction they represent, in the following sense. If ffm (z)g is a sequence ofanalytic functions that converges uniformly to f (z) for jzj > �, and

fm (z) =

1Xk=0

ak;m zk; f (z) =

mXk=0

ak zk;

then for each k � 0, we have ak;m ! ak as m!1.

63

V.7.11 2 3 P L K

LLLFind the zeros and orders of zeros of the following functions(a) z2+1

z2�1 (d) cos z � 1 (g) ez � 1(b) 1

z+ 1

z5(e) cos z�1

z(h) sinh2 z + cosh2 z

(c) z2 sin z (f) cos z�1z2

(i) Log zz

(principal value)

Solution(c)Let f (z) = z2 sin z. Then f has zeros at all integer multiples of � i.e. z0 = k�for k 2 Z. Since sin z has a zero of order 1 at 0, so sin z = zh (z) for someanalytic function h with h (0) 6= 0. Thus, f (z) = z3h (z) and so f has azero of order three at 0. Note that f 0 (z) = 2z sin z+ z2 cos z. Hence, for anyother zero z0 = k� where k 2 Zn f0g, we have

f 0 (k�) = 2k� sin k� + (k�)2 cos k� = (�1)k k� 6= 0:Thus, f has a simple zero at z0 = k� when k 6= 0.(d)Note that g (z) = cos z � 1 = 0 i¤ z = 2k� for k 2 Z. We will showthat all the zeros are double zeros. First observe that g0 (z) = � sin z andso g0 (2k�) = � sin 2k� = 0 for all k 2 Z. But g00 (z) = � cos z and sog0 (2k�) = � cos 2k� = �1 for all k 2 Z. Hence, all zeros are double zeros.

64

(a) simple zeros at �i, (b) simple zeros at �e�i=4, �e3�i=4, (c) triple zero at 0,simple zeros at n� , n = �1;�2; : : :, (d) double zeros at n� , n = �1;�2 : : :,(h) simple zeros at ��i=4 + n�i=2, n = �1;�2; : : :, (i) no zeros.(b) 1

z+ 1

z5= (z4 + 1) 1

z5= (z � z1) (z � z2) (z � z3) (z � z4) 1

z5, where zj =

ei(�=4+j=4�2�), these four points are simple zeros.(d) cos z = cos (x+ iy) = cos x cos (iy) � sin x sin (iy), cos (iy) = e�y+ey

2=

cosh y, sin (iy) = e�y�ey2i

= i�ey�e�y

2

�= i sinh y, cos z = 1 ) sin x sinh y =

0) y = 0 or x = n�. If x = n�, then cos (n�) cosh (n�) = (�1)n cosh (n�).This is = 1 only if n = 0, since cos (n�) > 1 for n 6= 0. Get x = 0, y = 0.If y = 0, then cosx = 1 only at x = 2n�, n = 0;�1;�2; : : :. Solutionsare z = 0;�2�;�4�; : : :. Since d

dz(cos z � 1) = � sin z = 0 at these points,

d2

dz2(cos z � 1) = � cos z 6= 0 , at these points. The zeros are double zeros.

(e) cos z�1z

, has only zeros at z = 2n� , n = 0;�1;�2; : : : . Simple poles atz = 0 . At the points d

dz= cos z�1

z= z(� sin z)�(cos(z�1))

z2= � cos(2n�)�1

(2n�)2for n 6= 0.

Other zeros are double. Double zeros at z = 2n� , n = 0;�1;�2; : : :.(g) ez � 1 = z+ z2

2!+ : : :, thus a simple zero at 0. Since ez � 1 is periodic, we

have simple zeros at i2�k, k 2 Z .(i) Log z = 0 when jzj = 1 and Arg z = 0, thus at z = 1. We writeLog z = Log (1� (1� z)) = 1� z� (1�z)2

2+ (1�z)3

3+ : : :, so we have a simple

zero at 1.

65

V.7.2 (From Hints and Solutions)1 2 3 P L K

Determine which of the functions in the preceding exercise are analytic at1, and determine the orders of any zeros at 1.

Solution(a) analytic at 1 , (b) analytic at 1, simple zero, (c) �(i) not analytic at1.

66

V.7.31 2 3 P L K

LLLShow that the zeros of sin z and tan z are all simple.

SolutionZeros of sin z are at n�, �1 < n < 1, since cos (n�) = d

dzsin zjz=n� =

�1 6= 0, the zeros are simple.

67

V.7.41 2 3 P L K

LLLShow that cos (z + w) = cos z cosw� sin z sinw, assuming the correspondingidentity for z and w real.

Solutioncos (z + w) = cos z cosw + sin z sinw = F (z; w) is entire in z for each �xedw and entire in w for each �xed z. F (z; w) = 0 for z; w real. Apply theorem,with D = C, E = R. Get F (z; w) = 0, so cos (z + w) = cos z cosw �sin z sinw.

68

V.7.51 2 3 P L K

Show that Z 1

�1e�zt

2+2wtdt =

r�

zew

2=z; z; w 2 C; Re z > 0;

where we take the principal branch of the square root. Compare the resultto Exercise IV.3.1. Hint. Show that the integral is analytic in z and w, andevaluate it for z = x > 0 and w real by making a change of variable andusing the known value

p� for z = 1 and w = 0.

Solution1R�1

eat2�2btdt, note the integral is improper, so a limit process may be used

for analytic. Re a < 0. x > 0,1R�1

eat2�2btdt = 1p

x

1R�1

e�s2� 2bp

xsds. F (�) =

1R�1

e�s2e�sds =

1R�1

e�(s��=2)e�2=4ds = e�

2=4

1Z�1

e�(s��=2)ds

| {z }1R�1

e�s2e�sds =

p�e�

2=4 ,1R�1

e�xt2e�2btdt = 1p

x

p�eb

2=x, x real

1R�1

e�zt2e�zbtdt =

p�zeb

2=z

69

V.7.61 2 3 P L K

LLLSuppose f (z) is analytic on a domain D and z0 2 D. Show that iff (m) (z0) = 0 for m � 1, then f (z) is constant on D.

SolutionSuppose that f (m) (z0) = 0 form � and de�ne g onD by g (z) = f (z)�f (z0).Then it su¢ ces to show that g is identically zero. Suppose not, then sinceg(m) (z0) for m � 0, the power series expansion for g is trivial and therefore gis identically zero on a disk on nonzero radius centered at z0 but this violatesthe �rst theorem on page 156 that asserts that an analytic function which isnot identically zero only has isolated zeros. This result may also be provedusing the Uniqueness Principle (the second theorem on page 156).

70

V.7.71 2 3 P L K

LLLShow that if u (x; y) is a harmonic function on a domain D such that all thepartial derivatives of u (x; y) vanish at the same point of D, then u (x; y) isconstant on D.

Solution

71

V.7.81 2 3 P L K

With the convention that the function that is identically zero has a zero ofin�nite order at each point, show that if f (z) and g (z) have zeros of order nandm respectively at z0, then f (z) + g (z) has a zero of order k � min (n;m).Show that strict inequality can occur here, but that equality holds wheneverm 6= n.

Solution

72

V.7.91 2 3 P L K

LLLShow that the analytic function f (z) has a zero of order N at z0, thenf (z) = g (z)N for some function g (z) analytic near z0 and satisfying g0 (z) 6=0.

Solution(From Hints and Solutions)Write f (z) = (z � z0)N h (z), where h (z) has a convergent power series andh (z0) 6= 0. Take g (z) = (z � z0) e(log(h(z)))=N for an appropriate branch ofthe logarithm.

73

V.7.101 2 3 P L K

Show that if f (z) is a continuous function on a domain D such that f (z)N

is analytic on D for some integer N , then f (z) is analytic on D.

Solutionf (z)N analytic. Zeros of f (z) are isolated, for this need maximum principle.f (z) is analytic, except possibly at the isolated points where f (z) = 0. (Bythe Riemann�s theorem, f (z) is analytic ????? there but don�t have this yet)Write f (z)N = (z � z0)m h (z), h (z) analytic, h (z) 6= 0. Then h (z) has N th

root near z0, by preceding exercise. f (z)N = g (z)N (z � z0)m, (f (z) =g (z))N =

(z � z0)m. (z � z0)m=N is continuous in a neighborhood of z0. f (z) =g (z) re-turns to original value, doing circle around z � z0. m is an integral multipleof N .

74

V.7.111 2 3 P L K

Show that if f (z) is a nonconstant analytic function on a domain D, thenthe image under f (z) of any open set is open. Remark. This is the openmapping theorem for analytic functions. The proof is easy when f 0 (z) 6= 0,since the Jacobian of f (z) coincides with jf 0 (z)j2. Use Exercise 9 to dealwith the points where f 0 (z) is zero.

SolutionIf f 0 (z0) 6= 0, then f maps open disks centred at z0 onto open sets, so f (D)contains a disk centred at f (z0).If f 0 (z0) = 0, assume f (z0) = 0, write f (z) = (z � z0)N h (z), h (z0) 6= 0.Then f has a N th root near z0, f (z) = g (z)

N , g0 (z0) 6= 0. g covers a ?????disk, so f cover disk centred at 0.

75

V.7.121 2 3 P L K

Show that the open mapping theorem for analytic functions implies the max-imum principle for analytic functions.

SolutionsClearly open mapping theorem) strict maximum principle for analytic func-tions, since a non constant analytic function can�t attain maximum at z0 andcover a disk centred at z0 . Then strict maximum principle ) maximumprinciple.

76

V.7.13 (From Hints and Solutions)1 2 3 P L K

Let fn (z) be a sequence of analytic functions on a domain D such thatfn (D) � D, and suppose that fn (z) converges normally to f (z) on D.Show that either f (D) � D, or else f (D) consists of a single point on @D.

Solutionf (D) � D [ @D. If f (z) is not constant, then f (D) is open, and f (D)cannot contain any point of @D.

77

V.7.141 2 3 P L K

A set E is discrete if every point of E is isolated. Show that a closed discretesubset of a domain D either is �nite or can be arranged in a sequence fzkgthat accumulates only on f1g [ @D.

SolutionLet Kn = fz 2 D : d (z; @D) > 1=n; jzj 6 ng. En is compact, only �nitelymany points of E belongs to Kn ????? points, shading with those in E \K1,the E \ (K2nK1) ,?????

78

V8.11 2 3 P L K

Suppose that the principal branch ofpz2 � 1 is continued analytically from

z = 2 around the �gure eight path indicated above. What is the analytic con-tinuation of the function at the end of the path? Answer the same questionform the functions (z3 � 1)1=3 and (z6 � 1)1=3.

Solutionpz2 � 1, phase change i at +1, �1 at �1, i at +1, returns to initial value.

3pz3 � 1, phase change ei�=3 at +1, ei�=3 at +1, returns to e2�i=3 times initial

value. 3pz6 � 1, phase change ei�=3 at +1, e�2�i=3 at �1, ei�=3 at +1, returns

to initial value.

79

V8.21 2 3 P L K

Show that f (z) = Log z = (z � 1)� 12(z � 1)2 + � � � has an analytic contin-

uation around the unit circle (t) = eit, 0 � t � 2�. Determine explicitlythe power series ft for each t. How is f2� related to f0?

Solutiondfdz= 1

z=P

, f (m) (z) = (�1)m�1zm

(m� 1)!

Series is1Pm=0

f (m)(eit)m!

(z � eit)m = f (eit) +1Pm=1

( �1)m�1 e�itmm

(z � eit)m

ft (z) = it+1Pm=1

(�1)m�1 eitmm(z � eit)m , get f2� (z) = f0 (z) + 2�i .

80

V.8.31 2 3 P L K

Show that each branch ofpz can be continued analytically along any path

in Cn f0g, and show that the radius of convergence of the power series ft (z)representing the continuation is j (t)j. Show that

pz cannot be analytically

along path containing 0.

Solution

81

V.8.41 2 3 P L K

Let f (z) be analytic on a domainD, �x z0 2 D, and let f (z) =Pan (z � z0)n

be the expansion of f (z) about z0. Let

F (z) =

Z z

z0

f (�) d� =1Xn=0

ann+ 1

(z � z0)n+1

be the inde�nite integral of f (z) for z near z0. Show that F (z) can becontinued analytically along any path in D starting at z0. What happens inthe case D = Cn f0g, z0 = 1, and f (z) = 1=z? What happens in the casethat D is star-shaped?

Solution

82

V.8.51 2 3 P L K

Show that the function de�ned by

f (z) =X

z2n

= z + z2 + z4 + z8 + � � �

is analytic on the open unit disk fjzj < 1g, and that it cannot be extendedanalytically to any larger open set. Hint. Observe that f (z) = z +f (z2),and that f (r)! +1 as r ! 1.

Solution

83

V.8.61 2 3 P L K

Suppose f (z) =Panz

n, where an = 0 except for n in a sequence nk thatsatis�es nk+1=nk � 1+ � for some � > 0. Suppose further that the series hasradius of convergence R = 1. Show that f (z) does not extend analyticallyto any point of the unit circle. Remark. Such a sequence with large gapsbetween successive nonzero terms is called a lacunary sequence. This resultis the Hadamard gap theorem. There is a slick proof. If f (z) extends ana-lytically across z = 1, consider g (w) = f (wm (1 + w) =2), where m is a largeinteger.Show that the power series for g (w) has radius of convergence r > 1, andthat this implies that the power series of f (z) converges for jz � 1j < ".

Solution

84

V.8.71 2 3 P L K

Suppose f (z) =Panz

n, where the series has radius of convergence R <1.Show that there is an angle � such that f (z) does not have an analyticcontinuation along the path (t) = tei�, 0 � t � R. Determine the radius ofconvergence of the power series expansion of f (z) about tei�.

Solution

85

V.8.81 2 3 P L K

Let f (z) be analytic at z0, and let (t), a � t � b, be a path such that (a) = z0. If f (z) cannot be continued analytically along , show that thereis a parameter value t1 such that there is an analytic continuation ft (z) fora � t < t1, and the radius of convergence of the power series ft (z) tends to0 as t! t1.

Solution

86

V.8.91 2 3 P L K

Let P (z; w) be a polynomial in z and w, of degree n in w. Suppose thatf (z) is analytic at z0 and satis�es P (z; f (z)) = 0. Show that if ft (z)is any analytic continuation of f (z) along any path starting at z0, thenP (z; ft (z)) = 0 for all t. Remark. An analytic function f (z) satis�es apolynomial equation P (z; f (z)) = 0 is called an algebraic function. Forinstance, the branches of n

pz are algebraic functions, since the satisfy z�wn =

0.

Solution

87

V.8.101 2 3 P L K

Let D be the punctured disk f0 < jzj < "g, suppose f (z) is analytic at z0 2D, and ew0 = z0. Show that f (z) has an analytic continuation along anypath inD starting at z0 if and only if there is an analytic function g (w) in thehalf-plane fRew < log "g such that f (ew) = g (w) for w near w0. Remark.If f (z) does not extend analytically to D but has an analytic continuationalong any path in D, we say that f (z) has a branch point at z = 0. For theproof, use the fact that any path in D starting at z0 is the composition ofa unique path in the half-plane starting at w0 and the exponential functionew.

Solution

88

VI 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 1912345678

1

VI.1.11 2 3 P L K111 LLLFind all possible Laurent expansions centered at 0 of the followingfunctions.(a) 1

z2�z (b) z�1z+1

(c) 1(z2�1)(z2�4)

Solution(a)In the region 0 < jzj < 1, we have

1

z2 � z = �1

z

1

1� z = �1

z

1Xk=0

zk = �1Xk=0

zk�1 = [k � 1 = n] = �1X

n=�1zn:

In the regionjzj > 1, we have

1

z2 � z =1

z21

1� 1z

=1

z2

1Xk=0

�1

z

�k=1

z2

1Xk=0

z�k�2 = [�k � 2 = n] =�2X

n=�1zn:

(b)In the region jzj < 1 we have

z � 1z + 1

= 1� 2

z + 1= 1� 2

1� (�z) = 1�21Xn=0

(�1)n zn = �1�21Xn=1

(�1)n zn:

In the region jzj > 1 we have

z � 1z + 1

= 1� 2

z + 1= 1� 2

z

1

1���1z

� = 1� 2z

1Xk=0

(�1)k

zk=

= 1� 21Xk=0

(�1)k

zk+1= [k + 1 = n] = 1� 2

1Xn=1

(�1)n�1

zn= 1 + 2

1Xn=1

(�1)n

zn:

(c)From computation, we have

2

1

(z2 � 1) (z2 � 4) = �1

3

1

z2 � 1 +1

3

1

z2 � 4In the region jzj < 1, we have

1

(z2 � 1) (z2 � 4) = �1

3

1

z2 � 1 +1

3

1

z2 � 4 =1

3

1

1� z2 �1

12

1

1� z2

4

=

=1

3

1Xn=0

z2n � 1

12

1Xn=0

z2n

4n=1

12

1Xn=0

�4� 4�n

�z2n:

In the region 1 < jzj < 2, we have

1

(z2 � 1) (z2 � 4) = �1

3

1

z2 � 1+1

3

1

z2 � 4 = �1

3

1

z2�1� 1

z2

�� 13

1

4�1� z2

4

� == � 1

3z2

1Xk=0

1

z2n� 1

12

1Xn=0

z2n

4n= �1

3

�1Xk=�1

z2k � 1

12

1Xn=0

z2n

4n

In the region jzj > 2, we have

1

(z2 � 1) (z2 � 4) = �1

3

1

z2 � 1 +1

3

1

z2 � 4 =

= �13

1

z2�1� 1

z2

� + 13

1

z2�1� 4

z2

� == � 1

3z2

1Xn=0

1

z2n+

1

3z2

1Xn=0

4n

z2n=

= �13

1Xn=0

1

z2(n+1)+1

3

1Xn=0

4n

z2(n+1)=1

3

1Xn=0

(4n � 1) z�2(n+1) = [n+ 1 = �k] =

=1

3

�1Xk=�1

�4�1�k � 1

�z2k:

3

VI.1.21 2 3 P L K111 LLLFor each of the functions in Exercise 1, �nd the Laurent expansion centeredat z = �1 that converges at z = 1

2. Determine the largest open set on which

each series converges.

Solution(a)Partial fractions give us

1

z2 � z = �1

z+

1

z � 1Laurent series for �1=z

� 1z= � 1

z + 1� 1 = �1

z + 1

1

1� 1z+1

= � 1

z + 1

1Xk=0

1

(z + 1)k=

= �1Xk=0

1

(z + 1)k+1= [k + 1 = �n] = �

�1Xn=�1

(z + 1)n

Laurent series for 1=z � 1

1

z � 1 =1

z + 1� 2 = �1

2

1

1� z+12

= �12

1Xn=0

(z + 1)n

2n= � 1

2n+1

1Xn=0

(z + 1)n

Thus we have

1

z2 � z =1X

k=�1

an (z + 1)n ; where an =

��1 for n � �1� 12n+1

for n � 0

����The function 1= (z2 � z) converges on the set 1 < jz + 1j < 2.(b)We have

z � 1z + 1

= 1� 2

z + 1

The function (z � 1) = (z + 1) converges on the set 0 < jz + 1j <1.

4

(c)Partial fractions give us

1

(z2 � 1) (z2 � 4) = �1

6

1

z � 1 +1

6

1

z + 1+1

12

1

z � 2 �1

12

1

z + 2=

= �16

1

z + 1� 2 +1

6

1

z + 1+1

12

1

z + 1� 3 �1

12

1

z + 1 + 1=

=1

12

1

1� z+12

+1

6

1

z + 1� 1

36

1

1� z+13

� 1

12 (z + 1)

1

1 + 1z+1

=

=1

12

1Xn=0

(z + 1)n

2n+1

6

1

z + 1� 1

36

1Xn=0

(z + 1)n

3n� 1

12

1Xk=0

(�1)k

(z + 1)k+1=

=1

12

1Xn=0

(z + 1)n

2n+1

6

1

z + 1� 1

36

1Xn=0

(z + 1)n

3n+1

12

n=�1X�1

(�1)n (z + 1)n =

1

12

1Xn=0

(�1)n (z + 1)n+�1

6� 1

12

�(z + 1)�1+

1

12

1Xn=0

�1

12 � 2n �1

36 � 3n

�(z + 1)n ;

The function 1= ((z2 � 1) (z2 � 4)) converges on the set 1 < jz + 1j < 2.

5

VI.1.31 2 3 P L K

LLLRecall the powe series for the Bessel function Jn (z), n � 0, given in ExerciseV.4.11, and de�ne J�n (z) = (�1)n Jn (z). For �xed w 2 C, establish theLaurent series expansion

exphw2(z � 1=z)

i=

1X�1

Jn (w) zn; 0 < jzj <1:

From the coe¢ cient formula (1.4), deduce that

Jn (z) =1

2�

Z 2�

0

ei(z sin ��n�); z 2 C:

Remark. This Laurent expansion is called Schlömlich formula.

Solution(a) f (z) = tan z, 3 < jzj < 4f (z) = tan z = sin z

cos z, singularities at �

2+ n�; n = 0;�1;�2; : : :

Simple poles at �2+ n�

limz!�=2

(z � �=2) sin zcos z

= sin �2limz!�=2

z��=2cos z�cos�=2 =

sin�=2� sin�=2 = �1

limz!��=2

(z + �=2) sin zcos z

= sin���2

�lim

z!��=2z+�=2

cos z�cos(��=2) =sin(��=2)� sin(��=2) = �1

) Principal parts at ��2are � 1

z��=2 .) f0 (z) = f (z) + 1

z��=2 +1

z+�=2is analytic for jzj < 3�

2

f1 (z) =�1

z��=2 �1

z+�=2is analytic for jzj > �

2;! 0 at ?????

) f (z) = f0 (z)+ 6 f1 (z) is the Laurent decomposition for 3 < jzj < 4, alsofor �

2< jzj < 3�

2.

(b) f1 (z) = ��z+�

2+z��

2

z2��2=4

�= � 2z

z2��2=4 = �2zz2

P1k=0

��2

4� 1z2

�n=

�2P1

n=0

��2

4

�n� 1z2n+1

, converges for jzj > �=2.(c) f (z) is odd, so f0 (z) is odd, and a0 = a2 = 0.

a1 =12�i

Hjzj=3

tan zz2dz = 1

2�i

�Hjzj=" +

Hjz��=2j="

Hjz+�=2j="

�tan zz2dz

At z � 0, tan zz2� sin z

z cos z� 1z,Hjzj=" = 2�i

At z � �2, tan z

z2� 1

z2� �1z��=2 � �

4�2

1z��=2 ,

Hjz��=2j=" =

�4�2� 2�i

At z � ��2, tan z

z2� 1

z2� �1z+�=2

� � 4�2

1z+�=2

,Hjz+�=2j=" =

�4�2� 2�i

6

(d) Since f0 (z) = f (z) � f1 (z) has poles at �3�2, otherwise is analytic for

jzj < 3�2, the series converges for jzj < 3�

2.

7

VI.1.41 2 3 P L K

LLLSuppose that f (z) = f0 (z) + f1 (z) is the Laurent decomposition of an an-alytic function f (z) on the annulus fA < jzj < Bg. Show that if f (z) isan even function, then f0 (z) and f1 (z) are even functions, and the Laurentseries expansion of f (z) has only even powers of z. Show that if f (z) is anodd function, then f0 (z) and f1 (z) are odd functions, ant the Laurent seriesexpansion of f (z) has only odd powers of z.

SolutionUse an =

Hjzj=r

f(z)zn+1

dz, or de�nition of f0 (z)+f1 (z), f (z) =1P�1anz

n, f (�z) =P(�1)n anzn. f even ) an = �an for n odd ) an = 0 for n odd. )

f0 (z) + f1 (z) have only even ????? ?????.

8

VI.1.51 2 3 P L K

LLLSuppose f (z) is analytic on the punctured plane D = Cn f0g. Showthat there is a constant c such that f (z)� c=z has a primitive in D.Give a formula for c in terms of an integral of f (z).

SolutionBy the Laurent Expansion Theorem we have for all z 2 D and r > 0

f (z) =1X

k=�1

akzk where ak =

1

2�i

ZCr(0)

f (�)

(� � z0)k+1d�:

Setting c = a�1 = 12�i

RCr(0)

f (�) d� we have

f (z)� c=z =Xk 6=�1

akzk:

Recall that an analytic function g has a primitive on a regionD i¤R g (z) dz =

0 for every piecewise smooth closed path in D. Let be such a path in Dand observe the series

Pk 6=�1 akz

k converges uniformly on any closed annulusfz : r � jzj � sg with 0 < r < s and hence on . Hence, by the �rst theoremon page 126, we may integrate the series termwise we obtain:Z

f (z)� c=zdz =Z

Xk 6=�1

akzk =

Xk 6=�1

ak

Z

zkdz =Xk 6=�1

ak � 0 = 0;

since zk has a primitive if k 6= �1. Hence, f (z)� c=z has a primitive in D.

9

VI.1.61 2 3 P L K

Fix an annulus D = fa < jzj < bg, and let f (z) be a continuous function onits boundary @D. Show that f (z) can be approximated uniformly on @D bypolynomials in z and 1=z if and only if f (z) has continuous extension to theclosed annulus D [ @D that is analytic on D.

SolutionUse the conformity theorem for polynomial approximation, V5 14. If f isanalytic, continuous ????? values, unit f = f0+f1, its Laurent decomposing.Then f0 + f1 have continuous boundary values. Can approximate f0 (z)uniformly by polynomials in z for f1 (z) uniformly by polynomials in 1=z.(?????? change of variable w = 1=z ). That does it. If f = lim gn (z)uniformly on @D, then by the maximum principle jgk � gjj ! 0 uniformlyon �D, so gk ! f uniformly on �D, and clearly Fn@D = f .

10

VI.1.71 2 3 P L K

Show that a harmonic function u on an annulus fA < jzj < Bg has a uniqueexpansion

u�rei��=

1Xn=�1

anrn cos (n�) +

Xn6=0

bnrn sin (n�) + c log r;

which is uniformly convergent on each circle in the annulus. Show that foreach r, A < r < B, the coe¢ cients an; bn and c satisfy

anrn + a�nr

�n =1

�

Z �

��u�rei��cos (n�) d�; n 6= 0;

bnrn + b�nr

�n =1

�

Z �

��u�rei��sin (n�) d�; n 6= 0;

a0 + c log r =1

2�

Z �

��u�rei��d�:

Hint. Use a decomposition of the form u = Re f+c log jzj, where f is analyticon the annulus. (See Exercise III.3.4.)

SolutionBy III.3.4, we have u = Re f +c log jzj for some e:Write its Laurent seriesf (z) =

P1k=�a ckz

k, A < jzj < B.Then for ck = �k�k, we haveRe f =

P1k=�1 �k Re z

k + �k Im zk

u =P1

k=�1 �krk cos (k�) + �kr

k sin (k�) + c logThe series converging uniformly on ????? The term w=�0 ?????Multiply by cosn�, ????? use orthogonality, get for n 6= 0,R ��� u (r; �) cosn�d� = �nr

nR ��� cos

2 n�d� + �nr�n R �

�� cos2 n�d� =

� (�nrn + ��nr

�n)) �nrn + ��nr�n = 1

�

R ��� u (r; �) cosn�d�, n 6= 0.

The formula for �nrn � ��nr�n = 1

�

R ��� u (r; �) sinn�d�, n 6= 0, is similar.

If we just integrate, we get 1�

R ��� u (r; �) =

12�

R ��� c log rd� +

�02�

Rd� = �0 +

c log rNote:

11

VI.1.8 (ej)Multiply f by a unimodular constant, assume jf (z)j 6 M ,

Hjzj=1

f (z) dz =

2�R0

f�ei��d� = 2�M . Take real parts, have

2�R0

Re�f�ei��iei��d� = 2�M ,

Re�f�ei��iei��d� 6

��f �ei���� 6 M . If have strict inequality somewhere,

then2�R0

< 2�M . We conclude thatRe�f�ei��iei��d� �M . Since

���f �ei�� iei���� 6M , we have Im

�f�ei��iei��d� � 0. ) f

�ei��iei� � M , f

�ei��� �ie�i�M ,

f (z) � �iM �z. ) f (z) = c�z for a constant, jcj =M .

VI.1.9 (ej)For the analyticity, di¤erentiate by hand. (See Exercise III.1.6). The Deriv-

ative is H 0 (w) = limw!1

1�w

R

hh(z)

z�(w+�w) �h(z)z�w

idz

= limw!1

1�w

R

h(z)�w(z�(w+�w))(z�w)dz =

R

h(z)dz

(z�w)2

12

VI.2.11 2 3 P L K

LLLFind the isolated singularities of the following functions, and de-termine where they are removable, essential, or poles. Determinethe order of any pole, and �nd the principal part at each pole.(a) z= (z2 � 1)2 (d) tan z = sin z

cos z(g) Log

�1� 1

z

�(b) zez

z2�1 (e) z2 sin�1z

�(h) Log z

(z�1)3

(c) e2z�1z

(f) cos zz2��2=4 (i) e1=(z

2+1)

Solution(a) (J.R.J Groves)We have that z = 1 is a pole of order 2 since z

(z2�1)2 =g(z)

(z�1)2 with g (z) =z

(z+1)2and g (z) is analytic near z = 1. Similarly we have, z = �1 is a pole

of order 2 since z(z�1)2 =

h(z)

(z+1)2with h (z) = z

(z�1)2 and h (z) is analytic nearz = �1.(b) (J.R.J Groves)We have that z = 1 is a pole of order 1 since zez

z2�1 =g(z)z�1 with g (z) =

zez

z+1,

and g (z) is analytic near z = 1. Similarly, z = �1 is a pole of order 1 sincezez

z2�1 =h(z)z+1

with h (z) = zez

z�1 , and h (z) is analytic near z = �1.(c) (A. Kumjian)Note that the given function is analytic on the punctured plane D = Cn f0g.It has an isolated singularity at z = 0. For z 6= 0 we have

e2z � 1z

=1

z

�1 +

1Xk=0

(2z)k

k!

!=1

z

1Xk=1

2kzk

k!=1

z

1Xj=0

2j+1

(j + 1)!zj:

Since the function has a power series expansion for all z 6= 0, it extends to ananalytic function at z = 0. Hence, the function has a removable singularityat 0.(d) (J.R.J Groves)We have that cos z has simple zeros at z = �=2 + k� for k 2 Z. So 1

cos zhas

simple poles at z = �=2 + k� for k 2 Z. Since sin z is analytic and di¤erentto 0 at these points, tan z = sin z

cos zhas simple poles at z = �=2+ k� for k 2 Z.

(e) (A. Kumjian)

13

The function is analytic on the punctured plane D = Cn f0g. It has anisolated singularity at z = 0. For z 6= 0 we have

z2 sin

�1

z

�= z2

1Xk=0

(�1)k

(2k + 1)!

�1

z

�2k+1=

=

1Xk=0

(�1)k

(2k + 1)!

1

z2k�1= z +

1Xk=1

(�1)k

(2k + 1)!

1

z2k�1:

Hence z = 0 is an essential singularity for the function since there are an in�-nite number of nonzero terms in the Laurent series with negative exponents.(e) (J.R.J Groves)The function z2 sin

�1z

�has an isolated singularity at z = 0. Since

z2 sin

�1

z

�= z2

1Xn=0

(�1)n

(2n+ 1)!

1

z2n+1=

1Xn=0

(�1)n

(2n+ 1)!

1

z2n�1;

z = 0 is an essential singularity.(f) (J.R.J Groves)The function z2�(�=2)2 = (z � �=2) (z + �=2) has simple zeros at z = ��=2and z = �=2. The function cos z has a simple zero at z = �=2. So we suspecta removable singularity. Set w = z � �=2. Then

cos (z)

z2 � (�=2)2=

cos (z)

(z � �=2) (z + �=2) =cos (w + �=2)

w (w + �)= � 1

w + �� sinww

:

Since we know that sin (w) =w = 1�w2=3!+ : : :, we deduce that our functionhas a removable singularity at w = 0 and so at z = �=2. In a similar fashion,it also has a removable singularity at z = ��=2.(g) (J.R.J Groves)The function Log (z) is analytic on the region Cn (�1; 0], it follows thatLog

�1� 1

z

�is analytic on Cn [0; 1]. There are no isolated singularities.

(h) (A. Kumjian)First observe that for z 2 C with jz � 1j < 1 we have

1

z=

1

1 + (z � 1) =1Xk=0

(�1)k (z � 1)k :

14

Since Log z is a primitive of 1=z, its power series expansion centered at z0 = 1my be obtained from that of 1=z by integrating termwise: we obtain (noteLog 1 = 0)

Log z =1Xk=1

(�1)k+1

k(z � 1)k ;

for jz � 1j < 1. Hence, we have for 0 < jz � 1j < 1

Log z

(z � 1)3=

1

(z � 1)31Xk=1

(�1)k+1

k(z � 1)k =

1Xk=1

(�1)k+1

k(z � 1)k�3 =

=1

(z � 1)2� 1

2 (z � 1) +1Xj=0

(�1)j

k + 3(z � 1)j

Hence, z = 0 is a pole of order 2. The principal part is given by 1(z�1)2�

12(z�1) .

(h) (J.R.J Groves)We have that z = 1 is a pole of order 2. This follows from the fact thatLog z = (z � 1)� (z�1)2

2+ (z�1)3

3� � � � for jz � 1j < 1 so that

Log z

(z � 1)3=

1

(z � 1)2� 12� 1

z � 1 +1

3� � � � :

(i) (J.R.J Groves)There are isolated singularities at z = �i and z = i. Moreover,

1

z2 + 1=i

2

�1

z + i� 1

z � i

�so that

e1

z2+1 = ei2� 1z+i e�

i2� 1z�i = e

i2� 1z+i

X1

n=0

(�i)n

2nn!� 1

(z � i)n :

since ei2� 1z+i is analytic and non-zero at z = i, e

1z2+1 has an essential singularity

at z = i. Similarly, it has an essential singularity at z = �i.

15

(a) Double poles at �1 , principal parts � (1=4) (z � 1)2.(b) Singularity at z = �1, simple poles zez

z2�1 =12

�ez

z�1 +ez

z+1

�. Principal part

1=2z�1 at z = 1,

1=2z+1

at z = �1.(d) tan z = sin z= cos z, poles at z = �=2 + n�, n = 0;�1;�2; : : :. cos z =� sin (�=2 + n�) (z � �=2� n�) +O

�(z � �=2� n�)3

�tan z =

sin(�=2+n�)+O((z��=2+n�)2)� sin(�=2+n�)(z��=2�n�)+O((:::)3)

= �1z��=2�n� + O (z � �=2� n�) Poles

are simple, principal part �1= (z � �=2� n�) .(f) cos z= (z2 � �2=4), singularities at ��=2 are both removable.(g) De�ned and analytic for 1 � 1=z 2 Cn [�1; 0), 1=z � 1 2 Cn [0;1),1=z 2 Cn [1;1), z 2 Cn [0; 1]. No isolated singularities.

(h) Isolated singularity at z = 1, Log z =zRdww=

2R1

dw1+w�1 =

R P(�1)n (w � 1)n dw =

1Pn=0

(�1)n(z�1)n+1n+1

Log z = (z � 1)� (z�1)22

+ (z�1)33

� : : :Log z

(z�1)3 =1

(z�1)2 �1

z(z�1) + analytic

Double pole, principal part 1(z�1)2 �

1z(z�2) at z = 1. (i) e

1=(z2+1), singularities

at z = �i. Values of 1= (z2 + 1) approach 1 from all directions as z ! �i,so values of e1=(z

2+1) do not converge, and singularities are essential.?????check e1(1�t

2) = f (it) ! +1 as t " 1 or t # �1, e1(1�t2) = f (it) ! 0 ast # 1 or t " �1, so no limit at �i.

16

VI.2.21 2 3 P L K

LLLFind the radius of convergence of the power series for the following functions,expanded about the indicated point.(a) z�1

z4�1 ; about z = 3 + i; (c) zsin z; about z = �i;

(b) cos zz2��2=4 ; about z = 0; (d) z2

sin3 z; about z = �i:

Solution(a) Removable singularity at z = 1, poles at �i;�1. R = 3 = ?????? to i(nearest singularity).(b) singularities at ��=2 are removable.(c) removable at z = 0, poles at ��;�2�. R = j�i� �j = �

p2.

(d) z2= sin2 z, simple poles at z = 0, R = �.

17

VI.2.31 2 3 P L K

Consider the function f (z) = tan z in the annulus f3 < jzj < 4g. Let f (z)= f0 (z) + f1 (z) be the Laurent decomposition of f (z), so that f0 (z) isanalytic for jzj < 4, and f1 (z) is analytic for jzj > 3 and vanishes at 1. (a)Obtain an explicit expression for f1 (z).(b)Write down the series expansion for f1 (z), and determine the largest domainon which it converges.(c)Obtain the coe¢ cients a0; a1 and a2 of the power series expansion of f0 (z).(d)What is the radius of convergence of the power series expansion for f0 (z)?

Solution(a) tan z has two poles in the disk jzj < 4, simple poles at ��=2, principalparts �1= (z � �=2). If f1 (z) = �1= (z � �=2) � 1 (z + �=2), then f0 (z) =f (z) � f1 (z) is analytic for jzj < 4, and f1 (z) is analytic for jzj > 3 and! 0 as z ! 1. By uniqueness, f (z) = f1 (z) + f2 (z) is the Laurentdecomposition.(b) Use geometric series. Converges for jzj > �=2. f1 (z) = �

�z+�=2+z��=2z2��2=4

�=

� 2zz2��2=4 = �

�2zz2

1Pk=0

��2

41z2

�n= , converges for jzj > �=2.

�21Pn=0

��2

4

�n1

z2n+1

(c) a0 = a2 = 0; a1 = 1 + 8=�2. (d) Since f0 (z) = f (z)� f1 (z) has poles at�3�=2, otherwise is analytic for jzj < 3, the series converges for jzj < 3�=2,and R = 3�=2.

18

VI.2.41 2 3 P L K

Suppose f (z) is meromorphic on the disk fjzj < sg, with only a �nite numberof poles in the disk. Show that the Laurent decomposition of f (z) withrespect to the annulus fs� " < jzj < sg has the form f (z) = f0 (z) + f1 (z),where f1 (z) is the sum of the principal parts of f (z) at its poles

SolutionSince f1 (z) is a sum of principal parts, with poles inside fjzj 6 � � "g, f1 (z)is analytic for jzj > � � ", and f1 (z) ! 0 as z ! 1. Further f0 (z) =f (z)�f1 (z) is analytic at ????? ??????, hence for jzj < 5. By the uniquenessof the Laurent decomposition, f is f0 (z)+f1 (z) = f (z). Use the uniquenessof the Laurent decomposition.

19

VI.2.51 2 3 P L K

By estimating the coe¢ cients of the Laurent series, prove that if z0 is anisolated singularity of f , and if (z � z0) f (z) ! 0 as z ! z0, then z0 isremovable. Give a second proof based on Morea�s theorem.

SolutionSuppose that z0 is an isolated singularity of f and that (z � z0)N f (z) isbounded near z0. Then by Riemann�s Theorem on removable singularities(see p. 172), (z � z0)N f (z) has a removable singularity at z0, the Laurentexpansion is of the following form: there is a r > 0 so that

(z � z0)N f (z) =1Xk=0

ak (z � z0)k

for all 0 < jz � z0j < r. And hence, we have

f (z) =1Xk=0

ak (z � z0)k�N =1X

j=�Naj+N (z � z0)j :

It follows that z0 is a pole of order at most N unless ak = 0 for all k =0; 1; : : : ; N � 1, in which case z0 is a removable singularity.

20

VI.2.61 2 3 P L K

LLLShow that if f (z) is continuous on a domain D, and if f (z)8 is analytic onD, then f (z) is analytic on D.

Solution

21

VI.2.71 2 3 P L K

LLLShow that if z0 is an isolated singularity of f (z), and if (z � z0)N f (z)is bounded near z0, then z0 is either removable or a pole of orderat most N .

SolutionSuppose that z0 is an isolated singularity of f and that (z � z0)N f (z) isbounded near z0. Then by Riemann�s Theorem on removable singularities(see p. 172), (z � z0)Nf (z) has a removable singularity at z0, the Laurentexpansion is of the following form: there is r > 0 so that

(z � z0)N f (z) =1Xk=0

ak (z � z0)k

for all 0 < jz � z0j < r. And hence, we have

f (z) =1Xk=0

ak (z � z0)k�N =1X

j=�Naj+N (z � z0)j :

It follows that z0 is a pole of order at most N unless ak = 0 for all k =0; 1; : : : ; N � 1, in which case z0 is a removable singularity.

22

VI.2.81 2 3 P L K

A meromorphic function f at z0 is said to have order N at z0 if f (z) =(z � z0)N g (z) for some analytic function g at z0 such that g (z0) 6= 0. Theorder of the function 0 is de�ned to be +1. Show that(a) order (fg; z0) = order (f; z0) + order (g; z0),(b) order (1=f; z0) = � order (f; z0),(c) order (f + g; z0) � min forder (f; z0) ; order (g; z0)g.Show that the inequality can occur in (c), but that equality hold in (c)whenever f and g have di¤erent orders at z0.

Solution

23

VI.2.91 2 3 P L K

Recall that "f (z) = O (h (z)) as z ! z0" means that there is a constantC such that jf (z)j � C jh (z)j for z near z0. Show that if z0 is an isolatedsingularity of an analytic function f (z), and if f (z) = O ((z � z0)m) asz ! z0, then the Laurent coe¢ cients of f (z) are 0 for k < m, that is theLaurent series of f (z) has the form

f (z) = am (z � z0) + am+1 (z � z0)m+1 + � � � :Remark. This shows that the use of notation O (zm) in section V.6 is con-sistent.

Solution

24

VI.2.101 2 3 P L K

LLLShow that if f (z) and g (z) are analytic functions that both have the sameorder N � 0 at z0, then

limz!z0

f (z)

g (z)=f (N) (z0)

g(N) (z0):

Solution

25

VI.2.111 2 3 P L K

Suppose f (z) =Pakz

k is analytic for jzj < R, and suppose that f (z)extends to be meromorphic for jzj < R + ", with only one pole z0 on thecircle jzj = R. Show that ak=ak+1 ! z0 as k !1.

SolutionIf f (z) =

Panz

n is analytic for jzj < R, and is meromorphic for jzj < R+",with poles of ?????6 N on the circle jzj = R, then an = O

�nN�1

�. If

f (z) =Panz

n as only one pole on the circle jzj = R, of ????? N , with

f (z) = A= (z � z0)N + lower order, then ak = A(�1)N+1kN(N�1)!zN+k+10

(1 +O (1=k)), so

that ak=ak+1 ! z0 as n ! 1. Proof: Write f (z) =NPl=0

Al(z�z0)l

+Pbkz

k

analytic, h (z) =Pbkz

k analytic. We have jbkj 6 1=sk for some s >

R, bk = O�1= jz0jk

�. 1

z�z0 = � 1z0

11�z=z0 = � 1

z0

1Pj=0

zj

zj0, (�1)d (l�1)!

(z�z0)l=

� 1z0

1Pj=l

j(j�1):::(j�l+1)zj�lzl0

1

(z�z0)l= (�1)l+1

(l�1)!

1Pk=0

1

zk+l+10

(k + l) : : : (k + 1) zk

1

(z�z0)l= (�1)l+1

(l�1)!

1Pk=0

1

zk+l+10

(k + l) : : : (k + 1) zk

26

VI.2.121 2 3 P L K

LLLShow that if z0 is an isolated singularity of f (z) that is not removable, thenz0 is an essential singularity for ef(z).

SolutionApply Arzela�-Weierstrass theorem. Suppose z0 = 0, an isolated singularityof f (z), not removable. 0 essential ) values of f (z) ???? on C as z ! 0,! values of ef(z) ????? on C as z ! 0. 0 a pole ) f (z) ! 1, as z ! 0,monotone, values of f (z) ?????? of 0 cover the extremum of some ??????fjwj > Rg. In any neighbourhood of 0 there are values of ef(z) = ew that arenear 0 (e�m) and that are near to (e+m), so ef(z) does not have a limit asz ! 0. 0 is essential singularity of ef(z).

27

VI.2.13 (From Hints and Solutions)1 2 3 P L K

Let S be a sequence converging to a point z0 2 C, and let f (z) be analyticon some disk centered at z0 except possibly at the points of S and z0. Showthat either f (z) extends to be meromorphic on some neighborhood of z0, orelse for any complex number L there is a sequence fwjg such that wj ! z0and f (wj)! L.

SolutionSuppose values of f (z) do not cluster at L as z ! z0. Then g (z) =1= (f (z)� L) is bounded for jz � z0j < "; z 6= zj. Apply Riemann�s the-orem �rst to the zj`s for j large, then to z0, to see that g (z) extends to beanalytic for jz � z0j < ", and f (z) is meromorphic there

28

VI.2.141 2 3 P L K

Suppose u�rei��is harmonic on a punctured disk f0 < r < 1g, with Laurent

series as in Exercise 7 of Section 1. Suppose � > 0 is such that r�u�rei��! 0

as r ! 0. Show that an = 0 = bn for n � ��.

Solutionu�rei��harmonic, 0 < r < 1. u =

1Pn=0

anrn cosn� +

1Pn6=0

bnrn sinn� + c log r.

Let m > �, m � � > 0. Have amrm + a�mr�m = O (r��), amr2m + a�m =O (r��+m) as r ! 0. Let r ! 0, get a�m = 0 for m > �. Similarly b�m = 0for m > �.

29

VI.2.151 2 3 P L K

Suppose u (z) is harmonic on the punctured disk f0 < jzj < �g.Show that if

u (z)

log (1= jzj) ! 0

as z ! 0, then u (z) extends to be harmonic at 0. What can you say ifyou know only that ju (z)j � C log (1= jzj) for some �xed constant C and0 < jzj < �?

SolutionAssume u (z) = O (log (1= jzj)) as z ! 0. Consider Laurent series as in14. Have. For m > 1, get

Ru�rei��d� = O (log (1=r)) amr

m + a�mr�m =

O (log (1=r)), amr2m + a�m = O (rm log (1=r)), a�m = 0 for all m > 0, andb�m = 0 for m > 0. u

�rei��= Re f

�rei��+ c log r, when f is analytic, c =

constant. u(r)log(1=r)

=Re f(rei�)log(1=r)

+ c log rlog(1=r)

, c = 0, and u = Re f is analytic atz = 0.

30

VI.3.11 2 3 P L K

Consider the functions in Exercise 1 of Section 2 above. Determine whichhave isolated singularities at 1, and classify them.

Solution

31

VI.3.21 2 3 P L K

Suppose that f (z) is an entire function that is not a polynomial. What kindof singularity can f (z) have at 1?

Solution

32

VI.3.31 2 3 P L K

Show that if f (z) is nonconstant entire function, then ef(z) has an essentialsingularity at z =1.

Solution

33

VI.3.41 2 3 P L K

Show that each branch of the following functions is meromorphic at 1, andobtain the series expansion for each branch at 1.(a) (z2 � 1)5=2 (b) 3

p(z3 � 1) (c)

pz2 � 1=z

Solution

34

VI.4.11 2 3 P L K

LLLFind the partial fractions decompositions of the following functions.(a) 1

z2�z (c) 1(z+1)(z2+2z+2)

(e) z�1z+1

(b) z2+1z(z2�1) (d) 1

(z2+1)2(f) z2�4z+3

z2�z�6(a)

1

z2 � z =A

z+

B

z � 1

A = limz!0z f (z) =

1

z � 1

����z=0

= �1

B = limz!1

(z � 1) f (z) = 1

z

����z=1

= 1

1

z2 � z = �1

z+

1

z � 1(b)

z2 + 1

z (z2 � 1) =A

z+

B

z � 1 +C

z + 1

A = limz!0z f (z) =

z2 + 1

z2 � 1

����z=0

= �1

B = limz!1

(z � 1) f (z) = z2 + 1

z (z + 1)

����z=1

= 1

C = limz!�1

(z + 1) f (z) =z2 + 1

z (z � 1)

����z=�1

= 1

z2 + 1

z (z2 � 1) = �1

z+

1

z � 1 +1

z + 1

(c)

1

(z + 1) (z2 + 2z + 2)=

A

z + 1+

B

z + 1 + i+

C

z + 1� i

35

A = limz!�1

z f (z) =1

(z2 + 2z + 2)

����z=�1

= 1

B = limz!�1�i

(z + 1 + i) f (z) =1

(z + 1) (z + 1� i)

����z=�1�i

= �12

C = limz!�1+i

(z + 1� i) f (z) = 1

(z + 1) (z + 1 + i)

����z=�1+i

= �12

1

(z + 1) (z2 + 2z + 2)=

1

z + 1� 1=2

z + 1 + i� 1=2

z + 1� i(d) (/snider 106) man så?

1

(z2 + 1)2=

A

(z � i)2+

B

(z � i) +C

(z + i)2+

D

z � i

A = limz!i

(z � i)2 f (z) = 1

(z + i)2

����z=i

= �14

B = limz!i

(z � i)2 f (z) = d

dz

1

(z + i)2

����z=i

= � i4

C = limz!�i

(z + i)2 f (z) =1

(z � i)2

����z=�i

= �14

D = limz!�i

(z + i)2 f (z) =d

dz

1

(z � i)2

����z=�i

=i

4

1

(z2 + 1)2= � 1=4

(z � i)2� i=4

(z � i) �1=4

(z + i)2+i=4

z � i(e)

z � 1z + 1

=A

z + 1+B

A = limz!�1

(z + 1) f (z) = z � 1jz=�1 = �2

B = limz!1

f (z) = 1

36

z � 1z + 1

= 1� 2

z + 1

(f)

z2 � 4z + 3z2 � z � 6 =

A

z + 2+

B

z � 3 + C

A = limz!�2

(z + 2) f (z) =z2 � 4z + 3z � 3

����z=�2

= �3

B = limz!3

(z � 3) f (z) = z2 � 4z + 3z (z + 1)

����z=3

= 0

C = limz!1

f (z) = 1

z2 � 4z + 3z2 � z � 6 =

�3z + 2

+ 1

37

VI.4.21 2 3 P L K

LLLUse the division algorithm to obtain the partial fractions decomposition ofthe following functions(a) z3+1

z2+1(b) z9+1

z6�1 (c) z6

(z2+1)(z�1)2

(a)

z3 + 1

z2 + 1= z +

�z + 1z2 + 1

= z +A

z + i+

B

z � i

A = limz!�i

(z + i) f (z) =�z + 1z � i

����z=�i

= �12+1

2i

B = limz!i

(z � i) f (z) = �z + 1z + i

����z=i

= �12� 12i

z3 + 1

z2 + 1= z +

�z + 1z2 + 1

= z +(�1 + i) =2z + i

� (1 + i) =2z � i

(b)

z9 + 1

z6 � 1 = z3+z3 + 1

z6 � 1 = z3+

1

z3 � 1 = z3+

1

(z � 1) (z2 + z + 1) = z3+

A

z � 1+B

z � w+C

z � �w

Because z3 � 1 = 0 have tree roots z1 = 1, z2 = e2�i=3, and z3 = e4�i=3. Setz2 = w, and we see that z3 = �w.

A = limz!1

(z � 1) f (z) = 1

z2 + z + 1

����z=1

=1

3

B = limz!w

(z � w) f (z) = 1

(z � 1) (z � e�2�i=3)

����z=e2�i=3

=w

3

C = limz! �w

(z � �w)2 f (z) =1

(z � 1) (z � e2�i=3)

����z=e�2�i=3

=�w

3

z9 + 1

z6 � 1 = z3 +

1=3

z � 1 +w=3

z � w +�w=3

z � �w; where w = e2�i=3

38

Konstigt svar i boken?�1=3z�1 +

w=3z�w +

�w=3z� �w

���� 1=3z�1 +

w=3z�w �

w=3z+w

�=

(c)

z6

(z2 + 1) (z2 � 2z + 1) =z6

z4 � 2z3 + 2z2 � 2z + 1

z6

z4 � 2z3 + 2z2 � 2z + 1 = z2 + 2z + 2 +

2z3 � z2 + 2z � 2(z2 + 1) (z � 1)2

2z3 � z2 + 2z � 2(z2 + 1) (z � 1)2

=A

z + i+

B

z � i +C

(z � 1)2+

D

z � 1

A = limz!�i

(z + i)2 f (z) =2z3 � z2 + 2z � 2(z � i) (z � 1)2

����z=�i

= �14

B = limz!i

(z � i) f (z) = 2z3 � z2 + 2z � 2(z + i) (z � 1)2

����z=i

= �14

C = limz!1

(z � 1)2 f (z) = 2z3 � z2 + 2z � 2(z2 + 1)

����z=1

=1

2

D = limz!1

(z � 1)2 f (z) = d

dz

2z3 � z2 + 2z � 2(z2 + 1)

����z=1

=5

2

39

VI.4.31 2 3 P L K

LLLLet V be the complex vector space of functions that are analytic on theextended complex plane except possibly at the points 0 and i, where theyhave poles of order at most two. What is the dimension of V ? Write downexplicitly a vector space basis for V .

Solution

40

VI.5.11 2 3 P L K

Show that if f (z) and g (z) have period w, then so do f (z) +g (z) andf (z) g (z).

Solution

41

VI.5.21 2 3 P L K

Expand 1= cos (2�z) in a series of powers of e2�iz that converges in the upperhalf-plane. Determine where the series converge absolutely and where itconverges uniformly.

Solution

42

VI.5.31 2 3 P L K

Expand tan z in a series of powers of exponentials eikz, �1 < k < 1, thatconverges in the upper half-plane. Also �nd an expansion of tan z as in anexponential series that converges in the lower half-plane.

Solution

43

VI.5.41 2 3 P L K

Let f (z) be an analytic function in the upper half-plane that is periodic,with real period 2�� > 0. Suppose that there are A;C > 0 such thatjf (x+ iy)j � CeAy for y > 0. Show that

f (z) =X

n��A�

aneinz=�;

where the series converges uniformly in each half-plane fy � "g, for �xed" > 0.

Solution

44

VI.5.51 2 3 P L K

Suppose that �1 are periods of a nonzero doubly periodic function f (z), andsuppose that there are no periods w of f (z) satisfying 0 < jwj < 1. Howmany periods of f (z) lie on the unit circle? Describe the possibilities, andsketch the set of periods for each possibility.

Solution

45

VI.5.61 2 3 P L K

We say that w1 and w2 generate the periods of a double periodic functionif the periods of the function are precisely the complex numbers of the formmw1 + nw2 where m and n are integers. Show that if w1 and w2 generatethe periods of a doubly periodic function f (z), and if �1 and �2 are complexnumbers, then �1 and �2 generate the periods of f (z) if and only if thereis a 2� 2 matrix A with integer entries and with determinant �1 such thatA (w1; w2) = (�1; �2).

Solution

46

VI.5.71 2 3 P L K

Let w1and w2 be two complex numbers that do lie on the same line through0. Let k � 3. Show that the series

1Xm;n=�1

1

(z � (mw1 + nw2))k

converges uniformly on any bounded subset of the complex plane to dou-ble periodic meromorphic function f (z), whose periods are generated byw1 and w2. Strategy. Show that the number of periods in any annulusfN � jzj � N + 1g is bounded by CN for some constant C.

Solution

47

VI.6.11 2 3 P L K

Consider the continuous function f�ei��= j�j, �� � � � �. Find the

complex Fourier series f�ei��and show that it can be expressed as a cosine

series. Sketch the graphs of the �rst three partial sums of the cosine series.Discuss the convergence of the series. Does it converge uniformly? Partialanswer. The cosine series is

j�j = �

2� 4

�

�cos � +

1

32cos 3� +

1

52cos 5� + � � �

�:

Solution

48

VI.6.21 2 3 P L K

Let f�ei��, �� < � � � (the principal value of the argument). Find the

complex Fourier series of f�ei��and the sine series of f

�ei��. Show that the

complex Fourier series diverges at � = ��, while the sine series convergesat ��. Di¤erentiate the complex Fourier series term by term and determinewhere the di¤erentiated series converges.

Solution

49

VI.6.31 2 3 P L K

Consider the continuous function f�ei��= �2, �� � � � �. Find the

complex Fourier series of f�ei��and show that it can be expressed as a cosine

series. Discuss the convergence of the series. Does it converge uniformly?By substituting � = 0, show that

�2

12= 1� 1

22+1

32� 1

42+ � � � :

Solution

50

VI.6.41 2 3 P L K

Consider the continuous function f�ei��= �4 � 2�2�2, �� � � � �. Find

the complex Fourier series of f�ei��and show that it can be expressed as a

cosine series. Relate the Fourier series to the series of the function in Exercise3.

Solution

51

VI.6.51 2 3 P L K

Show that ifPck converges absolutely, then

Pcke

ik� converges absolutelyfor each �, and the series converges uniformly for �� � � � �.

Solution

52

VI.6.61 2 3 P L K

Show that any function f�ei��on the unit circle with absolutely convergent

Fourier series has the form f�ei��= g

�ei��+ h (ei�), where g (z) and h (z)

are continuous functions on the unit circle that extend continuously to beanalytic on the open unit disk.

Solution

53

VI.6.71 2 3 P L K

If f�ei���Pcke

ik�, and the series converges uniformly to f�ei��, thenZ �

��

��f �ei����2 d�2�=

1Xk=�1

jckj2 :

Remark. This is called Parseval�s identity. Formula (6.6) show Parseval�sidentity holds for a function f

�ei��if and only if the partial sums of the

Fourier series of f�ei��converge to f

�ei��in the sense of "mean-square" or

"L2� approximation".

Solution

54

VI.6.81 2 3 P L K

By applying Parseval�s identity to the picewise constant function with series(6.5), show that

�2

8= 1 +

1

32+1

52+1

72+ � � � :

Use this identity and som algebraic manipulation to show that

�2

6= 1 +

1

22+1

32+1

42+ � � � :

Solution

55

VI.6.91 2 3 P L K

By applying Parseval�s identity to the function of Exercise 1, show that

�4

96= 1 +

1

34+1

54+1

74+ � � � :

Use this identity and som algebraic manipulation to show that

�4

90= 1 +

1

24+1

34+1

44+ � � � :

Solution

56

VI.6.101 2 3 P L K

If f (z) is analytic in some annulus containing the unit circle jzj = 1, withLaurent expansion

Pakz

k, then

1

2�

Ijzj=1

jf (z)j2 jdzj =1X

k=�1

jakj2 :

Solution

57

VI.6.111 2 3 P L K

Let f�ei��be a continuous function on the unit circle, with Fourier seriesP

ckeik�. Show that f

�ei��extends to be analytic on some annulus con-

taining the unit circle if and only if there exist r < 1 and C > 0 such thatjckj � Crjkj for �1 < k <1.

Solution

58

VI.6.121 2 3 P L K

Using the convergence theorem for Fourier series, prove that every continti-nously function on the unit circle in the complex plane can be approximateduniformly there by trigonometric polynomials, that is, by �nite linear com-binations of exponentials eik�, �1 < k < 1. Strategy. First approximatef�eik��by a smooth function.

Solution

59

VI.6.131 2 3 P L K

Let D be a domain bounded by a smooth boundary curve of length 2�. Weparametrize the boundary of D by arc length s, sot the boundary is given bya smooth periodic function (s), 0 � s � 2�. Let

Pcke

iks be the Fourierseries of (s).(a)Show that

Pk2 jckj2 = 1. Hint. Apply Parseval�s identity to 0 (s) and use

j 0 (s)j = 1 for a curve parameterized by arc length.(b)Show that the area of D is �

Pk jckj2. Hint. Use Exercise IV.1.4.

(c)Show that the area of D is � �, with equality if and only if D is a disk.Remark. This proves the isoperimetric theorem. Among all smooth closedcurves of a given length, the curve that surrounds the largest area is a circle.

Solution

60

VI.1.141 2 3 P L K

Show that

Z �

��

�����f �ei���nX

k=�m

bkeik�

�����2d�

2�=

Z �

��

�����f �ei���nX

k=�m

ckeik�

�����2d�

2�+

nXk=�m

jbk � ckj2 ;

for any choice of complex numbers bk, � m � k � n. Remark. Thisshows that the best mean-square approximate to f

�ei��by exponential sumsPn

�m bkeik�, for �xedm and n, is the corresponding partial sum of the Fourier

series.

Solution

61

VI.1.151 2 3 P L K

Show that a continously di¤erentiable function on the unit circle has anabsolutely convergent Fourier series. Strategy. Write the Fourier coe¢ cientsck of f

�ei��as akbk, where ak = 1=ik and bk is the Fourier coe¢ cient of

the derivative. Use Bessel�s inequality and the Cauchy-Schwarz inequality��P�k�k�� �qP j�kj2

qPj�kj

2.

Solution

62

VI.6.161 2 3 P L K

Let f�ei��be a continuous function on the unit circle. Suppose that f

�ei��

is piecewise continuously di¤erentiable, in the sense that it has a continuousderivative except at a �nite number of points, at each of which the derivativehas limits from the left and from the right. Show that the Fourier seriesf�ei��is absolutely convergent. Strategy. Cancel the discontinuities of the

derivative using translates of the function in Exercise 3, whose Fourier seriesis absolutely convergent.

Solution

63

VI.6.171 2 3 P L K

Let f�ei��be a piecewise continuously di¤erentiable, in the sense that it is

continuously di¤erentiable except at a �nite number of points, at each ofwhich both the function and its derivative have limits from the left and fromthe right. Show that the Fourier series of f

�ei��converges at each point, to

f�ei��if the function is continuous at ei�, and otherwise to the average of

the limits of f�ei��from the left and from the right. Strategy. Show that

f�ei��= f1

�ei��+Pbjhj

�ei��, where f1

�ei��satis�es the hypotheses of

Exercise 15, and each hj�ei��is obtained from the function of Exercise 2 by

change of variable � 7�! � � �j.

Solution

64

VII 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 191 X X X X2 X X X X X X X X X X X X3 X X X X X X X4 X X X X X X X X X X5 X X X X X X6 X X X X X X X7 X X X8 X

1

VII.1.1Evaluate the following residues.(a) Res

�1

z2+4; 2i�

(d) Res�sin zz2; 0�(g) Res

hz

Log z; 1i

(b) Res�

1z2+4

;�2i�(e) Res

�cos zz2; 0�(h) Res

�ez

z5; 0�

(c) Res�

1z5�1 ; 1

�(f) Res [cot z; 0] (i) Res

�zn+1zn�1 ; wk

�Solutiona)By rule 4,

Res

�1

z2 + 4; 2i

�=1

2z

����z=2i

=1

4i= � i

4:

b)By rule 4,

Res

�1

z2 + 4;�2i

�=1

2z

����z=�2i

= � 14i=i

4:

c)By rule 4,

Res

�1

z5 � 1 ; 1�=

1

5z4

����z=1

=1

5:

d)By rule 1,

Res

�sin z

z2; 0

�= lim

z!0

sin z

z= 1:

e)By rule 2,

Reshcos zz2

; 0i= lim

z!0

d

dzcos z = � sin zjz=0 = 0:

f)By rule 3,

Res [cot z; 0] = Reshcos zsin z

; 0i=cos z

cos z

���z=0

= 1:

2

g)By rule 3,

Res

�z

Log z; 1

�=

z

1=z

����z=1

= 1:

h)By Laurent expansion

ez

z5=1 + z

1!+ z2

2!+ z3

3!+ z4

4!+ o (z5)

z5=1

z5+

1

1!z4+

1

2!z3+

1

3!z2+1

4!z+ o (1) ;

hence

Res

�ez

z5; 0

�=1

4!=1

24:

i)We have poles of the form wk = e

2�ik=n, where k = 0; 1; 2; : : : n � 1, remarkthat wnk = 1, by rule 3,

Res

�zn + 1

zn � 1 ; wk�=zn + 1

nzn�1

����z=wk

=wnk + 1

nwn�1k

=2

nwn�1k

=2wkn=2e2�ik=n

n:

3

VII.1.2Calculate the residue at each singularity in the complex plane ofthe following functions.(a) e1=z (b) tan z (c) z

(z2+1)2(d) 1

z2+z

Solution(a)The function e1=z has a singularity at z = 0, by Laurent expansion

e1=z = 1 +1

z+

1

2!z2+ : : : :

Thus by de�nition

Res�e1=z; 0

�= 1:

(b)The function tan z = sin z

cos zhas isolated singularities at z = �

2+ �n, �1 <

n <1, they are simple poles.By rule 3,

Res [tan z; �=2 + n�] = Res

�sin z

cos z; �=2 + n�

�=

sin z

� sin z

����z=�=2+n�

= �1:

(c)The function z

(z2+1)2has isolated singularities at z = �i, they are double

poles.By rule 2,

Res

�z

(z2 + 1)2; i

�=d

dz

z

(z + i)2

����z=i

=(z + i)2 � 2z (z + i)

(z + i)4

�����z=i

= 0;

Res

�z

(z2 + 1)2;�i�=d

dz

z

(z � i)2

����z=�i

=(z � i)2 � 2z (z � i)

(z � i)4

�����z=�i

= 0:

(d)

4

The function 1z(z+1)

has isolated singularities at z = 0 and z = �1 they aresimple poles.By rule 1,

Res

�1

z (z + 1); 0

�= lim

z!0

1

z + 1=

1

z + 1

����z=0

= 1;

Res

�1

z (z + 1);�1

�= lim

z!�1

1

z=1

z

����z=�1

= �1:

5

VII.1.3Evaluate the following integrals using residue theorem.(a)

Rjzj=1

sin zz2dz (c)

Rjzj=2

zcos z

dz (e)Rjz�1j=1

1z8�1dz

(b)Rjzj=2

ez

z2�1dz (d)Rjzj=1

z4

sin zdz (f)

Rjz�1=2j=3=2

tan zzdz

Solution(a)Using the residue theorem and rule 1,

Zjzj=1

sin z

z2dz = 2�iRes

�sin z

z2; 0

�= 2�i

�limz!0

sin z

z

�= 2�i (1) = 2�i:

(b)Using the residue theorem and rule 3,

Zjzj=2

ez

z2 � 1dz =

= 2�i

�Res

�ez

z2 � 1 ; 1�+Res

�ez

z2 � 1 ;�1��

=

= 2�i

�ez

2z

����z=1

+ez

2z

����z=�1

�= 2�i

�e1 � e�12

�=

= 2�i sinh 1:

(c)Using the residue theorem and rule 3,

Zjzj=2

z

cos zdz = 2�i

�Res

h z

cos z;�

2

i+Res

h z

cos z;��2

i�=

= 2�i

z

� sin z

����z=�=2

+z

� sin z

����z=��=2

!= 2�i

���2� �2

�=

= �2�2i:

(d)Using the residue theorem and rule 3,

6

Zjzj=1

z4

sin zdz = 2�iRes

�z4

sin z; 0

�= 2�i

�z4

cos z

����z=0

�= 0:

(e)

|z ­ 1| = 1

VII.1.3e

Using the residue theorem with the contour in Figure VII.1.3e and rule 4,

Zjz�1j=1

1

z8 � 1dz = 2�i�Res

�1

z8 � 1 ; 1�+Res

�1

z8 � 1 ; ei�=4

�+Res

�1

z8 � 1 ; e�i�=4

��=

= 2�i

�1

8z7

����z=1

+1

8z7

����z=ei�=4

+1

8z7

����z=e�i�=4

�=�i

4

�1 + e��i=4 + e�i=4

�=

=�i

4

�1 +

1� ip2+1 + ip2

�=�i

4

�1 +

p2�:

(f)The function tan z

z= sin z=z

cos zhas removable singularity at z = 0, and a isolated

singularity at z = �2inside the domain jz � 1=2j = 3=2. Using the residue

theorem and rule 3,

7

Zjz�1=2j=3=2

tan z

zdz =

= 2�iRes

�sin z=z

cos z;�

2

�= 2�i

�limz!�=2

sin z=z

� sin z

�= 2�i

sin z=z

� sin z

����z=�=2

!= (2�i)

�� 2�

�=

= �4i:

8

VII.1.41 2 3 P L K

Suppose P (z) and Q (z) are polynomials such that the zeros of Q (z)are simple zeros at the points z1; : : : ; zm; and degP (z) < degQ (z).Show that the partial fractions decomposition P (z) =Q (z) is givenby

P (z)

Q (z)=

mXj=1

P (zj)

Q0 (zj)

1

z � zj:

SolutionSince the zeros of Q (z) are simple, the polynomial quotient P (z) =Q (z) hasat most simple poles. By Rule 3, the residue at zj is

P (z) =Q0 (z) :

Hence the principal part of P (z) =Q (z) at zj is

P (zj)

Q0 (zj)

1

z � zj:

Since deg P (z) < degQ (z), then

P (z)

Q (z)! 0

as z !1, and P (z) =Q (z) has no principal part at 1.By partial fraction decomposition our polynomial quotient P (z) =Q (z) orcan be written as the sum of its principal parts at its pole,

P (z)

Q (z)=

mXj=1

P (zj)

Q0 (zj)

1

z � zj:

9

VII.1.51 2 3 P L K

Let f (z) be a meromorphic function on the complex plane that isdoubly periodic, and suppose that none of the poles of f (z) lie onthe boundary of the period parallelogram P constructed in SectionVI.5. By integrating f (z) around of P , show that the sum of theresidues at the poles of f (z) in P is zero. Conclude that there is nodoubly periodic meromorphic function with only one pole, a simplepole, in the period parallelogram.

SolutionFollow directions.

R@P= 0, the curve integrals over opposite sides cancel.

Thus the sum of the residues is zero. In particular, there can not be a singlenonzero residue.

10

VII.1.6Consider the integral Z

@DR

e�i(z�1=2)2

1� e�2�iz dz;

where DR is the parallelogram with vertices �12� (1 + i)R.

(a)Use the residue theorem to show that the integral is (1 + i) =

p2.

(b)By parameterizing the sides of the parallelogram, show that theintegral tends to

(1 + i)

Z 1

�1e�2�t

2

dt

as R!1.(c)Use (a) and (b) to show thatZ 1

�1e�s

2

ds =p�:

Solution(a)

γ

γ

γ

γ

2

1

3

4

z1

VII.1.6

Set

11

I =

Z@DR

e�i(z�1=2)2

1� e�2�iz dz;

i.e, we integrate

f (z) =e�i(z�1=2)

2

1� e�2�izalong the parallelogram contour in Figure VII.1.6. The parallelogram hasvertices at �1=2� (1 + i)R.Residue at a simple pole at z1 = 0, where by Rule 3,

Res

"e�i(z�1=2)

2

1� e�2�iz ; 0#=e�i(z�1=2)

2

2�ie�2�iz

�����z=0

=e�i=4

2�i:

Using the Residue Theorem, we obtain thatZ@DR

e�i(z�1=2)2

1� e�2�iz dz = 2�i �e�i=4

2�i;

i.e., Z@DR

e�i(z�1=2)2

1� e�2�iz dz =1 + ip2:

(b)Integrate along 1 parametrized by z =

12+ (1 + i) t with �R � t � R, and

let R!1. This gives

Z 1

f (z) dz =

Z R

�R

e�i(z�1=2)2

1� e�2�iz dz =�z = 1

2+ (1 + i) t

dz = (1 + i) dt

�=

=

Z R

�R

e�i((1+i)t)2

(1 + i)

1� e�2�i(12+(1+i)t)

dt = (1 + i)

Z R

�R

e�2�t2

1 + e�2�i(1+i)tdt!

! (1 + i)

Z 1

�1

e�2�t2

1 + e�2�i(1+i)tdt:

Now integrate along 2 parameterized by z = �t + 12+ (1 + i)R with 0 �

t � 1, and let R!1.

12

����Z 2

f (z) dz

���� ==

�����Z 2

e�i(z�1=2)2

1� e�2�iz dz����� =

�z = �t+ 1

2+ (1 + i)R

dz = �dt

�=

=

�����Z 1

0

e�i(�t+(1+i)R)2

1� e�2�i(�t+1=2+(1+i)R)dt����� =

�����Z 1

0

e�(t2�2t)ie�2�(R

2�Rt)

1� e�(2t�1�2R)ie2�R dt����� �

� sup0�t�1

�����e�(t2�2t)ie�2�(R

2�Rt)

1� e�(2t�1�2R)ie2�R

����� � 1 � sup0�t�1

e�2�(R2�Rt)

j1� je�(2t�1�2R)ie2�Rjj =

= sup0�t�1

e�2�(R2�Rt)

j1� e2�Rj = sup0�t�1

e�2�(R2�Rt)

e2�R � 1 ! 0:

Integrate along 3, parameterized by z = �12+ (1 + i) t with �R � t � R,

and let R!1. This gives

Z 3

f (z) dz =

Z �R

R

e�i(z�1=2)2

1� e�2�iz dzdz =�z = �1

2+ (1 + i) t

dz = (1 + i) dt

�=

= �Z R

�R

e�i(�1+(1+i)t)2

(1 + i) dt

1� e�2�i(�12+(1+i)t)

= (1 + i)

Z R

�R

e�2�i(1+i)te�2�t2dt

1 + e�2�i(1+i)t!

! (1 + i)

Z 1

�1

e�2�i(1+i)te�2�t2dt

1 + e�2�i(1+i)tdt:

Now integrate along 4 parameterized by z = t� 12�(1 + i)R with 0 � t � 1,

and let R!1.

13

����Z 4

f (z) dz

���� ==

�����Z 4

e�i(z�1=2)2

1� e�2�iz dz����� =

�z = t� 1

2� (1 + i)R

dz = dt

�=

=

�����Z 1

0

e�i(t�1�(1+i)R)2

1� e�2�i(t�1=2�(1+i)R)dz����� =

�����Z 1

0

e�(2R�2Rt+t2�2t+1)ie�2�(R

2+R�Rt)

1� e��(2t�1�2R)ie�2�R dz

����� =� sup

0�t�1

�����e�(2R�2Rt+t2�2t+1)ie�2�(R

2+R�Rt)

1� e��(2t�1�2R)ie�2�R

������1 � sup0�t�1

e�2�(R2+R�Rt)

j1� je��(2t�1�2R)ie�2�Rjj �

� sup0�t�1

e�2�(R2+R�Rt)

j1� e�2�Rj ! 0:

Using the Residue Theorem and letting R!1, we obtain that

(1 + i)

Z 1

�1

e�2�t2

1 + e�2�i(1+i)tdt+ 0 + (1 + i)

Z 1

�1

e�2�i(1+i)te�2�t2dt

1 + e�2�i(1+i)tdt+ 0 =

= 2�i � e�i=4

2�i;

and therefore

(1 + i)

Z 1

�1

e�2�t2 �1 + e�2�i(1+i)t

�1 + e�2�i(1+i)t

dt =1 + ip2;

i.e.,

(1 + i)

Z 1

�1e�2�t

2

dt =1 + ip2:

(c)Comparing real parts thus yields,Z 1

�1e�2�t

2

dt =1p2;

and changing variables as follows gives the desired result

14

1p2=

Z 1

�1e�2�t

2

dt =

�s =

p2�t

ds =p2�dt

�=

1p2�

Z 1

�1e�s

2

ds;

i.e., Z 1

�1e�s

2

dt =p�:

15

VII.2.11 2 3 P L K

Show using residue theory thatZ 1

�1

dx

x2 + a2=�

a; a > 0:

Remark. Check the result by evaluating the integral directly, usingthe arctangent function.

Solution

γ

γ1

2

­R R

z 1

z 2

VII.2.1

Set

I =

Z 1

�1

dx

x2 + a2;

and integrate

f (z) =1

z2 + a2=

1

(z � ia) (z + ia)along the contour in Figure VII.2.1.Residue at a simple pole at z1 = ia, where by Rule 2,

Res

�1

z2 + a2; ia

�=1

2z

����z=ia

= � i

2a:

16

Integrate along 1, and let R!1. This givesZ 1

f (z) dz =

Z R

�R

dx

x2 + a2!Z 1

�1

dx

x2 + a2= I:

Integrate along 2, and let R!1. This gives����Z 2

f (z) dz

���� � 1

R2 � a2 � �R ��

R! 0:

Using the Residue Theorem and letting R!1, we obtain that

I + 0 = 2�i ��� i

2a

�;

and therefore Z 1

�1

dx

x2 + a2=�

a; a > 0:

Remark.Check of the result by evaluating the integral directly gives, because theintegrand is positive the integral can be computed taking a single limit,namely,

Z 1

�1

dx

x2 + a2= lim

b!1

Z b

�b

dx

x2 + a2= lim

b!1

�1

aarctan

x

a

�b�b=�

a; a > 0:

17

VII.2.21 2 3 P L K

Show using residue theory thatZ 1

�1

dx

(x2 + a2)2=

�

2a3:

Remark. Check the result by di¤erentiating the formula in the pre-ceding exercise with respect to the parameter.

Solution

γ

γ

2

1­R R

z 1

z 2

VII.2.2

Set

I =

Z 1

�1

dx

(x2 + a2)2;

and integrate

f (z) =1

(z2 + a2)2=

1

(z � ia)2 (z + ia)2

along the contour in Figure VII.2.2.Residue at a double pole at z1 = ia, where by Rule 2,

Res

�1

(z2 + a2)2; ia

�= lim

z!ia

d

dz

1

(z + ia)2=

�2(z + ia)3

����z=ia

= � i

4a3:

18

Integrate along 1, and let R!1. This givesZ 1

f (z) dz =

Z R

�R

dx

(x2 + a2)2!Z 1

�1

dx

(x2+a2)2= I:

Integrate along 2, and let R!1. This gives����Z 2

f (z) dz

���� � 1

(R2 � a2)2� �R � �

R3! 0:

Using the Residue Theorem and letting R!1, we obtain that that

I + 0 = 2�i ��� i

4a3

�;

and therefore Z 1

�1

dx

(x2 + a2)2=�

2a3:

Remark.From Exercise VII.2.1 we have the formula

(1)Z 1

�1

1

x2 + a2dx =

�

a; a > 0:

It is allowed to di¤erentiate both sides in the formula, because for everycompact interval with a 6= 0, both the integrand and its derivative are con-tinuous, and the primitive of the derivative is uniformly bounded by a a �independent function h (x) such that

Rh (x) <1 on every such like interval.

We start by di¤erentiate the left hand side in the formula with respect to theparameter a

(2)d

da

�Z 1

�1

1

x2 + a2dx

�=

Z 1

�1

�2a(x2 + a2)2

dx = �2aZ 1

�1

dx

(x2 + a2)2;

and di¤erentiate the right hand side in the formula in the same way

(3)d

da

�

a= � �

a2:

By (1) - (3), we have that

19

Z 1

�1

dx

(x2 + a2)2=�

2a3:

20

VII.2.3Show using residue theory thatZ 1

�1

x2dx

(x2 + 1)2=�

2:

Remark. Check the result by combining the preceding two exercises.

Solution

γ

γ

2

1­R R

z 1

z 2

VII.2.3

Set

I =

Z 1

�1

x2dx

(x2 + 1)2;

and integrate

f (z) =z2

(z2 + 1)2=

z2

(z � i)2 (z + i)2

along the contour in Figure VII.2.3.Residue at a double pole at z1 = i, where by Rule 2,

Res

�z2

(z2 + 1)2; i

�= lim

z!i

d

dz

z2

(z2 + i)2=2z (z + i)2 � 2z2 (z + i)

(z + i)4

�����z=i

= � i4:

Integrate along 1, and let R!1. This gives

21

Z 1

f (z) dz =

Z R

�R

x2dx

(x2 + 1)2!Z 1

�1

x2dx

(x2 + 1)2= I:

Integrate along 2, and let R!1. This gives����Z 2

f (z) dz

���� � R2

(R2 � 1)2� �R � �

R! 0:

Using the Residue Theorem and letting R!1, we obtain that

I + 0 = 2�i ��� i4

�;

and therefore Z 1

�1

x2dx

(x2 + 1)2=�

2:

Remark.We split the integrand using partial fraction, and put a = 1 in the formulasfrom the preceding two exercisesZ 1

�1

x2dx

(x2 + 1)2=

Z 1

�1

dx

x2 + 1�Z 1

�1

dx

(x2 + 1)2= � � �

2=�

2:

22

VII.2.4Using residue theory, show thatZ 1

�1

dx

x4 + 1=�p2:

Solution

γ

γ

2

1­R R

zz

z z

12

3 4

VII.2.4

Set

I =

Z 1

�1

dx

x4 + 1;

and integrate

f (z) =1

z4 + 1=

1�z �

�1+ip2

���z �

��1+ip2

���z �

��1�ip2

���z �

�1�ip2

��along the contour in Figure VII.2.4.Residue at a simple pole at z1 = 1+ip

2, where by Rule 2,

Res

�1

z4 + 1;1 + ip2

�=

1

4z3

����z=(1+i)=

p2

=

p2

8(�1� i)

Residue at a simple pole at z2 = �1+ip2, where by Rule 2,

23

Res

�1

z4 + 1;�1 + ip

2

�=

1

4z3

����z=(�1+i)=

p2

=

p2

8(1� i)

Integrate along 1, and let R!1. This givesZ 1

f (z) dz =

Z R

�R

dx

x4 + 1!Z 1

�1

dx

x4 + 1= I:

Integrate along 2, and let R!1. This gives����Z 2

f (z) dz

���� � 1

R4 � 1 � �R ��

R3! 0:

Using the Residue Theorem and letting R!1, we obtain that

I + 0 = 2�i � p

2

8(�1� i)+

p2

8(1� i)

!;

and therefore Z 1

�1

dx

x4 + 1=�p2:

24

VII.2.5Using residue theory, show thatZ 1

0

x2

x4 + 1dx =

�

2p2:

Solution

γ

γ

2

1­R R

zz

z z

12

3 4

VII.2.5

Set

I =

Z 1

0

x2

x4 + 1dx) 2I =

Z 1

�1

x2

x4 + 1dx

and integrate

f (z) =z2

z4 + 1=

z2�z �

�1+ip2

���z �

��1+ip2

���z �

��1�ip2

���z �

�1�ip2

��along the contour in Figure VII.2.5.Residue at a simple pole at z1 = 1+ip

2, where by Rule 3,

Res

�z2

z4 + 1;1 + ip2

�=1

4z

����z=(1+i)=

p2

=

p2

8(1� i) :

Residue at a simple pole at z2 = �1+ip2, where by Rule 3,

25

Res

�z2

z4 + 1;�1 + ip

2

�=1

4z

����z=(�1+i)=

p2

=

p2

8(�1� i) :

Integrate along 1, and let R!1. This givesZ 1

f (z) dz =

Z R

�R

x2

x4 + 1dx!

Z 1

�1

x2

x4 + 1dx = 2I:

Integrate along 2, and let R!1. This gives����Z 2

f (z) dz

���� � R2

R4 � 1 � �R ��

R! 0:

Using the Residue Theorem and letting R!1, we obtain that

2I + 0 = 2�i � p

2

8(1� i) +

p2

8(�1� i)

!;

and therefore Z 1

0

x2

x4 + 1dx =

�

2p2:

26

VII.2.6Show that Z 1

�1

x

(x2 + 2x+ 2) (x2 + 4)dx = � �

10:

Solution

γ

γ

2

1­R R

z

z

z

z

1

2

3

4

VII.2.6

Set

I =

Z 1

�1

x

(x2 + 2x+ 2) (x2 + 4)dx

and integrate

f (z) =z

(z2 + 2z + 2) (z2 + 4)=

z

(z � (�1 + i)) (z � 2i) (z � (�1� i)) (z + 2i)

along the contour in Figure VII.2.6.Residue at a simple pole at z1 = 2i, where by Rule 3,

Res

�z

(z2 + 2z + 2) (z2 + 4); 2i

�=

z

(z2 + 2z + 2) 2z

����z=2i

= � 120� 1

10i:

Residue at a simple pole at z2 = �1 + i, where by Rule 3,

27

Res

�z

(z2 + 2z + 2) (z2 + 4);�1 + i

�=

z

(2z + 2) (z2 + 4)

����z=�1+i

=1

20+3

20i:

Integrate along 1, and let R!1. This gives

Z 1

f (z) dz =

Z R

�R

x

(x2 + 2x+ 2) (x2 + 4)dx!

!Z 1

�1

x

(x2 + 2x+ 2) (x2 + 4)dx = I:

Integrate along 2, and let R!1. This gives����Z 2

f (z) dz

���� � R

(R2 � 2R� 2) (R2 � 4) � �R ��

R2! 0:

Using the Residue Theorem and letting R!1, we obtain that

I + 0 = 2�i ��� 120� 1

10i+

1

20+3

20i

�and therefore Z 1

�1

x

(x2 + 2x+ 2) (x2 + 4)dx = � �

10:

28

VII.2.71 2 3 P L K

LLLShow thatZ 1

�1

cos (ax)

x4 + 1dx =

�p2e�a=

p2

�cos

ap2+ sin

ap2

�; a > 0:

Solution

γ

γ

2

1­R R

zz

z z

12

3 4

VII.2.7

Set

I =

Z 1

�1

cos (ax)

x4 + 1dx a > 0;

and integrate

f (z) =eiaz

z4 + 1=

eiaz�z �

�1+ip2

���z �

��1+ip2

���z �

��1�ip2

���z �

�1�ip2

��along the contour in Figure VII.2.7.Residue at a simple pole at z1 = 1+ip

2, where by Rule 3,

29

Res

�eiaz

z4 + 1;1 + ip2

�=eiaz

4z3

����z=(1+i)=

p2

=

= �p2e�a=

p2 cos

�a=p2�

8�p2e�a=

p2 sin

�a=p2�

8i�p2e�a=

p2 cos

�a=p2�

8i+

p2e�a=

p2 sin

�a=p2�

8

Residue at a simple pole at z2 = �1+ip2, where by Rule 3,

Res

�eiaz

z4 + 1;�1 + ip

2

�=eiaz

4z3

����z=(�1+i)=

p2

=

=

p2e�a=

p2 cos

�a=p2�

8�p2e�a=

p2 sin

�a=p2�

8i�p2e�a=

p2 cos

�a=p2�

8i�p2e�a=

p2 sin

�a=p2�

8

Integrate along 1, and let R!1. This gives

Z 1

f (z) dz =

Z R

�R

eax

x4 + 1dx!

Z 1

�1

cos (ax)

x4 + 1dx+i

Z 1

�1

sin (ax)

x4 + 1dx = I+i

Z 1

�1

sin (ax)

x4 + 1dx

Integrate along 2, and let R ! 1. Because jeiazj � 1 in the upper halfplane if a > 0, this gives����Z

2

f (z) dz

���� � 1

R4 � 1 � �R ��

R3! 0:

Using the Residue Theorem and letting R!1, we obtain that

I+i

Z 1

�1

sin (ax)

x4 + 1dx = 2�i�

�p2e�a=

p2 cos

�a=p2�

4i�

p2e�a=

p2 sin

�a=p2�

4i

!;

and thereforeZ 1

�1

cos (ax)

x4 + 1dx =

�p2e�a=

p2

�cos

ap2+ sin

ap2

�; a > 0:

30

VII.2.8Show that Z 1

�1

cosx

(x2 + 1)2dx =

�

e:

Solution

γ

γ

2

1­R R

z 1

z 2

VII.2.8

Set

I =

Z 1

�1

cosx

(x2 + 1)2dx

and integrate

f (z) =eiz

(z2 + 1)2=

eiz

(z � i)2 (z + i)2

along the contour in Figure VII.2.8.Residue at a double pole at z1 = i, where by Rule 2,

Res

�eiz

(z2 + 1)2; i

�= lim

z!i

d

dz

eiz

(z + i)2==

ieiz (z + i)2 � 2eiz (z + i)(z + i)4

�����z=i

= �e�1

2i

Integrate along 1, and let R!1. This gives

31

Z 1

f (z) dz =

=

Z 1

�1

eix

(x2 + 1)2dx!

Z 1

�1

cosx

(x2 + 1)2dx+ i

Z 1

�1

sin x

(x2 + 1)2dx =

= I + i

Z 1

�1

sin x

(x2 + 1)2dx:

Integrate along 2, and let R ! 1. Because jeizj � 1 in the upper halfplane, this gives ����Z

2

f (z) dz

���� � 1

(R2 � 1)2� �R � �

R3! 0:

Using the Residue Theorem and letting R!1, we obtain that

I + i

Z 1

�1

sin x

(x2 + 1)2dx+ 0 = 2�i �

��e

�1

2i

�;

and therefore Z 1

�1

cosx

(x2 + 1)2dx =

�

e:

32

VII.2.9Show that Z 1

�1

sin2 x

x2 + 1dx =

�

2

�1� 1

e2

�:

Solution

γ

γ

2

1­R R

z 1

z 2

VII.2.9

Set

I =

Z 1

�1

sin2 x

x2 + 1dx =

Z 1

�1

1� cos 2x2 (x2 + 1)

dx

and integrate

f (z) =1� ei2z2 (z2 + 1)

=1� ei2z

2 (z � i) (z + i)along the contour in Figure VII.2.9.Residue at a simple pole at z1 = i, where by Rule 3,

Res

�1� ei2z2 (z2 + 1)

; i

�=1� ei2z4z

����z=i

=1

4ie�2 � 1

4i

Integrate along 1, and let R!1. This gives

33

Z 1

f (z) dz =

=

Z R

�R

1� ei2x2 (x2 + 1)

dx!Z 1

�1

1� cos 2x2 (x2 + 1)

dx+ i

Z 1

�1

sin 2x

2 (x2 + 1)dx =

= I + i

Z 1

�1

sin 2x

2 (x2 + 1)dx

Integrate along 2, and let R ! 1. Because jeizj � 1 in the upper halfplane, this gives ����Z

2

f (z) dz

���� � 2

2(R2 � 1) � �R ��

R! 0:

Using the Residue Theorem and letting R!1, we obtain that

I + i

Z 1

�1

sin 2x

2 (x2 + 1)dx+ 0 = 2�i �

�1

4ie�2 � 1

4i

�;

and therefore Z 1

�1

sin2 x

x2 + 1dx =

�

2

�1� 1

e2

�:

34

VII.2.10Show thatZ 1

�1

cos (ax)

x2 + b2dx =

�e�jajb

b; �1 < a <1; b > 0:

For which complex values of the parameters a and b does the in-tegral exist? Where does the integral depend analytically on theparameters?

SolutionThe integral exists only for a real, and Re b 6= 0 or cos (iab) = 0. It dependsanalytically on b for Re b 6= 0. It depends analytically on a in the intervals�1 < a < 0 and 0 < a <1.Case 1: a < 0

γ

γ

2

1­R Rz 2

z 1

VII.2.10a

Set

I =

Z 1

�1

cos ax

x2 + b2dx; a < 0; b > 0

and integrate

f (z) =eiaz

z2 + b2=

eiaz

(z � ib) (z + ib)along the contour in Figure 7.2.10a. (The lower half-plane)Residue at a simple pole at z2 = �ib, where by Rule 3,

35

Res

�eiaz

z2 + b2;�ib

�=eiaz

2z

����z=�ib

=eab

2bi

Integrate along 1 and let R!1. This gives

Z 1

f (z) dz =

=

Z R

�R

eiax

x2 + b2dx!

Z 1

�1

cos (ax)

x2 + b2dx+ i

Z 1

�1

sin (ax)

x2 + b2dx =

= I + i

Z 1

�1

sin (ax)

x2 + b2dx

Integrate along 2 and let R!1. Because jeiazj � 1 in the lower half planeif a < 0, this gives����Z

2

f (z) dz

���� � 1

R2 � b2� �R � �

R! 0:

Using the Residue Theorem and letting R!1, we obtain that

I + 0 = �2�i ��eab

2bi

�;

and hence

(1) I =�eab

b; a < 0:

Case 2: a > 0

36

γ

γ

4

3­R R

z 1

z 2

VII.2.10b

Set

I =

Z 1

�1

cos ax

x2 + b2dx; a > 0; b > 0;

and integrate

f (z) =eiaz

z2 + b2=

eiaz

(z � ib) (z + ib)along the contour in Figure VII.2.10b. (The upper half-plane)Residue at a simple pole at z1 = ib, where by Rule 3,

Res

�eiaz

z2 + b2; ib

�=eiaz

2z

����z=ib

= �e�ab

2bi

Integrate along 3 and let R!1. This gives

Z 3

f (z) dz =

=

Z R

�R

eiax

x2 + b2dx!

Z 1

�1

cos (ax)

x2 + b2dx+ i

Z 1

�1

sin (ax)

x2 + b2dx =

= I + i

Z 1

�1

sin (ax)

x2 + b2dx

37

Integrate along 4 and let R!1. Because jeiazj � 1 in the upper half planeif a > 0, this gives����Z

4

f (z) dz

���� � 1

R2 � b2� �R � �

R! 0:

Using the Residue Theorem and letting R!1, we obtain that

I + 0 = 2�i ���e

�ab

2bi

�;

and hence

(2) I =�e�ab

b; a > 0:

By (1) and (2) we can concluede thatZ 1

�1

cos (ax)

x2 + b2dx =

�e�jajb

b; �1 < a <1; b > 0:

38

VII.2.111 2 3 P L K

Evaluate Z 1

�1

cosx

(x2 + a2) (x2 + b2)dx:

Indicate the range of the parameters a and b.

SolutionThe integral converges absolutely if a2; b2 62 (�1; 0], a; b 62 iR. It alsoconverges absolutely for certain points on the imaginary axis, so that �iaare zeros of cosx, �ib are zeros of cosx, but they are not the same zeros.The integrand, is an even function of a and b, so it su¢ ces to evaluate it fora; b in the right half-plane. For this, we can assume that a; b > 0, and extendthe formula by analyticity. We can also assume that a 6= b, and get othercase in this limit. So we may as well assume that 0 < a < b.

γ

γ

2

1­R R

z

z

1

2

z

z

3

4

VII.2.11

The integralR1�1

cosx(x2+a2)(x2+b2)

dx converges absolutely if a2; b2 =2 (�1; 0] i.e.a; b =2 iR. It also converges absolutely for certain points on imaginary axis,so that �ia are zeros of cosx, �ib are zeros of cosx, but they are not samezeros. (Problem should may be simpli�ed to avoid thus points.)The integrand, is an even function of a and b, so it su¢ ces to evaluate it fora; b in the right half-plane. For this, we can assume that a; b > 0, and extendthe formula by analyticity. We can assume that a 6= b, and get other case inthis limit. So we may as well assume 0 < a < b.

39

Case 1: (a; b > 0; a 6= b)Set

I =

Z 1

�1

cosx

(x2 + a2) (x2 + b2)dx

and integrate

f (z) =eiz

(z2 + a2) (z2 + b2)=

eiz

(z � ia) (z � ib) (z + ia) (z + ib)along the contour in Figure VII.2.11.Residue at a simple pole at z1 = ai, where by Rule 3,

Res

�eiz

(z2 + a2) (z2 + b2); ia

�=

eiz

2z (z2 + b2)

����z=ai

=e�a

2a (b2 � a2) iResidue at a simple pole at z1 = ib, where by Rule 3,

Res

�eiz

(z2 + a2) (z2 + b2); ib

�=

eiz

2z (z2 + a2)

����z=bi

= � e�b

2b (b2 � a2) iIntegrate along 1 and let R!1. This gives

Z 1

f (z) dz =

Z R

�R

eix

(x2 + a2) (x2 + b2)dx!

!Z 1

�1

cosx

(x2 + a2) (x2 + b2)dx+ i

Z 1

�1

sin x

(x2 + a2) (x2 + b2)dx =

= I + i

Z 1

�1

sin x

(x2 + a2) (x2 + b2)dx

Integrate along 2 and let R!1. Because jeizj � 1 in the upper half plane,this gives ����Z

2

f (z) dz

���� � 1

(R2 � a2) (R2 � b2) � �R ��

R3! 0:

Using the Residue Theorem and letting R!1, we obtain that

I + 0 = 2�i ��

e�a

2a (b2 � a2) i�e�b

2b (b2 � a2) i

�;

40

and therefore Z 1

�1

cosx

(x2 + a2) (x2 + b2)dx =

��be�a � ae�b

�ab (b2 � a2) :

Case 2: (a; b > 0; a = b)From Case 1 we have

Z 1

�1

cosx

(x2 + a2) (x2 + b2)dx =

��be�a � ae�b

�ab (b2 � a2) =

�

ab (b+ a)

(b� a) e�a + a�e�a � e�b

�(b� a) :

If we let b! a, we getZ 1

�1

cosx

(x2 + a2)2dx =

�

2a3(1 + a) :

For a = 1, we have Z 1

�1

cosx

(x2 + 1)2dx =

�

e;

which thus yields the answer from Exercise 8.

41

VII.2.12Let Q (z) be a polynomial of degree m with no zeros on the realline, and let f (z) be a function that is analytic in the upper half-plane and across the real line. Suppose there is b < m� 1 such thatjf (z)j � jzjb for z in the upper half-plane, jzj > 1. Show thatZ 1

�1

f (x)

Q (x)dx = 2�i

XRes

�f (z)

Q (z); zj

�;

summed over the zeros zj of Q (z) in the upper half-plane.

Solution

γ

γ

2

1­R R

z z

zz

z

12

3j ­ 1

j

VII.2.12

Set

I =

Z 1

�1

f (x)

Q (x)dx

and integrate

f (z) =f (z)

Q (z)

along the contour in Figure VII.2.12.The sum of the residues at the poles z1; z2; : : : ; zj in the upper half plane,X

Res

�f (z)

Q (z); zj

�:

42

Integrate along 1, and let R!1. This givesZ 1

f (z) dz =

Z R

�R

f (x)

Q (x)dx!

Z 1

�1

f (x)

Q (x)dx = I:

Integrate along 2, and let R!1. This gives

����Z 2

f (z) dz

���� = � z = Rei�0 � � � �

��Z �

0

��f �Rei����jQ (Rei�)jd� �

��f �Rei����jQ (Rei�)j � �R �

� Rb

Rm (C +O (1=R))� �

R (C +O (1=R))� �

CR! 0:

Using the Residue Theorem and letting R!1, we obtain that

I + 0 = 2�iX

Res

�f (z)

Q (z); zj

�and therefore Z 1

�1

f (x)

Q (x)dx = 2�i

XRes

�f (z)

Q (z); zj

�Q.E.D.

43

VII.3.1Show using residue theory thatZ 2�

0

cos �

2 + cos �d� = 2�

�1� 2p

3

�:

Solution

|z| = 1

zzz 123

VII.3.1

I =

Z 2�

0

cos �

2 + cos �d�:

A change of variables as in book page 203-204 gives

I =

Z 2�

0

cos �

2 + cos �d� =

Ijzj=1

12

�z + 1

z

��2 + 1

2

�z + 1

z

�� dziz=1

i

Ijzj=1

z2 + 1

z3 + 4z2 + zdz:

Integrate

f(z) =z2 + 1

z3 + 4z2 + z=

z2 + 1

z�z �

��2�

p3�� �

z ���2 +

p3��

along the unit circle contour in Figure VII.3.1.Residue at a simple pole at z1 = 0, where by Rule 3,

Res

�z2 + 1

z3 + 4z2 + z; 0

�=

z2 + 1

3z2 + 8z + 1

����z=0

= 1:

44

Residue at a simple pole at z2 = �2 +p3, where by Rule 3,

Res

�z2 + 1

z3 + 4z2 + z;�2 +

p3

�=

z2 + 1

3z2 + 8z + 1

����z=�2+

p3

= �2p3

3:

Using the Residue Theorem, we thus obtain

I = 2�i � 1i

1�2

p3

3

!;

and therefore Z 2�

0

cos �

2 + cos �d� =2�

�1� 2p

3

�:

45

VII.3.2Show using residue theory thatZ 2�

0

d�

a+ b sin �=

2�pa2 � b2

; a > b > 0:

Solution

|z | = 1

z

z

1

2

VII.3.2 (a = 7, b = 6)

Set

I =

Z 2�

0

1

a+ b sin �d�:

A change of variables as in book page 203-204 gives

I =

Z 2�

0

1

a+ b sin �d� =

Ijzj=1

1�a+ b � 1

2i

�z � 1

z

�� dziz=

Ijzj=1

2

bz2 + 2azi� bdz

Integrate

f(z) =2

bz2 + 2azi� b =2

b�z �

��a�

pa2�b2b

i���

z ���a+

pa2�b2b

i��

along the unit circle contour in Figure VII.3.2.Residue at a simple pole at z1 = �a+

pa2�b2b

i, where by Rule 3,

46

Res

�2

bz2 + 2azi� b;�a+

pa2 � b2b

i

�=

2

2bz + 2ai

����z=�a+

pa2�b2b

i

= � ipa2 � b2

:

Using the Residue Theorem, we thus obtain

I = 2�i ��� 1p

a2 � b2i

�;

and therefore Z 2�

0

1

a+ b sin �d� =

2�pa2 � b2

; a > b > 0:

47

VII.3.3Show using residue theory thatZ �

0

sin2 �

a+ cos �d� = �

ha�

pa2 � 1

i; a > 1:

Solution

|z| = 1

zzz 123

VII.3.3 (a = 2)

Set

I =

Z �

0

sin2 �

a+ cos �d� =

�� = 2� � td� = �dt

�= �

Z �

2�

sin2 (2� � t)a+ cos(2� � t)dt =

=

Z 2�

�

sin2 (2� � t)a+ cos(2� � t)dt =

Z 2�

�

sin2 t

a+ cos tdt:

We have

2I =

Z 2�

0

sin2 �

a+ cos �d�:

A change of variables as in book page 203-204 gives

2I =

Z 2�

0

sin2 �

a+ cos �d� =

Ijzj=1

�12i

�z � 1

z

��2�a+ 1

2

�z + 1

z

�� dziz= � 1

2i

Ijzj=1

(z2 � 1)2

z2 (z2 + 2az + 1)dz:

48

Integrate

f(z) =(z2 � 1)2

z2 (z2 + 2az + 1)=

(z2 � 1)2

z2�z���a+

pa2 � 1

�� �z���a�

pa2 � 1

��along the unit circle contour in Figure VII.3.3.Residue at a double pole at z1 = 0 , where by Rule 2,

Res

"(z2 � 1)2

z2 (z2 + 2az + 1); 0

#= lim

z!0

d

dz

(z2 � 1)2

(z2 + 2az + 1)= �2a:

Residue at a simple pole at z2 = �a+pa2 � 1, where by Rule 3,

Res

"(z2 � 1)2 =z2

z2 + 2az + 1;�a+

pa2 � 1

#=(z2 � 1)2 =z2

2z + 2a

�����z=�a+

pa2�1

= 2pa2 � 1:

Using the Residue Theorem, we thus obtain

2I = 2�i ��� 12i

���2a+ 2

pa2 � 1

�;

and therefore Z �

0

sin2 �

a+ cos �d� =�

�a�

pa2 � 1

�; a > 1:

49

VII.3.4Show using residue theory thatZ �

��

d�

1 + sin2 �= �

p2:

Solution

|z | = 1

z zzz 1 234

VII.3.4

Set

I =

Z �

��

1

1 + sin2 �d� =

�� = � � td� = �dt

�=

= �Z 0

2�

1

1+ sin2(� � t)dt =

Z 2�

0

1

1+ sin2 tdt:

We have

I=

Z 2�

0

1

1 + sin2 �d�:

A change of variables as in book page 203-204 gives

I=

Z 2�

0

1

1 + sin2 �d� =

Ijzj=1

1

1 +�12i

�z � 1

z

��2 dziz = �4iIjzj=1

z

z4 � 6z2 + 1 dz:

50

Integrate

f(z) =z

z4 � 6z2 + 1 =

=z�

z ��p2� 1

�� �z �

�p2 + 1

�� �z �

��p2 + 1

�� �z �

��p2� 1

��along the unit circle contour in Figure VII.3.4Residues at a simple poles at z1;2 = �

�p2� 1

�, where by Rule 3,

Res

�z

z4 � 6z2 + 1 ;��p2� 1

��=

z

4z3 � 12z

����z=�(

p2�1)

= � 1

8p2:

Using the Residue Theorem, we thus obtain

I = 2�i � �4i

�� 1

8p2� 1

8p2

�;

and therefore Z �

��

1

1 + sin2 �d� = �

p2:

51

VII.3.5Show using residue theory thatZ �

��

1� r21� 2r cos � + r2

d�

2�= 1; 0 � r < 1:

Solution

|z | = 1

zz 12

VII.3.5 (r = 1/2)

Set

I =

Z �

��

1� r21� 2r cos � + r2

d�

2�=

�� = � � td� = �dt

�=

= �Z 0

2�

1� r21� 2r cos (� � t) + r2

dt

2�=

=

Z 2�

0

1� r21 + 2r cos t+ r2

dt

2�

A change of variables as in book page 203-204 gives

52

I =

Z 2�

0

1� r21 + 2r cos t+ r2

dt

2�=

=1

2�

Ijzj=1

1� r2

1 + 2r�12

�z + 1

z

��+ r2

dz

iz=

=1

2�i

Ijzj=1

1� r2(r + z) (rz + 1)

dz:

Integrate

f(z) =1� r2

(r + z) (rz + 1)=

1� r2r (z + r) (z + 1=r)

along the unit circle contour in Figure VII.3.4Residue at a simple pole at z1 = �r, where by Rule 3,

Res

�1� r2

r (z + r) (z + 1=r);�r

�=

1� r2�2rz + r2 + 1

������z=�r

= 1:

Using the Residue Theorem, we thus obtain

I = 2�i � 12�i

(1) ;

and therefore Z �

��

1� r21� 2r cos � + r2

d�

2�= 1; 0 � r < 1:

53

VII.3.6By expanding both sides of the identity

1

2�

Z 2�

0

d�

w + cos �=

1pw2 � 1

in a power series at 1, show that

1

2�

Z 2�

0

cos2k �d� =(2k)!

22k (k!)2; k � 0:

SolutionRewrite both sides in the given identity

(1) I =1

2�

Z 2�

0

d�

1 + cos �w

=

�1� 1

w2

��1=2;

where the integrand in the left side in the identity can be rewritten as

1

1 + cos �w

=1Xk=0

(�1)k cosk �

wk:

Now, we work on the left side of the identity

I =1

2�

Z 2�

0

X1

k=0(�1)k cos

k �

wkd� =

1

2�

X1

k=0

(�1)k

wk

Z 2�

0

cosk �d�:

BecauseR 2�0cosk �d� = 0 if k odd, we have

(2) I = 1 +1Xk=1

1

2�

�Z 2�

0

cos2k �d�

�1

w2k:

Now, we use the power series expansion (MH p. 192.) for real � and �1 <x < 1,

(1 + x)� = 1+�x+� (�� 1)

2!x2+

� (�� 1) (�� 2)3!

x3+� � �+��n

�xn+� � � ;

we have

54

�1� 1

w2

��1=2=

= 1 +

��12

�1!

��1w2

�1+

��12

� ��32

�2!

��1w2

�2+

��12

� ��32

� ��52

�3!

��1w2

�3+

+

��12

� ��32

� ��52

� ��72

�4!

��1w2

�4+� � �+

��12

� ��32

� ��52

� ��72

�� � ��1�2k2

�k!

��1w2k

�k+� � � =

= 1 +X1

k=1

1 � 3 � � � � � (2k � 1)2kk!

1

w2k=

= 1 +X1

k=1

1 � 2 � 3 � 4 � 5 � : : : � (2k � 1) � 2k2kk! � 2 � 4 � 6 � : : : � 2k

1

w2k=

1 +X1

k=1

1 � 2 � 3 � 4 � 5 � : : : � (2k � 1) � 2k2kk! � 2k (1 � 2 � 3 � : : : � k)

1

w2k=

= 1 +X1

k=1

(2k)!

2kk!2kk!

1

w2k= 1 +

X1

k=1

(2k)!

22k (k!)21

w2k;

i.e.,

(2)

�1� 1

w2

��1=2= 1 +

X1

k=1

(2k)!

22k (k!)21

w2k:

Comparing the two power series expressions (2) and (3) of the both sides inthe identity (1) gives us together with the uniqueness theorem for the powerseries

1

2�

Z 2�

0

cos2k �d� =(2k)!

22k (k!)2; k � 0:

55

VII.3.7Show using residue theory thatZ 2�

0

d�

(w + cos �)2=

2�w

(w2 � 1)3=2; w 2 C n [�1; 1] :

Specify carefully the branch of the power function. Check youanswer by di¤erentiating the integral of 1= (w + cos �) with respectto the parameter w.

Solution

|z| = 1

zz 12

VII.3.7 (w = 2)

Set

I =

Z 2�

0

d�

(w + cos �)2

A change of variables as in book page 203-204 gives

I =

Z 2�

0

d�

(w + cos �)2=

Z 2�

0

1�w + 1

2

�z + 1

z

��2 dziz =4iIjzj=1

z

(z2 + 2wz + 1)2dz:

Integrate f(z) = 4z(z2+2wz+1)2

along the unit circle contour in Figure VII.3.7.

We choose branch ofpw2 � 1 on Cn [�1; 1] that is positive for w 2 (1;1).

It su¢ ces to check the identity for w > 1.Residue at a double pole at z1 = �w +

pw2 � 1, where by Rule 2,

56

Res

�z

(z2 + 2wz + 1)2;�w +

pw2 � 1

�=

= limz!�w+

pw2�1

d

dz

z�z �

��w �

pw2 � 1

��2 ==

�z �

��w �

pw2 � 1

��2 � 2z �z � ��w �pw2 � 1���z �

��w �

pw2 � 1

��4�����z=�w+

pw2�1

=

=

�z �

��w �

pw2 � 1

��� 2z�

z ���w �

pw2 � 1

��3�����z=�w+

pw2�1

=

=1

4

w

(w2 � 1)3=2:

Using the Residue Theorem, we thus obtain

I = 2�i � 4i

1

4

w

(w2 � 1)3=2

!;

and thereforeZ 2�

0

d�

(w + cos �)2=

2�w

(w2 � 1)3=2; w 2 C n [�1; 1] :

Check.From Exercise VII.3.6 we have the formula

(1)1

2�

Z 2�

0

d�

w + cos �=

1pw2 � 1

It is allowed to di¤erentiate both sides in the formula, because for everycompact interval with w 2 C n [�1; 1], both the integrand and its derivativeare continuous, and that the primitive of the derivative is uniformly boundedby an w� independent funtion h (�) such that

Rh (�) d� <1 on every such

like interval on w.We start by di¤erentiating the left hand side in the formula with respect tothe parameter w

57

(2)d

dw

�1

2�

Z 2�

0

d�

w + cos �

�= � 1

2�

Z 2�

0

d�

(w + cos �)2;

and di¤erentiate the right hand side in the formula in the same way

(3)d

dw

1pw2 � 1

= � w

(w2 � 1)3=2:

By (1) - (3), we thus have thatZ 2�

0

d�

(w + cos �)2=

2�w

(w2 � 1)3=2:

58

VII.4.1By integrating around the keyhole contour, show thatZ 1

0

x�a

1 + xdx =

�

sin (�a); 0 < a < 1:

Solution

­R Rz 1

γ

γ

γγ

1

2

34

VII.4.1

Set

I =

Z 1

0

x�a

1 + xdx

and integrate

f (z) =z�a

1 + z=jzj�a e�ia arg z

1 + z

along the keyhole contour in Figure VII.4.1. We make a branchcut for z�a

along the positive real axis, where 0 < arg z < 2�.Residue at a simple pole at z1 = �1, where by Rule 3,

Res

�jzj�a e�ia arg z

1 + z;�1

�= jzj�a e�ia arg z

��z=�1 = e

��ia:

59

Integrate along 1, and let R!1 and "! 0+. This gives

Z 1

f (z) dz =Z 1

jzj�a e�ia arg z1 + z

dz =

�z = xe0i

dz = dx

�=

Z R

"

x�a

1 + xdx!

!Z 1

0

x�a

1 + xdx = I:

Integrate along 2 and let R!1. This gives then 0 < a < 1����Z 2

f (z) dz

���� � R�a

R� 1 � 2�R �2�

Ra! 0:

Integrate along 3, and let R!1 and "! 0+. This gives

Z 3

f (z) dz =

=

Z 3

jzj�a e�ia arg z1 + z

dz =

�z = xe2�i

dz = dx

�=

Z "

R

x�ae�2�ia

1 + xdx!

!Z 0

1

x�ae�2�ia

1 + xdx = �e�2�ia

Z 1

0

x�a

1 + xdx = �e�2�iaI:

Integrate along 4, and let "! 0+. This gives then 0 < a < 1����Z 4

f (z) dz

���� � "�a

1� " � 2�" � 2�"1�a ! 0:

Using the Residue Theorem and letting R!1 and "! 0+, we obtain that

I + 0� e�2�iaI + 0 = 2�i ��e��ia

�:

Multiplying with e�ia, this yields�e�ia � e��ia

�I = 2�i;

and hence solving for I,

60

I =2�i

e�ia � e��ia =�

sin (�a);

i.e., Z 1

0

x�a

1 + xdx =

�

sin (�a); 0 < a < 1:

61

VII.4.2By integrating around the boundary of a pie-slice domain of aper-ture 2�=b, show thatZ 1

0

dx

1 + xb=

�

b sin (�=b); b > 1:

Remark. Check the result by changing variable and comparing with Exercise1.

Solution

Rγ1

γ2γ3

z 1

ε

VII.4.2 (b = 12)

Set

I =

Z 1

0

dx

1 + xb;

and integrate

f (z) =1

1 + zb=

1

1 + jzjb eib arg z

along the pie-slice of aperture 2�=b contour in Figure VII.4.2. We make abranch cut for zb along the negative imaginary axis, where ��=2 < arg z <3�=2.Residue at a simple pole at z1 = e�i=b, where by Rule 3,

Res

�1

1 + zb; e�i=b

�=

1

bzb�1

����z=e�i=b

=�1be�i=b:

62

Integrate along 1, and let R!1 and "! 0+. This gives

Z 1

f (z) dz =

=

Z 1

1

1 + jzjb eib arg zdz =

�z = xe0i

dz = dx

�=

Z R

"

1

1 + xbdx!

!Z 1

0

1

1 + xbdx = I:

Integrate along 2 and let R!1. This gives then b > 1����Z 2

f (z) dz

���� � 1

Rb � 1 �2�R

b� 2�

Rb�1! 0:

Integrate along 3, and let R!1 and "! 0+. This gives

Z 3

f (z) dz =

=

Z 3

1

1 + jzjb eib arg zdz =

�z = xe2�i=b

dz = e2�i=bdx

�=

Z "

R

1

1 + xbe2�i=bdx!

!Z 0

1

1

1 + xbe2�i=bdx = �e2�i=b

Z 1

0

1

1 + xbdx = �e2�i=bI:

Integrate along 4, and let "! 0+. This gives then b > 1����Z 4

f (z) dz

���� � 1

1� "b� 2�"b� 2�"

b! 0:

Using the Residue Theorem and letting R!1 and "! 0+, we obtain that

I + 0� e2�i=bI + 0 = 2�i ���1be�i=b

�:

Multiplying with e��i=b, this yields�e��i=b � e�i=b

�I =

�2�ib;

and hence solving for I,

63

I =�2�i

(e��i=b � e�i=b) =�

b sin (�=b)

i.e., Z 1

0

dx

1 + xb=

�

b sin (�=b); b > 1:

Remark.We begin with the �rst integral I, we make the substitution xb = t and usethe integral from Exercise VII.4.1 where we choose a = 1 � 1=b. (Becauseb > 1, then 0 < 1�1=b < 1 and the restriction that 0 < a < 1 for the integralin Exercise VII.4.1 is satis�ed.)

I =

Z 1

0

dx

1 + xbdx =

24 xb = tx = t1=b

dx = 1bt1=b�1dt

35 ==

Z 1

0

1

b

t1=b�1

1 + tdt =

1

b

Z 1

0

t�(1�1=b)

1 + tdt =

=1

b

�

sin (� (1=b� 1)) =�

b sin (�=b):

64

VII.4.3By integrating around the keyhole contour, show thatZ 1

0

log x

xa (x+ 1)dx =

�2 cos (�a)

sin2 (�a); 0 < a < 1:

Remark. Check the result by di¤erentiating the identity in Exercise 1.

Solution

­R Rz 1

γ

γ

γγ

1

2

34

VII.4.3

Set

I =

Z 1

0

log x

xa (x+ 1);

and integrate

f (z) =log x

xa (x+ 1)=

log jzj+ i arg zjzja eia arg z (z + 1)

along the keyhole contour in Figure VII.4.3, where the cimicircle 4 haveradius ". We make a branch cut for log z along the positive real axis, so that0 < arg z < 2�.Residue at a simple pole at z1 = �1, where by Rule 3,

65

Res

�log jzj+ i arg zjzja eia arg z (z + 1) ;�1

�=log jzj+ i arg zjzja eia arg z

����z=�1

= �ie��ia:

Integrate along 1, and let R!1 and "! 0+. This gives

Z 1

f (z) dz =

=

Z 1

log jzj+ i arg zjzja eia arg z (z + 1)dz =

�z = xe0i

dz = dx

�=

Z R

"

log x

xa (x+ 1)dx!

!Z 1

0

log x

xa (x+ 1)dx = I:

Integrate along 2 and let R!1. This gives then 0 < a < 1

����Z 2

f (z) dz

���� �qlog2R + (2�)2

Ra (R� 1) � 2�R � 2� logR

R�! 0:

Integrate along 3, and let R!1 and "! 0+. This gives

Z 3

f (z) dz =

=

Z 3

log jzj+ i arg zjzja eia arg z (z + 1)dz =

�z = xe2�i

dz = dx

�=

Z "

R

log x+ 2�i

xae2�ia (x+ 1)!

!Z 0

1

log x+ 2�i

xae2�ia (x+ 1)dx =

= �e�2�iaZ 1

0

log x

xa (x+ 1)dx� 2�ie�2�ia

Z 1

0

dx

xa (x+ 1)=

= �e�2�iaI � 2�ie�2�iaJ

Integrate along 4, and let "! 0+. This gives then 0 < a < 1����Z 4

f (z) dz

���� �plog2 "+ 2�

"a (1� ") � 2�" � 2�"1�a jlog "j ! 0:

Using the Residue Theorem and letting R!1 and "! 0+, we obtain that

66

I + 0� e�2�iaI � 2�ie�2�iaJ + 0 = 2�i ���ie��ia

�:

Multiplying with e�ia, this yields�e�ia � e��ia

�I � 2�ie��iaJ = �2�2;

and thus

2i sin (�a) I � 2�ie��iaJ = �2�2:Separating real and imaginary parts, we get simultaneous equations�

�2� sin (�a) J = �2�22 sin (�a) I � 2� cos (�a) J = 0;

and hence the solutions

I =

Z 1

0

log x

xa (x+ 1)dx =

�2 cos (�a)

sin2 (�a); J =

Z 1

0

dx

xa (x+ 1)=

�

sin (�a);

i.e. Z 1

0

log x

xa (x+ 1)dx =

�2 cos (�a)

sin2 (�a); 0 < a < 1:

Remark.From the Exercise VII.4.1 we have the formula

(1)Z 1

0

x�a

1 + xdx =

�

sin (�a); 0 < a < 1:

It is allowed to di¤erentiate both sides in the formula, because for everycompact interval with 0 < a < 1, both the integrand and its derivative iscontinuous. And that the primitive of the derivative is uniformly limited bya constant on every compact interval on a, there 0 < a < 1.We begin to di¤erentiate the left side in the formula with respect to theparameter a

(2)d

da

�Z 1

0

x�a

1 + xdx

�=

Z 1

�1

� log x � x�a1 + x

dx = �Z 1

�1

log x

xa (x+ 1)dx;

67

and di¤erentiate the right hand side in the formula in the same way

(3)d

da

�

sin (�a)= � cos (�a)

sin2 (�a):

By (1) - (3), we have thatZ 1

0

log x

xa (x+ 1)dx =

�2 cos (�a)

sin2 (�a):

68

VII.4.4For �xed m � 2, show that by integrating around the keyhole con-tour thatZ 1

0

x�a

(1 + x)mdx =

�a (a+ 1) � � � (a+m� 2)(m� 1)! sin (�a) ; 1�m < a < 1:

Remark. The result can be obtained also by integrating the formula in Ex-ercise 1 by parts.

Solution

­R Rz 1

γ

γ

γγ

1

2

34

VII.4.4

Set

I =

Z 1

0

x�a

(1 + x)mdx

and integrate

f (z) =z�a

(1 + z)m=jzj�a e�ia arg z(1 + z)m

along the keyhole contour in Figure VII.4.4. We make a branchcut for z�a

along the positive real axis, where 0 < arg z < 2�.

69

From Sa¤/Snider page 310 we have

Res [f(z); z0] = limz!z0

1

(m� 1)!dm�1

dzm�1[(z � z0)m f (z)] :

Residue at a pole of order m at z1 = �1

Res

�z�a

(1 + z)m;�1

�= lim

z!�1

1

(m� 1)!dm�1

dzm�1�z�a�=

= limz!�1

1

(m� 1)! (�a) (�a� 1) � � � (�a�m+ 2) z(�a�(m�1)) =

= limz!�1

1

(m� 1)! (�1)m�1 a (a+ 1) � � � (a+m� 2) jzj(�a�(m�1)) ei(�a�(m�1)) arg z =

=1

(m� 1)! (�1)m�1 a (a+ 1) � � � (a+m� 2) j�1j(�a�(m�1)) ei(�a�(m�1))� =

=1

(m� 1)! (�1)m�1 a (a+ 1) � � � (a+m� 2) (�1)m�1 e��ia =

=a (a+ 1) � � � (a+m� 2)

(m� 1)! e��ia:

Integrate along 1, and let R!1 and "! 0+. This gives

Z 1

f (z) dz =

Z 1

jzj�a e�ia arg z(1 + z)m

dz =

�z = xe0i

dz = dx

�=

Z R

"

x�a

(1 + x)mdx!

!Z 1

0

x�a

(1 + x)mdx = I:

Integrate along 2 and let R ! 1. This gives then 1 �m < a < 1, wherem � 2 ����Z

2

f (z) dz

���� � R�a

(R� 1)m � 2�R �2�

Rm+a�1! 0:

Integrate along 3, and let R!1 and "! 0+. This gives

70

Z 3

f (z) dz =Z 3

jzj�a e�ia arg z(1 + z)m

dz =

�z = xe2�i

dz = dx

�=

Z "

R

x�ae�2�ia

(1 + x)mdx!

!Z 0

1

x�ae�2�ia

(1 + x)mdx = �e�2�ia

Z 1

0

x�a

(1 + x)mdx = �e�2�iaI:

Integrate along 4, and let " ! 0+. This gives then 1 �m < a < 1, wherem � 2 ����Z

4

f (z) dz

���� � "�a

(1� ")m � 2�" � 2�"1�a ! 0:

Using the Residue Theorem and letting R!1 and "! 0+, we obtain that

I + 0� e�2�iaI + 0 = 2�i ��a (a+ 1) � � � (a+m� 2)

(m� 1)! e��ia�:

Multiplying with e�ia, this yields

�e�ia � e��ia

�I = 2�i

a (a+ 1) � � � (a+m� 2)(m� 1)! ;

and hence solving for I,

I =2i

(e�ia � e��ia)�a (a+ 1) � � � (a+m� 2)

(m� 1)! =�a (a+ 1) � � � (a+m� 2)

(m� 1)! sin (�a) ;

i.e. Z 1

0

x�a

(1 + x)mdx =

�a (a+ 1) � � � (a+m� 2)(m� 1)! sin (�a) ; 1�m < a < 1:

Remark.Set (this is our induction hypothesis)

Im (a) =

Z 1

0

x�a

(1 + x)mdx =

�a (a+ 1) � � � (a+m� 2)(m� 1)! sin (�a)

71

From Exercise VII.4.1 we haveZ 1

0

x�b

1 + xdx =

�

sin (�b); 0 < b < 1:

Use Exercise VII.4.1 and integrate by part

�

sin (�b)=

Z 1

0

x�b

1 + xdx =

Z 1

0

x�b1

1 + xdx =

=

�x�b+1

1� b1

1 + x

�10

�Z 1

0

x�b+1

1� b1

(1 + x)2dx =

=1

1� b

Z 1

0

x�b+1

(1 + x)2dx:

Set b = a+1 and solve for the integral in the left side in the inequality aboveZ 1

0

x�a

(1 + x)2dx = �a �

sin (� (a+ 1))=

�a

sin (�a);

that is the same as I2 (a), thus our induction hypothesis is valid for m = 2.Now suppose that our formula is valid for m = p, and integrate by parts,

�b (b+ 1) � � � (b+ p� 2)(p� 1)! sin (�b) =

=

Z 1

0

x�b

(1 + x)pdx =

Z 1

0

x�b1

(1 + x)pdx =

=

�x�b+1

1� b1

(1 + x)p

�10

�Z 1

0

x�b+1

1� b�p

(1 + x)p+1dx =

=p

1� b

Z 1

0

x�b+1

(1 + x)p+1dx:

Set b = a+1 and solve for the integral in the left side in the inequality above

72

Z 1

0

x�a

(1 + x)p+1dx =

=�ap

� (a+ 1) (a+ 2) � � � ((a+ 1) + p� 2)(p� 1)! sin (� (a+ 1)) =

=�a (a+ 1) (a+ 2) � � � (a+ (p+ 1)� 2)

p! sin (�a);

that is the same as Ip+1 (a), thus our induction hypothesis is valid for m =p+ 1.We have showed that the formula Im (a) is valid for all m > 2, that was tobe shown.

73

VII.4.5By integrating a branch of (log z)2

(z+a)(z+b)around the keyhole contour,

show thatZ 1

0

log x

(x+ a) (x+ b)dx =

(log a)2 � (log b)2

2 (a� b) ; a; b > 0; a 6= b:

Solution

­R Rz 1

γ

γ

γγ

1

2

34

z2

VII.4.5 (a = 1, b = 2)

Set

I =

Z 1

0

log x

(x+ a) (x+ b)dx

and integrate

f (z) =(log z)2

(z + a) (z + b)=(log jzj+ i arg z)2

(z + a) (z + b)

around the keyhole contour in Figure VII.4.5. We make a branchcut for log zalong the positive real axis, where 0 < arg z < 2�.Residue at a simple pole at z1 = �a, where by Rule 3,

74

Res

"(log jzj+ i arg z)2

(z + a) (z + b);�a

#=(log jzj+ i arg z)2

(z + b)

�����z=�a

=(log a+ i�)2

b� a :

Residue at a simple pole at z1 = �b, where by Rule 3,

Res

"(log jzj+ i arg z)2

(z + a) (z + b);�b

#=(log jzj+ i arg z)2

(z + a)

�����z=�b

=(log b+ i�)2

a� b :

Integrate along 1, and let R!1 and "! 0+. This gives

Z 1

f (z) dz =

=

Z 1

(log jzj+ i arg z)2

(z + a) (z + b)dz =

�z = xe0i

dz = dx

�=

Z R

"

(log x)2

(x+ a) (x+ b)dx!

!Z 1

0

(log x)2

(x+ a) (x+ b)dx = J:

Integrate along 2 and let R!1. This gives����Z 2

f (z) dz

���� � log2R + (2�)2

(R� a) (R� b) � 2�R �4� log2R

R! 0:

Integrate along 3, and let R!1 and "! 0+. This gives

Z 3

f (z) dz =

=

Z 3

(log jzj+ i arg z)2

(z + a) (z + b)dz =

�z = xe2�i

dz = dx

�=

Z "

R

(log x+ 2�i)2

(x+ a) (x+ b)dx!

!Z 0

1

(log x+ 2�i)2

(x+ a) (x+ b)dx = �

Z 1

0

(log x+ 2�i)2

(x+ a) (x+ b)dx =

= �Z 1

0

(log x)2

(x+ a) (x+ b)dx� 4�i

Z 1

0

log x

(x+ a) (x+ b)+

Z 1

0

4�2

(x+ a) (x+ b)dx =

= �J�4�iI+K

75

Integrate along 4, and let "! 0+. This gives����Z 4

f (z) dz

���� � log2 "+ (2�)2

(a� ") (b� ") � 2�" �2�" log2 "

ab! 0:

Using the Residue Theorem and letting R!1 and "! 0+, we obtain that

J + 0� J�4�iI+K + 0 = 2�i � (log jbj+ i�)2

b� a +(log jaj+ i�)2

a� b

!;

and hence, setting imaginary parts equal,

I =(log a)2 � (log b)2

2 (a� b) ;

i.e. Z 1

0

log x

(x+ a) (x+ b)dx =

(log a)2 � (log b)2

2 (a� b) ; a; b > 0; a 6= b:

76

VII.4.6Using residue theory, show thatZ 1

0

xa log x

(1 + x)2dx =

� sin (�a)� a�2 cos (�a)sin2 (�a)

; �1 < a < 1:

Solution

­R Rz 1

γ

γ

γγ

1

2

34

VII.4.6

Set

I =

Z 1

0

xa log x

(1 + x)2dx

and integrate

f (z) =za log z

(1 + z)2=jzja eia arg z (log jzj+ i arg z)

(1 + z)2

along the keyhole contour in Figure VII.4.6. We make a branchcut for za

and log z along the real axis, where 0 < arg z < 2�.Residue at a double pole at z1 = �1, where by Rule 3,

77

Res

�jzja eia arg z (log jzj+ i arg z)

(1 + z)2;�1

�=

=d

dz(za log z)

����z=�1

=aza log z

z+za

z

����z=�1

=

=za

z(a log z + 1)

����z=�1

=

=jzja eia arg z

z(a (log jzj+ i arg z) + 1)

����z=�1

=

= e�ia (�1� �ai) :

Integrate along 1, and let R!1 and "! 0+. This gives

Z 1

f (z) dz =

=

Z 1

jzja eia arg z (log jzj+ i arg z)(1 + z)2

dz =

�z = xe0i

dz = dx

�=

Z R

"

xa log x

(1 + x)2dx!

!Z 1

0

xa log x

(1 + x)2dx = I:

Integrate along 2 and let R!1. This gives

����Z 2

f (z) dz

���� � Raqlog2R + (2�)2

(R� 1)2� 2�R � 2� logR

R1�a! 0:

Integrate along 3, and let R!1 and "! 0+. This gives

78

Z 3

f (z) dz =

=

Z 3

jzja eia arg z (log jzj+ i arg z)(1 + z)2

dz =

�z = xe2�i

dz = dx

�=

Z "

R

xae2�ia (log jxj+ 2�i)(1 + x)2

dx!

!Z 0

1

xae2�ia (log jxj+ 2�i)(1 + x)2

dx = �e2�iaZ 1

0

xa (log jxj+ 2�i)(1 + x)2

dx =

= �e2�iaZ 1

0

xa log x

(1 + x)2dx� 2�ie2�ia

Z 1

0

xadx

(1 + x)2=

= �e2�iaI � 2�ie2�iaJ

Integrate along 4, and let "! 0+. This gives

����Z 4

f (z) dz

���� � "aqlog2 "+ (2�)2

(1� ")2� 2�" � 2�"1+a jlog "j ! 0:

Using the Residue Theorem and letting R!1 and "! 0+, we obtain that

I + 0� e2�iaI � 2�ie2�iaJ + 0 = 2�i � e�ia (�1� �ai) ;multiplying with e��ia, this yields�

e��ia � e�ia�I � 2�ie�iaJ = 2�i (�1� �ai) ;

and thus

�2i sin (�a) I � 2�ie�iaJ = 2�2a� 2�i:Separating real and imaginary parts, we get simultaneous equations�

2� sin (�a) J = 2�2a�2 sin (�a) I � 2� cos (�a) J = �2�;

and hence the solutions

I =

Z 1

0

xa log x

(1 + x)2=� sin (�a)� a�2 cos (�a)

sin2 (�a); J =

Z 1

0

xadx

(1 + x)2=

�a

sin (�a);

79

i.e. Z 1

0

xa log x

(1 + x)2dx =

� sin (�a)� a�2 cos (�a)sin2 (�a)

; �1 < a < 1:

80

VII.4.7Show that Z 1

0

xa�1

1 + xbdx =

�

b sin (�a=b); 0 < a < b:

Determine for which complex values of the parameter a the integral exists(in the sense that the integral of the absolute value is �nite), and evaluateit. Where does the integral depend analytically on the parameter a?

Solution

Rγ1

γ2γ3

z 1

ε

VII.4.7 (b = 12)

Set

I =

Z 1

0

xa�1

1 + xb;

and integrate

f (z) =za�1

1 + zb=jzja�1 ei(a�1) arg z

1 + jzjb eib arg z

along the pie-slice, of aperture 2�=b, contour in Figure VII.4.7. We makea branch cut for za�1 along the negative imaginary axis , where ��=2 <arg z < 3�=2.Residue at a simple pole at z1 = e�i=b, where by Rule 3,

81

Res

�za�1

1 + zb; e�i=b

�=za�1

bzb�1

����z=e�i=b

=�1be�ia=b:

Integrate along 1, and let R!1 and "! 0+. This gives

Z 1

f (z) dz =

=

Z 1

jzja�1 ei(a�1) arg z

1 + jzjb eib arg zdz =

�z = xe0i

dz = dx

�=

Z R

"

xa�1

1 + xbdx!

!Z 1

0

xa�1

1 + xbdx = I:

Integrate along 2 and let R!1. This gives then 0 < a < b����Z 2

f (z) dz

���� � Ra�1

Rb � 1 �2�R

b� 2�

bRb�a! 0:

Integrate along 3, and let R!1 and "! 0+. This gives

Z 3

f (z) dz =

=

Z 3

jzja�1 ei(a�1) arg z

1 + jzjb eib arg z=

�z = xe2�i=b

dz = e2�i=bdx

�=

Z "

R

xa�1e2�i(a�1)=b

1 + xbe2�i=bdx!

!Z 0

1

xa�1e2�i(a�1)=b

1 + xbe2�i=bdx = �e2�ia=b

Z 1

0

xa�1

1 + xbdx = �e2�ia=bI:

Integrate along 4, and let "! 0+. This gives then 0 < a < b����Z 4

f (z) dz

���� � "a�1

1� "b �2�"

b� 2�"a

b! 0:

Using the Residue Theorem and letting R!1 and "! 0+, we obtain that

I + 0� e2�ia=bI + 0 = 2�i ���1be�ia=b

�;

and multiplying with e��ia=b, this yields

82

�e��ia=b � e�ia=b

�I = �2�i

b

and hence solving for I,

I =2�i

b (e�ia=b � e��ia=b) =�

b sin (�a=b);

i.e., Z 1

0

xa�1

1 + xbdx =

�

b sin (�a=b); 0 < a < b:

Now we suppose a and b complex and rewrite the nominator in the integrand

xa�1 = e(a�1) log x = e(Re a�1) log xei Im a log x = xRe a�1ei Im a log x:

The absolute value of the integrand is���� xa�11 + xb

���� = ����xRe a�1ei Im a log x1 + xb

���� = xRe a�1

1 + xb;

thus Re a � 1 > �1 and Re a � 1 � b < �1. Thus the integrand dependsanalytically on the parameter a when 0 < Re a < b.

83

VII.4.8By integrating a branch of (log z) = (z3 + 1) around the boundary ofan indented sector of aperture 2�=3, show thatZ 1

0

log x

x3 + 1dx = �2�

2

27;

Z 1

0

1

x3 + 1dx =

2�

3p3:

Remark. Compare the results with those of Exercise 3 (after changing vari-ables) and Exercise 2.

Solution

ε R

γ2

γ1

γ3

z 2

z

z

1

3

γ4

VII.4.8

Set

I =

Z 1

0

log x

x3 + 1dx; J =

Z 1

0

1

x3 + 1dx

and integrate

f (z) =log z

z3 + 1=

log jzj+ i arg z�z � e�i=3

�(z � e�i)

�z � e5�i=3

�along the indented sector, of aperture 2�=3, contour in Figure VII.4.8. Wemake a branchcut for log z along the negative imaginary axis, where ��=2 <arg z < 3�=2.Residue at a simple pole at z1 = e�i=3, where by Rule 3,

84

Res

�log jzj+ i arg z

z3 + 1; e�i=3

�=log jzj+ i arg z

3z2

����z=e�i=3

=�i

9e�2�i=3:

Integrate along 1, and let R!1 and "! 0+. This gives

Z 1

f (z) dz =

=

Z 1

log jzj+ i arg zz3 + 1

dz =

�z = xe0i

dz = dx

�=

Z R

"

log x

x3 + 1dx!

!Z 1

0

log x

x3 + 1dx = I:

Integrate along 2, and let R!1. This gives

����Z 2

f (z) dz

���� �qlog2R +

�2�3

�2R3 � 1 � 2�R

3� 2� logR

3R2! 0:

Integrate along 3, and let R!1 and "! 0+. This gives

Z 3

f (z) dz =

=

Z 3

log jzj+ i arg zz3 + 1

dz =

�z = xe2�i=3

dz = e2�i=3dx

�=

Z "

R

log x+ 2�i=3

x3 + 1e2�i=3dx!

!Z 0

1

log x+ 2�i=3

x3 + 1e2�i=3dx = �

Z 1

0

log x+ 2�i=3

x3 + 1e2�i=3dx =

= �e2�i=3Z 1

0

log x

x3 + 1dx�2�i

3e2�i=3

Z 1

0

1

x3 + 1dx =� e2�i=3I�2�i

3e2�i=3J:

Integrate along 4, and let "! 0+. This gives

����Z 4

f (z) dz

���� �qlog2 "+

�2�3

�21� "3 � 2�"

3� 2�" jlog "j

3! 0:

Using the Residue Theorem and letting R!1 and "! 0+, we obtain that

85

I + 0�e2�i=3I�2�i3e2�i=3J + 0 = 2�i

��i

9e�2�i=3

�:

Multiplying with e��i=3, this yields

�2i sin (�=3) I � 2�3ie�i=3J =

2�2

9;

and thus

�p3Ii+

�p3

3J � �i

3J =

2�2

9:

Separating real and imaginary parts, we get the simultaneous equations��p3

3J = 2�2

9

�p3I � �

3J = 0;

and hence the solutions

I =

Z 1

0

log x

x3 + 1dx = �2�

2

27; J =

Z 1

0

1

x3 + 1dx =

2�

3p3:

Remark.We begin with the �rst integral I, we make the substitution x3 = t and usethe integral from Exercise VII.4.3, where we choose a = 2=3.

I =

Z 1

0

log x

x3 + 1dx =

24 x3 = tx = t1=3

dx = 13t�2=3dt

35 ==1

9

Z 1

0

log t

t2=3 (t+ 1)dt =

1

9

�2 cos�2�3

�sin2

�2�3

� = �2�2

27:

For the second integral J , we use the integral from Exercise VII.4.2, wherewe choose b = 3.

J =

Z 1

0

dx

1 + xb=

�

3 sin (�=3)=

2�

3p3:

86

VII.4.9By integrating around an appropriate contour and using the resultsof Exercise 8, show thatZ 1

0

(log x)2

x3 + 1dx =

10�3

81p3:

Solution

ε R

γ2

γ1

γ3

z 2

z

z

1

3

γ4

VII.4.9

Set

I =

Z 1

0

(log x)2

x3 + 1dx

Integrate

f (z) =(log z)2

z3 + 1=

(logjzj+i arg z)2�z � e�i=3

�(z � e�i)

�z � e5�i=3

�along the pie-slice contour with 0 < arg z < 2�=3 in Figure VII.4.9. Wemake a branchcut for log z along the negative imaginary axis, where ��=2 <arg z < 3�=2.Residue at a simple pole at z1 = e�i=3, where by Rule 3,

Res

"(log jzj+ i arg z)2

z3 + 1; e�i=3

#=(log jzj+ i arg z)2

3z2

�����z=e�i=3

= ��2

27e�2�i=3:

87

Integrate along 1, and let R!1 and "! 0+. This gives

Z 1

f (z) dz =Z 1

(log jzj+ i arg z)2

z3 + 1dz =

�z = xe0i

dz = dx

�=

Z R

"

(log x)2

x3 + 1dx!

!Z 1

0

(log x)2

x3 + 1dx = I:

Integrate along 2, and let R!1. This gives����Z 2

f (z) dz

���� � log2R +�2�3

�2R3 � 1 � 2�R

3� 2� log2R

3R2! 0:

Integrate along 3, and let R!1 and "! 0+. This gives

Z 3

f (z) dz =

=

Z 3

(log jzj+ i arg z)2

z3 + 1dz =

�z = xe2�i=3

dz = e2�i=3dx

�=

Z "

R

(log x+ 2�i=3)2

x3 + 1e2�i=3dx!

!Z 0

1

(log x+ 2�i=3)2

x3 + 1e2�i=3dx = �

Z 1

0

(log x+ 2�i=3)2

x3 + 1e2�i=3dx =

= �e2�i=3Z 1

0

(log x)2

x3 + 1dx�4�i

3e2�i=3

Z 1

0

log x

x3 + 1dx+

4�2

9e2�i=3

Z 1

0

1

x3 + 1dx =

= �e2�i=3I�4�i3e2�i=3

Z 1

0

log x

x3 + 1dx+

4�2

9e2�i=3

Z 1

0

1

x3 + 1:

Integrate along 4, and let "! 0+. This gives����Z 4

f (z) dz

���� � log2"+�2�3

�21� "3 � 2�"

3� 2�" log2 "

3! 0:

Using the Residue Theorem and letting R!1 and "! 0+, we obtain that

I+0�e2�i=3I�4�i3e2�i=3

Z 1

0

log x

x3 + 1dx+

4�2

9e2�i=3

Z 1

0

1

x3 + 1+0 = 2�i

���

2

27e�2�i=3

�:

88

multiplying with e��i=3, this yields

�e��i=3 � e�i=3

�I�4�i

3

Z 1

0

log x

x3 + 1e�i=3dx+

4�2

9

Z 1

0

1

x3 + 1e�i=3dx =

2�3

27i;

and thus, substituting the values for the integrals from Exercise 9,

�2i sin (�=3) I�4�i3e�i=3

��2�

2

27

�+4�2

9e�i=3

�2�

3p3

�=2�3

27i:

And hence, setting imaginary parts equal,

�p3I+

4�3

81+12�3

81=2�3

27;

and therefore Z 1

0

(log x)2

x3 + 1dx =

10�2

81p3:

89

VII.4.10By integrating a branch of (log z) = (z3 � 1) around the boundary ofan indented half-disk and using the result of Exercise 8, show thatZ 1

0

log x

x3 � 1dx =4�2

27:

Solution

­R R

γ4

γ1 γ3γ5

γ6 γ2

z 1

z 2

z 3

VII.4.10

Set

I =

Z 1

0

log x

x3 � 1dx

Integrate

f (z) =log z

z3 � 1 =log jzj+ i arg z

(z � 1)�z � e2�i=3

� �z � e4�i=3

�along the indented half-disk contour in Figure VII.4.10. We make a branchcutfor log z along the negative imaginary axis, where ��=2 < arg z < 3�=2.Residue at a simple pole at z1 = 1, where by Rule 3,

Res

�log jzj+ i arg z

z3 � 1 ; 1

�=log jzj+ i arg z

3z2

����z=1

= 0:

Residue at a simple pole at z2 = e2�i=3, where by Rule 3,

90

Res

�log jzj+ i arg z

z3 � 1 ; e2�i=3�=log jzj+ i arg z

3z2

����z=e2�i=3

=2�i

9e�4�i=3:

Integrate along 1 and 3, and let R!1 and "! 0+. This gives

Z 1

f (z) dz +

Z 3

f (z) dz =

=

Z 1

log jzj+ i arg zz3 � 1 dz +

Z 3

log jzj+ i arg zz3 � 1 dz =

=

�z = xe0i

dz = dx

�=

=

Z 1�"

"

log x

x3 � 1dx+Z R

1+"

log x

x3 � 1dx!

!Z 1

0

log x

x3 � 1dx = I:

Integrate along 2, and let "! 0+. This givesZ 2

f (z) dz ! �i (� � 0) � 0 = 0:

Integrate along 4, and let R!1. This gives����Z 4

f (z) dz

���� �plog2R + �2

R3 � 1 � �R � � logR

R2! 0:

Integrate along 5, and let R!1 and "! 0+. This gives

91

Z 5

f (z) dz =

=

Z 5

log jzj+ i arg zz3 � 1 dz =

�z = xe�i

dz = e�idx

�=

Z "

R

log x+ i�

�x3 � 1 e�id!

!Z 0

1

log x+ i�

�x3 � 1 e�idx = �

Z 1

0

log x+ i�

x3 + 1dx =

= �Z 1

0

log x

x3 + 1dx� i�

Z 1

0

1

x3 + 1dx =

= �Z 1

0

log x

x3 + 1dx� i�

Z 1

0

1

x3 + 1dx:

Integrate along 6, and let "! 0+. This gives����Z 6

f (z) dz

���� �plog2 "+ �

1� "3 � �" � �" jlog "j ! 0:

Using the Residue Theorem and letting R!1 and "! 0+, we obtain that

I + 0 + 0�Z 1

0

log x

x3 + 1dx� i�

Z 1

0

1

x3 + 1dx+ 0 = 2�i

�2�i

27e�4�i=3

�;

and thus, substituting the values for the integrals from Exercise 9,

I ���2�227

��i�

�2�

3p3

�= �4�

2

9e�4�i=3:

And hence, setting imaginary parts equal,

I+2�2

27=4�2

18;

and therefore Z 1

0

log x

x3 � 1dx =4�2

27:

92

VII.5.1Use the keyhole contour indented on the lower edge of the axis atx = 1 to show thatZ 1

0

log x

xa (x� 1)dx =2�2

1� cos (2�a) ; 0 < a < 1:

Solution

­R Rz

γ

γ

γγ

1

2

36

γ5 γ4

VII.5.1

Set

I =

Z 1

�1

log x

xa (x� 1)dx

and integrate

f (z) =log z

za (z � 1) =(log jzj+ i arg z)

ea(logjzj+i arg z) (z � 1)along the contour in Figure VII.5.1. We make a branchcut for log z alongthe positive imaginary axis, where 0 < arg z < 2�. Thus f (z) is analytic onthe keyhole domain. It extends analytically to (0;1) from above, and the

93

apparent singularity at z = 1 is removable. However, the extension to (0;1)from below has a simple pole at z = 1.Residue at a simple pole at z = 1, where by Rule 3,

Res

�(log jzj+ i arg z)

ea(logjzj+i arg z) (z � 1) ; 1�=(log jzj+ i arg z)ea(logjzj+i arg z)

����z=1

= 2�ie�2�ia:

Integrate along 1, and let R!1 and "! 0+. This gives

Z 1

f (z) dz =

=

Z 1

(log jzj+ i arg z)ea(logjzj+i arg z) (z � 1)dz =

�z = xe0i

dz = dx

�=

Z R

"

log x

xa (x� 1)dx!

!Z 1

0

log x

xa (x� 1)dx = I:

Integrate along 2, and let "! 0+. This gives when 0 < a < 1

����Z 2

f (z) dz

���� �qlog2R + (2�)2

Ra (R� 1) � 2�R � 2� logR

Ra! 0:

Integrate along 3 and 5, and let R!1 and "! 0+. This gives

Z 3

f (z) dz+

Z 5

f (z) dz =

=

Z 3

(log jzj+ i arg z)ea(logjzj+i arg z) (z � 1)dz+

Z 5

(log jzj+ i arg z)ea(logjzj+i arg z) (z � 1)dz =

=

�z = xe2�i

dz = dx

�=e�2�ia

Z 1+"

R

(log x+ 2�i)

xa (x� 1) dx+e�2�ia

Z "

1�"

(log x+ 2�i)

xa (x� 1) dx!

! e�2�iaZ 0

1

(log x+ 2�i)

xa (x� 1) dx = �e�2�ia

Z 1

0

(log x+ 2�i)

xa (x� 1) dx =

= �e�2�iaZ 1

0

log x

xa (x� 1)dx� 2�ie�2�ia

Z 1

0

1

xa (x� 1)dx = �e�2�iaI�2�ie�2�iaJ:

Integrate along 4, use fractional residue formula and let "! 0+. This gives

94

Z 4

f (z) dz ! �i (� � 0) � 2�ie�2�ia = 2�2e�2�ia:

Integrate along 6, and let "! 0+. This gives 0 < a < 1

����Z 6

f (z) dz

���� �qlog2 "+ (2�)2

"a (1� ") � 2�" � 2�"1�a jlog "j ! 0:

Using the Residue Theorem and letting R!1 and "! 0+, we obtain that

I � e�2�iaI � 2�ie�2�iaJ + 2�2e�2�ia + 0 = 0:Multiplying with e2�ia, this yields

Ie2�ia � I � 2�iJ + 2�2 = 0:We equate real parts

(cos (2�a)� 1) I + 2�2 = 0;and thereforeZ 1

0

log x

xa (x� 1)dx =2�2

1� cos (2�a) ; 0 < a < 1:

95

VII.5.2Show using residue theory thatZ 1

�1

sin (ax)

x (x2 + 1)dx = �

�1� e�a

�; a > 0:

Hint. Replace sin (az) by eiaz, and integrate around the boundary of a half-disk indented at z = 0.

Solution

­R R

γ4

γ3

γ2

γ1

z

z

z

1

2

3

VII.5.2

Set

I =

Z 1

�1

sin ax

x (x2 + 1)dx

and integrate

f (z) =eiaz

z (z2 + 1)=

eiaz

z (z � i) (z + i)along the contour in Figure VII.5.2.Residue at a simple pole at z1 = 0, where by Rule 3,

Res

�eiaz

z (z2 + 1); 0

�=

eiaz

3z2 + 1

����z=0

= 1:

Residue at a simple pole at z2 = i, where by Rule 3,

96

Res

�eiaz

z (z2 + 1); i

�=

eiaz

3z2 + 1

����z=i

= �e�a

2:

Integrate along 1 and 3, and let R!1 and "! 0+. This gives

Z 1

f (z) dz +

Z 3

f (z) dz =

=

Z 1

eiaz

z (z2 + 1)dz +

Z 3

eiaz

z (z2 + 1)dz =

�z = xdz = dx

�=

=

Z �"

�R

eiax

x (x2 + 1)dx+

Z R

"

eiax

x (x2 + 1)dx!

!Z 1

�1

eiax

x (x2 + 1)dx =

Z 1

�1

cos ax

x (x2 + 1)dx+ i

Z 1

�1

sin ax

x (x2 + 1)dx =

=

Z 1

�1

cos ax

x (x2 + 1)dx+ iI:

Integrate along 2, use fractional residue formula and let "! 0+. This givesZ 2

f (z) dz ! �i (� � 0) � 1 = ��i:

Integrate along 4, and let R ! 1. Because jeiazj � 1 in the upper halfplane if a > 0, this gives����Z

4

f (z) dz

���� � 1

R (R2 � 1) � �R ��

R2! 0:

Using the Residue Theorem and letting R!1 and "! 0+, we obtain thatZ 1

�1

cos ax

x (x2 + 1)dx+ iI � �i+ 0 = 2�i �

��e

�a

2

�;

and hence, setting imaginary parts equal,

I = � � �e�a;i.e.,

97

Z 1

�1

sin ax

x (x2 + 1)dx = �

�1� e�a

�; a > 0:

98

VII.5.3Show using residue theory thatZ 1

�1

sin (ax)

x (�2 � a2x2)dx =2

�; a > 0:

Solution

­R R

γ8

γ γ γ2 4 6

γ1 γ3 γ5 γ7z z z3 1 2

VII.5.3

Set

I =

Z 1

�1

sin (ax)

x (�2 � a2x2)dx

and integrate

f (z) =eiaz

z (�2 � a2z2) =eiaz

�a2z (z � �=a) (z + �=a)along the contour in Figure VII.5.3.Residue at a simple pole at z1 = 0, where by Rule 3,

Res

�eiaz

z (�2 � a2z2) ; 0�=

eiaz

�2 � 3a2z2

����z=0

=1

�2:

Residue at a simple pole at z2 = �=a, where by Rule 3,

Res

�eiaz

z (�2 � a2z2) ;�

a

�=

eiaz

�2 � 3a2z2

����z=�=a

=1

2�2:

99

Residue at a simple pole at z3 = ��=a, where by Rule 3,

Res

�eiaz

z (�2 � a2z2) ;��

a

�=

eiaz

�2 � 3a2z2

����z=��=a

=1

2�2:

Integrate along 1; 3; 5 and 7, and let R!1 and "! 0+. This gives

Z 1

f (z) dz +

Z 3

f (z) dz +

Z 5

f (z) dz +

Z 7

f (z) dz =

=

Z 1

eiaz

z (�2 � a2z2)dz +Z 3

eiaz

z (�2 � a2z2)dz+

+

Z 5

eiaz

z (�2 � a2z2)dz +Z 7

eiaz

z (�2 � a2z2)dz =

=

�z = xdz = dx

�=

=

Z ��=a�"

�R

eiax

x (�2 � a2x2)dx+Z �"

��=a+"

eiax

x (�2 � a2x2)dx+

+

Z �=a�"

"

eiax

x (�2 � a2x2)dx+Z R

�=a+"

eiax

x (�2 � a2x2)dx!

!Z 1

�1

eiax

x (�2 � a2x2)dx =

=

Z 1

�1

cos (ax)

x (�2 � a2x2)dx+ iZ 1

�1

sin (ax)

x (�2 � a2x2)dx =

=

Z 1

�1

cos (ax)

x (�2 � a2x2)dx+ iI:

Integrate along 2, use fractional residue formula and let "! 0+. This givesZ 2

f (z) dz ! �i (� � 0) � 12�2

= � 1

2�i:

Integrate along 4, use fractional residue formula and let "! 0+. This givesZ 4

f (z) dz ! �i (� � 0) � 1�2= � 1

�i:

100

Integrate along 6, use fractional residue formula and let "! 0+. This givesZ 6

f (z) dz ! �i (� � 0) � 12�2

= � 1

2�i:

Integrate along 8, and let R ! 1. Because jeiazj � 1 in the upper halfplane if a > 0, this gives����Z

2

f (z) dz

���� � 1

R (a2R2 � �2) � �R ��

a2R2! 0:

Using the Residue Theorem and letting R!1 and "! 0+, we obtain thatZ 1

�1

cos ax

x (x2 + 1)dx+ iI � 1

2�i� 1

�i� 1

2�i = 0;

and hence, setting imaginary parts equal

I � 1

2�� 1

�� 1

2�= 0;

i.e., Z 1

�1

sin (ax)

x (�2 � a2x2)dx =2

�; a > 0:

101

VII.5.4Show using residue theory thatZ 1

0

1� cosxx2

dx =�

2:

Solution

­R R

γ4

γ3

γ2

γ1z 1

VII.5.4

Set

I =

Z 1

0

1� cosxx2

dx) 2I =

Z 1

�1

1� cosxx2

dx

and integrate

f (z) =1� eizz2

along the contour in Figure VII.5.4.Residue at a double pole at z1 = 0, where by Rule 2,

Res

�1� eizz2

; 0

�= lim

z!0

d

dz

�1� eiz

�= �ieiz

��z=0

= �i

Integrate along 1 and 3, and let R!1 and "! 0+. This gives

102

Z 1

f (z) dz +

Z 3

f (z) dz =

=

Z 1

1� eizz2

dz +

Z 3

1� eizz2

dz =

=

�z = xdz = dx

�=Z �"

�R

1� eixx2

dx+

Z R

"

1� eixx2

dx!Z 1

�1

1� eixx2

dx =Z 1

�1

1� cosxx2

dx� iZ 1

�1

sin x

x2dx = 2I � i

Z 1

�1

sin x

x2dx:

Integrate along 2, use fractional residue formula and let "! 0+. This givesZ 2

f (z) dz ! �i (� � 0) � (�i) = ��:

Integrate along 4, and let R ! 1. Because jeiazj � 1 in the upper halfplane if a > 0, this gives����Z

4

f (z) dz

���� � 2

R2� �R � 2�

R! 0:

Using the Residue Theorem and letting R!1 and "! 0+, we obtain that

2I � iZ 1

�1

sin x

x2dx� � + 0 = 0;

and hence, setting imaginary parts equal

2I � � = 0;i.e., Z 1

0

1� cosxx2

dx =�

2:

103

VII.5.5By integrating (e�2iz � 1) =z2 over appropriate indented contoursand using Cauchy´s theorem, show thatZ 1

�1

sin2 x

x2dx = �:

Solution

­R R

γ2

γ1

γ4

γ3z 1

VII.5.5

Set

I =

Z 1

�1

sin2 x

x2dx =

Z 1

�1

1� cos 2x2x2

dx

and integrate

f (z) =ei2z � 1z2

along the contour in Figure VII.5.5.Residue at a double pole at z1 = 0, where by Rule 2,

Res

�ei2z � 1z2

; 0

�= lim

z!0

d

dz

�ei2z � 1

�= 2iei2z

��z=0

= 2i

Integrate along 1 and 3, and let R!1 and "! 0+. This gives

104

Z 1

f (z) dz +

Z 3

f (z) dz =

=

Z 1

ei2z � 1z2

dz +

Z 3

ei2z � 1z2

dz =

�z = xdz = dx

�=

=

Z �"

�R

ei2x � 1x2

dx+

Z R

"

ei2x � 1x2

dx!Z 1

�1

ei2x � 1x2

dx =

=

Z 1

�1

cos 2x� 1x2

dx+ i

Z 1

�1

sin 2x

x2dx = �2I + i

Z 1

�1

sin 2x

x2dx:

Integrate along 2, use fractional residue formula and let "! 0+. This givesZ 2

f (z) dz ! �i (� � 0) � 2i = 2�:

Integrate along 4, and let R ! 1. Because jeiazj � 1 in the upper halfplane if a > 0, this gives����Z

4

f (z) dz

���� � 2

R2� �R � 2�

R! 0:

Using the Residue Theorem and letting R!1 and "! 0+, we obtain that

�2I � iZ 1

�1

sin 2x

2x2dx+ 2� + 0 = 0;

and hence, setting real parts equal,

�2I + 2� = 0;i.e., Z 1

�1

sin2 x

x2dx = �:

105

VII.5.6By integrating a branch of (log z) = (z3 � 1) around the boundary ofan indented sector of aperture 2�=3, show thatZ 1

0

log x

x3 � 1dx =4�2

27:

Remark. See also Exercise 4.10.

Solution

R

γ4

γ1 γ3

γ8 γ2γ7

γ5 γ6

z 1

z 2

z 3

VII.5.6

Set

I =

Z 1

0

log x

x3 � 1dx

and integrate

f (z) =log z

z3 � 1 =log jzj+ i arg z

(z � e0i)�z � e2�i=3

� �z � e4�i=3

�along the contour in Figure VII.5.6. We make a branchcut for log z alongthe negative imaginary axis, where ��=2 < arg z < 3�=2.Residue at a simple pole at z1 = 1, where by Rule 3,

Res

�log jzj+ i arg z

z3 � 1 ; 1

�=log jzj+ i arg z

3z2

����z=1

= 0:

Residue at a simple pole at z2 = e2�i=3, where by Rule 3,

106

Res

�log jzj+ i arg z

z3 � 1 ; e2�i=3�=log jzj+ i arg z

3z2

����z=e2�i=3

=2�i

9e2�i=3:

Integrate along 1 and 3, and let R!1 and "! 0+. This gives

Z 1

f (z) dz+

Z 3

f (z) dz =Z 1

log jzj+ i arg zz3 � 1 dz+

Z 3

log jzj+ i arg zz3 � 1 dz =

�z = xe0i

dz = dx

�=

=

Z 1�"

"

log x

x3 � 1dx+Z R

1+"

log x

x3 � 1dx!Z 1

0

log x

x3 � 1dx = I:

Integrate along 2, use fractional residue formula and let "! 0+. This givesZ 2

f (z) dz ! �i (� � 0) � 0 = 0:

Integrate along 4, and let R!1. This gives

����Z 4

f (z) dz

���� �qlog2R +

�2�3

�2R3 � 1 � 2�R

3� 2� logR

R2! 0:

Integrate along 5 and 7, and let R!1 and "! 0+. This gives

Z 5

f (z) dz +

Z 7

f (z) dz =

=

Z 5

log jzj+ i arg zz3 � 1 dz+

Z 7

log jzj+ i arg zz3 � 1 dz =

�z = xe2�i=3

dz = e2�i=3dx

�=

=

Z 1+"

R

log x+ 2�i=3

x3 � 1 e2�i=3dx+

Z "

1�"

log x+ 2�i=3

x3 � 1 e2�i=3dx!

! e2�i=3Z 0

1

log x+ 2�i=3

x3 � 1 dx = �e2�i=3Z 1

0

log x

x3 � 1dx�2�i

3e2�i=3

Z 1

0

1

x3 � 1dx =

= �e2�i=3I�2�i3e2�i=3

Z 1

0

1

x3 � 1dx:

Integrate along 6, use fractional residue formula and let "! 0+. This gives

107

Z 6

f (z) dz ! �i (� � 0) ��2�i

9e2�i=3

�=2�2

9e2�i=3:

Integrate along 8, and let "! 0+. This gives

����Z 8

f (z) dz

�����qlog2 "+

�2�3

�21� "3 � 2�"

3� 2�" jlog "j

3! 0:

Using the Residue Theorem and letting R!1 and "! 0+, we obtain that

I + 0� e2�i=3I�2�i3e2�i=3

Z 1

0

1

x3 � 1dx+2�2

9e2�i=3 = 0:

Multiplying with e�2�i=3, this yields

I

�12�p3

2i

!� I � 2�i

3

Z 1

0

1

x3 � 1dx+2�2

9= 0:

We equate real parts

�32I +

2�2

9= 0;

and hence Z 1

0

log x

x3 � 1dx =4�2

27:

108

VII.6.1Integrate 1= (1� x2) directly, using partial fractions, and show that

PV

Z 1

0

dx

1� x2 = 0:

Show that Z 1

0

dx

1� x2 = +1;Z 1

1

dx

1� x2 = �1:

SolutionPartial fractions gives

1

1� x2 =1

2

�1

1 + x+

1

1� x

�

PV

Z 1

0

dx

1� x2 = lim"!0

�Z 1�"

0

dx

1� x2 +Z 1

1+"

dx

1� x2

�=

= lim"!0

�1

2[log (1 + x)� log (1� x)]1�"0 +

1

2[log j1 + xj � log j1� xj]11+"

�=

= lim"!0

1

2

�log

1 + x

1� x

�1�"0

+1

2

�log

����1 + x1� x

�����11+"

!=

= lim"!0

1

2

�log

�2� ""

�� log

�2 + "

"

��=

= lim"!0

1

2log

�2� "2 + "

�= 0:

Inspection gives

Z 1

0

dx

1� x2 = lim"!0+

Z 1�"

0

dx

1� x2 = lim"!0+

1

2[log (1 + x)� log (1� x)]1�"0 =

= lim"!0+

1

2

�log

1 + x

1� x

�1�"0

= lim"!0+

1

2log

�2� ""

�= +1 :

109

and the second integral

Z 1

1

dx

1� x2 = lim"!0+

Z 1

1+"

dx

1� x2 = lim"!0+

1

2[log (1 + x)� log (1� x)]11+" =

= lim"!0+

�1

2

�log

����1 + x1� x

�����11+"

�= lim

"!0+�12log

�2 + "

"

�! �1 :

110

VII.6.2Obtain the principal value in Exercise 1 by taking imaginary partsof the identity (5.4) p.211 in the preceding section and making achange of variable

SolutionIdentity (5.4)

Z 1

0

log x

x2 � 1dx+Z �1�"

�1

log jxj+ �ix2 � 1 dx+

Z 0

�1+"

log jxj+ �ix2 � 1 dx+

ZC"

log z

z2 � 1dz = 0:

Taking imaginary part of (5.4), getZ �1�"

�1

�

x2 � 1dx+Z 0

�1+"

�

x2 � 1dx+ ImZC"

log z

z2 � 1dz = 0:

Note

(�) Im

ZC"

log z

z2 � 1dz ! 0;

as "! 0, by estimation on p. 211.Making the change of variables, x! �x, and by using (�) we get

Z �1�"

�1

�

x2 � 1dx+Z 0

�1+"

�

x2 � 1dx =�x = �xdx = �dx

�=Z 1

1+"

�

x2 � 1dx+Z 1�"

0

�

x2 � 1dx! 0 as "! 0;

which gives that

PV

Z 1

0

dx

1� x2 = 0:

Note. For a linear change of variables, we can take the limit either before orafter change of variable. Otherwise we can not do this.

111

VII.6.3By integrating around the boundary of an indented half-disk in theupper half-plane, show that

PV

Z 1

�1

1

(x2 + 1) (x� a)dx = ��a

a2 + 1; �1 < a <1:

Solution

­R R

γ4

γ1 γ3

γ2

z 1

z 2

z 3

VII.6.3 (a = 2)

Set

I = PV

Z 1

�1

1

(x2 + 1) (x� a)dx

and integrate

f (z) =1

(z2 + 1) (z � a) =1

(z � i) (z + i) (z � a)along the contour in Figure VII.6.3.Residue at a simple pole at z1 = a, where by Rule 1,

Res

�1

(z2 + 1) (z � a) ; a�=

1

z2 + 1

����z=a

=1

a2 + 1:

Residue at a simple pole at z1 = i, where by Rule 3,

112

Res

�1

(z2 + 1) (z � a) ; i�=

1

3z2 � 2az + 1

����z=i

= � 1

2 (a2 + 1)+

a

2 (a2 + 1)i:

Integrate along 1 and 3, and let R!1 and "! 0+. This gives

Z 1

f (z) dz +

Z 3

f (z) dz =

=

a�"Z�R

1

(x2 + 1) (x� a)dx+RZ

a+"

1

(x2 + 1) (x� a)dx!

!Z 1

�1

1

(x2 + 1) (x� a)dx = I:

Integrate along 2, and let "! 0+. This givesZ 2

f (z) dz ! �i (� � 0) ��

1

a2 + 1

�= � �

a2 + 1i:

Integrate along 4, and let R!1. This gives����Z 4

f (z) dz

���� � 1

(R2 � 1) (R� jaj) � �R ��

R2! 0:

Using the Residue Theorem and letting R!1 and "! 0+, we obtain that

I � �

a2 + 1i = 2�i

�� 1

2 (a2 + 1)+

a

2 (a2 + 1)i

�;

i.e.,

PV

Z 1

�1

1

(x2 + 1) (x� a)dx = ��a

a2 + 1:

113

VII.6.4Suppose m � 2 and a1 < a2 < � � � < am. By integrating around theboundary of an indented half-disk in the upper half-plane, showthat

PV

Z 1

�1

1

(x� a1) (x� a2) � � � (x� am)dx = 0:

Solution

γR

­R Ra1 a2 am ­ 1 am

γ γ γ γ1 2 m ­ 1 m

VII.6.4

Set

I = PV

Z 1

�1

1

(x� a1) (x� a2) � � � (x� am)dx

and integrate

f (z) =1

(z � a1) (z � a2) � � � (z � am)=

1Qmk=1 (z � ak)

along the semicircular contour indented at z = aj with small semicircles ofradii ", see Figure VII.6.4.Residue at a simple pole at zj = aj; 1 � j � m, where by Rule 3,

Res

�1Qm

k=1 (z � ak); aj

�=

1Qmk=1;k 6=j (z � ak)

�����z=aj

=1Qm

k=1;k 6=j (aj � ak):

114

Integrate along the union of line segments on real axis denoted by �L, andlet R!1 and "! 0+. This gives

Z�L

f (z) dz =

Z a1�"

�Rf (z) dz+

m�1Xk=1

Z ak+1�"

ak+"

f (z) dz+

Z R

am+"

f (z) dz =

=

Z a1�"

�R

1Qmk=1 (z � ak)

dz+m�1Xk=1

Z ak+1�"

ak+"

1Qmk=1 (z � ak)

dz+

Z R

am+"

1Qmk=1 (z � ak)

dz =

=

�z = xdz = dx

�=

=

Z a1�"

�R

1Qmk=1 (x� ak)

dx+m�1Xk=1

Z ak+1�"

ak+"

1Qmk=1 (x� ak)

dx+

Z R

am+"

1Qmk=1 (x� ak)

dx!

!Z 1

�1

1Qmk=1 (x� ak)

dx = I:

Integrate along 1; 2; : : : m, and let "! 0+. This gives

mXj=1

Z j

f (z) dz ! �i (� � 0) �

mXj=1

1Qmk=1;k 6=j (aj � ak)

!:

Integrate along R, and let R!1, this gives����Z R

f (z) dz

���� � 1Qmk=1 (z � jakj)

� �R � �

Rm�1! 0:

Using the Residue Theorem and letting R!1 and "! 0+, we obtain that

I � i�mXj=1

1Qmk=1;k 6=j (aj � ak)

= 0:

Now, note that the the sums of the residues at a simple pole at zj = aj;1 � j � m, are real, and identifying the real part, we have

PV

Z 1

�1

1

(x� a1) (x� a2) � � � (x� am)dx = 0:

115

VII.6.5Show that

PV

Z 1

�1

xa�1

xb � 1dx = ��

bcot��ab

�; 0 < a < b:

Hint. For b > 1 one can integrate a branch of za�1=�zb � 1

�around a sector

of aperture 2�=b, indented at z = 1 and z = e2�i=b.

Solution

R

γ4

γ1 γ3

γ8 γ2γ7

γ5 γ6

z 1

z 2

z 3

VII.6.5 (b = 3)

Set

I = PV

Z 1

�1

xa�1

xb � 1dx

and integrate

f (z) =za�1

zb � 1 =jzja�1 ei(a�1) arg z

jzjb eib arg z � 1along the contour in Figure VII.6.5. We make a branchcut for za�1 along thenegative imaginary axis, where ��=2 < arg z < 3�=2.Residue at a simple pole at z1 = 1, where by Rule 3,

Res

�za�1

zb � 1 ; 1�=jzja�1 ei(a�1) arg z

bzb�1

�����z=1

=1

b:

116

Residue at a simple pole at z2 = e2�i=b, where by Rule 3,

Res

�za�1

zb � 1 ; e2�i=b

�=jzja�1 ei(a�1) arg z

bzb�1

�����z=e2�i=b

=e2�i=be2�i(a�1)=b

b:

Integrate along 1 and 3, and let R!1 and "! 0+. This gives

Z 1

f (z) dz+

Z 3

f (z) dz =

=

Z 1

jzja�1 ei(a�1) arg z

jzjb eib arg z � 1dz +

Z 3

jzja�1 ei(a�1) arg z

jzjb eib arg z � 1dz =

�z = xe0i

dz = dx

�=

=

Z 1�"

"

xa�1

xb � 1dx+Z R

1+"

xa�1

xb � 1dx!Z 1

0

xa�1

xb � 1dx = I:

Integrate along 2, and let "! 0+. This givesZ 2

f (z) dz ! �i (� � 0) ��1

b

�= ��i

b:

Integrate along 4, and let R!1. This gives when 0 < a < b����Z 4

f (z) dz

���� � Ra�1

Rb � 1 �2�R

b� 2�

bRb�a! 0:

Integrate along 5 and 7, and let R!1 and "! 0+. This gives

Z 5

f (z) dz+

Z 7

f (z) dz =Z 5

jzja�1 ei(a�1) arg z

jzjb eib arg z � 1dz +

Z 7

jzja�1 ei(a�1) arg z

jzjb eib arg z � 1dz =

�z = xe2�i=b

dz = e2�i=bdx

�=

=

Z 1+"

R

xa�1e2�i(a�1)=b

xb � 1 e2�i=bdx+

Z "

1�"

xa�1e2�i(a�1)=b

xb � 1 e2�i=bdx!

!Z 0

1

xa�1e2�i(a�1)=b

xb � 1 e2�i=bdx = �e2�i=be2�i(a�1)=bZ 1

0

xa�1

xb � 1dx =

= �e2�i=be2�i(a�1)=bI

117

Integrate along 6, and let "! 0+. This gives

Z 6

f (z) dz ! �i (� � 0) ��e2�i=be2�i(a�1)=b

b

�= ��ie

2�i=be2�i(a�1)=b

b:

Integrate along 8, and let "! 0+. This gives when 0 < a < b����Z 8

f (z) dz

���� � "a�1

1� "b �2�"

b� 2�"a

b! 0:

Using the Residue Theorem and letting R!1 and "! 0+, we obtain that

I � �ib+ 0� e2�i=be2�i(a�1)=bI � �ie

2�i=be2�i(a�1)=b

b= 2�i � 0:

Hence solving for I,

I =�i

b

1 + e2�i=be2�i(a�1)=b

1� e2�i=be2�i(a�1)=b =�i

b

1 + e2�ia=b

1� e2�ia=b = ��

b

cos��ab

�sin��ab

� = ��bcot��ab

�;

i.e.,

PV

Z 1

�1

xa�1

xb � 1dx = ��

bcot��ab

�; 0 < a < b:

118

VII.6.6By integrating a branch of (log z) =

�zb � 1

�around an indented sec-

tor of aperture 2�=b, show that for b > 1,

Z 1

0

log x

xb � 1dx =�2

b2 sin2 (�=b); PV

Z 1

0

1

xb � 1dx = ��

bcot (�=b) :

Solution

R

γ4

γ1 γ3

γ8 γ2γ7

γ5 γ6

z 1

z 2

z 3

VII.6.6 (b = 3)

Set

I =

Z 1

0

log x

xb � 1dx J = PV

Z 1

0

1

xb � 1dx

and integrate

f (z) =log z

zb � 1 =log jzj+ i arg zjzjb eib arg z � 1

along the contour in Figure VII.6.5. We make a branchcut for log z alongnegative imaginary axis, where ��=2 < arg z < 3�=2.Residue at a simple pole at z1 = 1, where by Rule 3,

Res

�log z

zb � 1 ; 1�=log jzj+ i arg z

bzb�1

����z=1

= 0:

Residue at a simple pole at z2 = e2�i=b, where by Rule 3,

119

Res

�log z

zb � 1 ; e2�i=b

�=log jzj+ i arg z

bzb�1

����z=e2�i=b

=2�ie2�i=b

b2:

Integrate along 1 and 3, and let R!1 and "! 0+. This gives

Z 1

f (z) dz+

Z 3

f (z) dz =

=

Z 1

log jzj+ i arg zjzjb eib arg z � 1

dz +

Z 3

log jzj+ i arg zjzjb eib arg z � 1

dz =

�z = xe0i

dz = dx

�=

=

Z 1�"

"

logx

xb � 1dx+Z R

1+"

logx

xb � 1dx!Z 1

0

logx

xb � 1dx = I:

Integrate along 2, and let "! 0+. This givesZ 2

f (z) dz ! �i (� � 0) � (0) = 0:

Integrate along 4, and let R!1. This gives when b > 1

����Z 4

f (z) dz

���� �qlog2R +

�2�b

�2Rb � 1 � 2�R

b� 2� logR

bRb�1! 0:

Integrate along 5 and 7, and let R!1 and "! 0+. This gives

Z 5

f (z) dz+

Z 7

f (z) dz =

=

Z 5

log jzj+ i arg zjzjb eib arg z � 1

dz +

Z 7

log jzj+ i arg zjzjb eib arg z � 1

dz =

�z = xe2�i=b

dz = e2�i=bdx

�=

=

Z 1+"

R

logx+ 2�i=b

xb � 1 e2�i=bdx+

Z "

1�"

logx+ 2�i=b

xb � 1 e2�i=bdx!

!Z 0

1

logx+ 2�i=b

xb � 1 e2�i=bdx = �e2�i=bZ 1

0

logx

xb � 1dx�2�i

be2�i=b

Z 1

0

1

xb � 1dx =

= �e2�i=bI � 2�ibe2�i=bJ:

120

Integrate along 6, and let "! 0+. This givesZ 6

f (z) dz ! �i (� � 0) ��2�ie2�i=b

b2

�=2�2e2�i=b

b2:

Integrate along 8, and let "! 0+. This gives when b > 1

����Z 8

f (z) dz

���� �qlog2 "+

�2�b

�21� "b � 2�"

b� 2�" jlog "j

b! 0:

Using the Residue Theorem and letting R!1 and "! 0+, we obtain that

I + 0� e2�i=bI � 2�ibe2�i=bJ +

2�2e2�i=b

b2= 2�i � 0:

Multiplying with e�2�i=b, this yields

�e�2�i=b � 1

�I � 2�i

bJ +

2�2

b2= 0:

Separating real and imaginary parts, we get simultaneous equations� �cos�2�b

�� 1�I + 2�2

b2= 0

� sin�2�b

�I � 2�

bJ = 0;

i.e.,

I =

Z 1

0

log x

xb � 1dx =�2

b2 sin2 (�=b); J = PV

Z 1

0

1

xb � 1dx = ��

bcot (�=b) :

121

VII.6.7Suppose that P (z) and Q (z) are polynomials, degQ (z) � degP (z)+2,and the zeros of Q (z) on the real axis are all simple. Show that

PV

Z 1

�1

P (z)

Q (z)dx = 2�i

XRes

�P (z)

Q (z); zj

�+ �i

XRes

�P (z)

Q (z); xk

�;

summed over the poles zj of P (z) =Q (z) in the open upper half-plane and the poles xk of P (z) =Q (z) on the real axis. Remark.In other words, the principal value of the integral is 2�i times thesum of the residues in the upper half-plane, where we count thepoles on the real axis as being half in and half out of the upperhalf-plane.

Solution

γR

­R R

z z

zz

z12

3j ­ 1

j

a1 a2 am ­ 1 am

γ γ γ γ1 2 m ­ 1 m

VII.6.7

Use a semicircular contour indented at ak = xk with small semicircles of radii", see Figure VII.6.7.Get Z

k

P (z)

Q (z)! �i (� � 0)Res

�P (z)

Q (z); xk

�= ��iRes

�P (z)

Q (z); xk

�The condition degQ � degP + 2 guarantees that the integral converges at1: Otherwise we would have to use a residue at 1 (in this case degQ =degP + 1).

122

Use the Residue Theorem, take limit, and the result follows

Set

I = PV

Z 1

�1

P (x)

Q (x)dx

and integrate

f (z) =P (z)

Q (z)

along the contour in Figure VII.6.7.The sum of the residues at the simple poles ak = xk; 1 � k � m on the realaxis is

mXk=1

Res

�P (z)

Q (z); xk

�The sum of the residues at the simple poles zk; 1 � k � j inside the integra-tion contour is

jXk=1

Res

�P (z)

Q (z); zj

�Integrate along segments on real axis, and let R ! 1 and " ! 0+. Thisgives

Zf (z) dz =

Z a1�"

�Rf (z) dz+

m�1Xk=1

Z ak+1�"

ak+"

f (z) dz+

Z R

am+"

f (z) dz =

=

Z a1�"

�R

P (z)

Q (z)dz+

m�1Xk=1

Z ak+1�"

ak+"

P (z)

Q (z)dz+

Z R

am+"

P (z)

Q (z)dz =

=

�z = xdz = dx

�=

=

Z a1�"

�R

P (x)

Q (x)dx+

m�1Xk=1

Z ak+1�"

ak+"

P (x)

Q (x)dx+

Z R

am+"

P (x)

Q (x)dx!

!Z 1

�1

P (x)

Q (x)dx = I:

123

Integrate along R, where we have degQ (z) � degP (z)+2 and let R!1.This gives ����Z

R

f (z) dz

���� � RdegP

RdegQ� �R � �

R! 0:

Integrate along 1; 2; : : : m, and let "! 0+. This gives

mXk=1

Z k

f (z) dz ! �i (� � 0) �mXk=1

�Res

�P (z)

Q (z); xk

��:

Using the Residue Theorem and letting R!1 and "! 0+, we obtain that

I � i�mXk=1

�Res

�P (z)

Q (z); xk

��= 2�i

jXk=1

Res

�P (z)

Q (z); zk

�!;

i.e.,

PV

Z 1

�1

P (x)

Q (x)dx = 2�i

mXj=1

Res

�P (z)

Q (z); zj

�+ �iRes

mXk=1

�P (z)

Q (z); xk

�:

124

VII.7.11 2 3 P L K

LLLShow that Z 1

0

jsin xjx

dx = +1:

Hint : Show that the area under the mth arc of jsin xj =x is � 1=m.

Solution

π 2π 3π 4π 5π

VII.7.1

For x in the interval (m� 1)� + �4� x � m� � �

4, we have

jsin xj � sin �4=

1p2;

and

1

x� 1

m�:

We estimate the integrand

jsin xjx

� 1p2

1

m�:

We have Z m���=4

(m�1)�+�4

jsin xjx

dx � 1p2

1

m�� =

C

m:

Hence

125

Z m�

0

jsin xjx

dx �mXk=1

Z k���=4

(k�1)�+�4

jsin xjx

dx = C

�1 +

1

2+ : : :+

1

m

�! +1

as m!1.We have used the fact that the harmonic series 1+ 1

2+ 1

3+ 1

4+ � � �+ 1

mtends

to in�nity as as m!1, by taking the oversum for the integralR m+11

dxx,

1 +1

2+1

3+ � � �+ 1

m�Z m+1

1

dx

x= log (m+ 1)! +1

as m!1.

126

VII.7.2Show that

limR!1

Z R

�R

x3 sin x

(x2 + 1)2dx =

�

2e:

Solution

γ

γ

2

1­R R

z 1

z 2

VII.7.2

Set

I =

Z 1

�1

x3 sin x

(x2 + 1)2dx

and integrate

f (z) =z3eiz

(z2 + 1)2=

z3eiz

(z � i)2 (z + i)2

along the contour in Figure VII.7.2.Residue at a double pole at z1 = i, where by Rule 2,

Res

�z3eiz

(z2 + 1)2; i

�= lim

z!i

d

dz

z3eiz

(z + i)2=1

4e:

Integrate along 1 , and let R!1. This gives

127

Z 1

f (z) dz =

=

Z 1

z3eiz

(z2 + 1)2dz =

�z = xdz = dx

�=

Z R

�R

x3eix

(x2 + 1)2dx!

!Z 1

�1

x3eix

(x2 + 1)2dx =

Z 1

�1

x3 cosx

(x2 + 1)2dx+ i

Z 1

�1

x3 sin x

(x2 + 1)2dx =

=

Z 1

�1

x3 cosx

(x2 + 1)2dx+ iI:

Integrate along 2, and let R!1. Use Jordan�s Lemma this gives����Z 2

f (z) dz

���� � R3

(R2 � 1)2Z 2

��eiz�� jdzj < �R3

(R2 � 1)2� �

R! 0:

Using the Residue Theorem and letting R!1, we obtain thatZ 1

�1

x3 cosx

(x2 + 1)2dx+ iI = 2�i �

�1

4e

�;

and hence, setting imaginary parts equal,i.e., Z 1

�1

x3 sin x

(x2 + 1)2dx =

�

2e:

128

VII.7.3Evaluate the limits

limR!1

Z R

�R

x sin (ax)

x2 + 1dx; �1 < a <1:

Show that they do not depend continuously on the parameter a.

Solution

γ

γ

2

1­R Rz 2

z 1

VII.7.3a

Case 1 a < 0:Set

I =

Z 1

�1

x sin (ax)

x2 + 1dx

and integrate

f (z) =zeiaz

z2 + 1=

zeiaz

(z � i) (z + i)along the contour in Figure VII.7.3a.Residue at a simple pole at z2 = �i, where by Rule 3,

Res

�zeiaz

z2 + 1;�i�=zeiaz

2z

����z=�i

=ea

2:

Integrate along 1, and let R!1. This gives

129

Z 1

f (z) dz =

Z 1

zeiaz

z2 + 1dz =

�z = xe0i

dz = dx

�=

Z R

�R

xeiax

x2 + 1dx!

!Z 1

�1

xeiax

x2 + 1dx =

Z 1

�1

x cos (ax)

x2 + 1dx+ i

Z 1

�1

x sin (ax)

x2 + 1dx =

=

Z 1

�1

x cos (ax)

x2 + 1dx+ iI:

Integrate along 2, and let R!1. Use Jordan�s Lemma this gives����Z 2

f (z) dz

���� � R

R2 � 1

Z 2

��eiaz�� jdzj < �R

R2 � 1 ��

R! 0:

Using the Residue Theorem and letting R!1, we obtain thatZ 1

�1

x cos (ax)

x2 + 1dx+ iI + 0 = �2�i �

�ea

2

�;

and hence, setting imaginary parts equal,

I = ��ea:

Case 2 a = 0:

limR!1

Z R

�R

x sin (ax)

x2 + 1dx = lim

R!1

Z R

�R

0

x2 + 1= 0

Case 3 a > 0:

130

γ

γ

4

3­R R

z 1

z 2

VII.7.3b

Set

J =

Z 1

�1

x sin (ax)

x2 + 1dx

and integrate

f (z) =zeiaz

z2 + 1=

zeiaz

(z � i) (z + i)along the contour in Figure VII.7.3bResidue at a simple pole at z1 = i, where by Rule 3,

Res

�zeiaz

z2 + 1; i

�=zeiaz

2z

����z=i

=e�a

2:

Integrate along 3, and let R!1. This gives

Z 3

f (z) dz =

Z 3

zeiaz

z2 + 1dz =

�z = xe0i

dz = dx

�=

Z R

�R

xeiax

x2 + 1dx!

!Z 1

�1

xeiax

x2 + 1dx =

Z 1

�1

x cos (ax)

x2 + 1dx+ i

Z 1

�1

x sin (ax)

x2 + 1dx =

=

Z 1

�1

x cos (ax)

x2 + 1dx+ iJ:

Integrate along 4, and let R!1. Use Jordan�s Lemma this gives

131

����Z 4

f (z) dz

���� � R

R2 � 1

Z 4

��eiaz�� jdzj < �R

R2 � 1 ��

R! 0:

Using the Residue Theorem and letting R!1, we obtain thatZ 1

�1

x cos (ax)

x2 + 1dx+ iI + 0 = 2�i �

�e�a

2

�;

and hence, setting imaginary parts equal,

I = �e�a:

and we have the solution

Z 1

�1

x sin (ax)

x2 + 1dx =

8<:��ea; a < 0;0 a = 0�e�a; a > 0:

Not continuous at a = 0 becauce sin 0 = 0.

132

VII.7.4By integrating za�1eiz around the boundary of a domain in the �rstquadrat bounded by the real and imaginary axes and a quater-circle, show that

limR!1

Z R

0

xa�1 cosxdx = � (a) cos (�a=2) ; 0 < a < 1;

limR!1

Z R

0

xa�1 sin xdx = � (a) sin (�a=2) ; 0 < a < 1;

where � (a) is the gamma function de�ned by

� (a) =

Z 1

0

ta�1e�tdt:

Remark : The formula for the sine integral holds also for �1 < a < 0. To seethis, integrate by parts.

Solution

Rγ1

γ2

γ3

VII.7.4

Set

I = limR!1

Z R

0

xa�1 cosxdx J = limR!1

Z R

0

xa�1 sin xdx

where 0 < a < 1, and integrate

133

f (z) = za�1eiz = jzja�1 ei(a�1) arg zeiz

along the contour in Figure VII.7.4Integrate along 1, and let R!1 and "! 0+. This gives

Z 1

f (z) dz =

=

Z 1

jzja�1 ei(a�1) arg zeizdz =�z = xe0i

dz = dx

�=

Z R

0

xa�1eixdx!

!Z 1

0

xa�1eixdx =

Z 1

0

xa�1 cosxdx+ i

Z 1

0

xa�1 sin xdx = I + iJ:

Integrate along 2, and let R!1. Use Jordan�s Lemma this gives����Z 2

f (z) dz

���� � Ra�1Z 2

��eiz�� jdzj < �Ra�1 � �

R1�a! 0:

Integrate along 3, and let R!1 and "! 0+. This gives

Z 3

f (z) dz =

=

Z 3

jzja�1 ei(a�1) arg zeizdz =�z = xe�i=2

dz = e�i=2dx

�= �

Z R

0

xa�1e�ia=2e�xdx!

! �Z 1

0

xa�1e�ia=2e�xdx = �e�ia=2Z 1

0

xa�1e�xdx = �e�ia=2� (a) =

= �� (a) cos(�a=2)� � (a) sin (�a=2) :

Using the Residue Theorem and letting R!1, we obtain that

I + iJ � � (a) cos(�a=2)� � (a) sin (�a=2) = 0:We equate real and imaginary parts

I = � (a) cos(�a=2) J = � (a) sin (�a=2) :

Formulas for the Gamma function, Mathematics Handbook page 278

134

� (z + 1) = z� (z)

Z R

0

xa�1 cosxdx =�xa�1 sin x

�R0�Z R

0

(a� 1)xa�2 sin xdx

(a� 1)Z R

0

xa�2 sin xdx =�xa�1 sin x

�R0�Z R

0

xa�1 cosxdx

(a� 1)Z R

0

xa�2 sin xdx = Ra�1 sinR�Z R

0

xa�1 cosxdx

Z R

0

xa�2 sin xdx =Ra�1 sinR

a� 1 � 1

a� 1

Z R

0

xa�1 cosxdx

Z R

0

xa�2 sinxdx =sinR

(1� a)R1�a �1

a� 1� (a) cos (�a=2)

limR!1

Z R

0

xa�1 sin xdx =�� (a+ 1)

acos��2(a+ 1)

�=

= �� (a) [cos (�a=2) cos (�=2)� sin (�a=2) sin (�=2)] == � (a) sin (�a=2) :

135

VII.7.5Show that

limR!1

Z R

0

sin�x2�dx = lim

R!1

Z R

0

cos�x2�dx =

p�

2p2;

by integrating eiz around the boundary of the pie-slice domain determinedby 0 < arg z < �=4 and jzj < R. Remark. These improper integrals arecalled the Fresnel integrals. The identities can also be deduced from thepreceding exercise by changing variable.

Solution

Rγ1

γ2

γ3

VII.7.5

Set

I = limR!1

Z R

0

sin�x2�dx J = lim

R!1

Z R

0

cos�x2�dx

where �1 < a < 0, and integrate

f (z) = eiz2

along the contour in Figure VII.7.5Integrate along 1, and let R!1. This gives

136

Z 1

f (z) dz =

Z 1

eiz2

dz =

�z = xe0i

dz = dx

�=

Z R

0

eix2

dx!

!Z 1

0

eix2

dx =

Z 1

0

cos�x2�dx+ i

Z 1

0

sin�x2�dx = J + iI:

Integrate along 2 as we can parametrize as z = Reit where 0 � t � �=4,and let R!1. Use that 4x=� � sin 2x for 0 � x � �=4, this gives

����Z 2

f (z) dz

���� = ����Z 2

eiz2

dz

���� = � z = Reit

dz = iReitdt

�=

=

�����Z �=4

0

eiR2e2itiReitdt

����� � RZ �=4

0

���eiR2(cos(2t)+i sin(2t))��� dt == R

Z �=4

0

e�R2 sin 2tdt � R

Z �=4

0

e�4R2t=�ds =� �

4R

he�4R

2t=�i�=40=

=�

4R

�1� e�R2

�� �

4R! 0:

Integrate along 3 as we can parametrize as z = xe�i=4 where 0 � x � R,and let R!1. This gives

Z 3

f (z) dz =

Z 3

eiz2

dz =

�z = xe�i=4

dz = e�i=4dx

�= �

Z R

0

e�x2

e�i=4dx!

! �Z 1

0

e�x2

e�i=4dx = �1 + ip2

Z 1

0

e�x2

dx =

= � 1p2

Z 1

0

e�x2

dx� i 1p2

Z 1

0

e�x2

dx:

We will use integral (41) from Mathematics Handbook page 177 with a = 1,Z 1

0

e�ax2

dx =1

2

r�

a:

and hence

137

Z 3

f (z) dz = � 1p2

Z 1

0

e�x2

dx� i 1p2

Z 1

0

e�x2

dx = �p�

2p2� i

p�

2p2:

Using the Residue Theorem and letting R!1, we obtain that

J + iI + 0�p�

2p2� i

p�

2p2= 0:

We equate real and imaginary parts

I =

p�

2p2

J =

p�

2p2:

138

VII.8.11 2 3 P L K

Evaluate the residue at 1 of the following functions.(a) z

z2�1 (c) z3+1z2�1 (e) zne1=z, n = 0;�1; : : :

(b) 1(z2+1)2

(d) z9+1z6�1 (f)

qz�az�b

Note. There are two possibilities for (f), one for each branch of the squareroot.a)

z

z2 � 1 =1

z

1

1� 1z2

=1

z

�1 +

1

z2+1

z4+ : : :

�;

hence

Res

�z

z2 � 1 ;1�= �1:

b)

1

(z2 + 1)2= O

�1

z4

�as z !1;

hence

Res

�1

(z2 + 1)2;1�= 0:

c)

z3 + 1

z2 � 1 =z (z2 � 1) + z + 1

z2 � 1 = z +z

z2 � 1 +1

z2 � 1 =

= z +1

z

1

1� 1z2

+1

z2 � 1 = z +1

z

�1 +

1

z2+1

z4+ : : :

�+

1

z2 � 1 ;

hence

Res

�z3 + 1

z2 � 1 ;1�= �1:

139

d)

z9 + 1

z6 � 1 =z3 (z6 � 1) + z3 + 1

z6 � 1 = z3 +z3 + 1

z6 � 1 ;

hence

Res

�z9 + 1

z6 � 1 ;1�= 0:

e)

zne1=z = zn�1 +

1

z+

1

2!z2+

1

3!z3+

1

3!z3+

1

4!z4+ : : :

�=

= zn + zn�1 +1

2!zn�2 +

1

3!zn�3 +

1

4!zn�4 + :

We can see that the coe¢ cient of 1zis

1

(n+ 1)!;

hence

Res�zne1=z;1

�= � 1

(n+ 1)!; n = 0;�1; : : : :

f) rz � az � b =

s1� a=z1� b=z =

r1� aw1� bw ;

analytic at w = 0:coe¢ cient of w is g0(0) = 1

2

�1�aw1�bw

��1=2 ��a(1�bw)+b(1�aw)(1�bw)2

����w=0

= 121p1(b� a) =

� b�a2:

140

7.8.21 2 3 P L K

Show by integrating around the dogbone contour thatZ 1

0

x4px (1� x)

dx =35�

128:

Solution

γ4γ3

γ2γ1

VII.8.2

Set

I =

Z 1

0

x4px (1� x)

dx

and integrate

f (z) =z4p

z (1� z)=

z4pjzjei arg� z=2

pj1� zjei arg0(1�z)=2

along the contour in Figure VII.8.2.

141

VIII.8.2arg� z arg0 z

π π π 0 0 0

−π −π −π 0 0 0­2 ­1 1 2x

y

π π π 0 0 0

π π π 2π 2π 2π­2 ­1 1 2x

y

arg� z arg0 (1� z)

π π π 0 0 0

−π −π −π 0 0 0­2 ­1 1 2x

y

2π 2π 2π 2π π π

0 0 0 0 π π­2 ­1 1 2x

y

The argument for the function f (z) = z4pjzjei arg� z=2

pj1�zjei arg0(1�z)=2

3π/2 3π/2 3π/2 π π/2 π/2

−π/2 −π/2 −π/2 0 π/2 π/2­2 ­1 1 2x

y

Residue at a simple pole at z1 =1 , where by result from Exercise VII.8.13and Rule 1,

Res [f (z) ;1] = �Res�1

w2f

�1

w

�; 0

�= �Res

24 1w2

1w4q

1w

�1� 1

w

� ; 035 =

= �Res�1

w51pw � 1

�= � lim

w!0

1

4!

d4

dz4

�1pw � 1

�=

= � 35

128 (w � 1)9=2= � 35

128 (jw � 1j)9=2 ei9 arg(w�1)=2= � 35

128e��i=2 = � 35

128i

142

Integrate along 1, and let "! 0+. This gives

Z 1

f (z) dz =

=

Z 1�"

"

z4pjzjei arg� z=2

pj1� zjei arg0(1�z)=2

dz =

Z 1�"

"

x4pjxjei�0=2

pj1� xjei�0=2

dx =

=

Z 1�"

"

x4px (1� x)

dx!Z 1

0

x4px (1� x)

dx = I:

Integrate along 2, and let "! 0+. This gives����Z 2

f (z) dz

���� � (1 + ")4p(1 + ") "

� 2�" � 2�"9=2 ! 0:

Integrate along 3, and let "! 0+. This gives

Z 3

f (z) dz =

=

Z "

1�"

z4pjzjei arg� z=2

pj1� zjei arg0(1�z)=2

dz =

Z "

1�"

1pjxjei�0=2

pj1� xjei2�=2

dx =

=

Z 1�"

"

x4px (1� x)

dx!Z 1

0

x4px (1� x)

dx = I:

Integrate along 4, and let "! 0+. This gives����Z 4

f (z) dz

���� � "4p" (1� ")

� 2�" � 2�"9=2 ! 0:

Using the Residue Theorem and letting R!1 and "! 0+, we obtain that

I + 0 + I + 0 = 2�i ��35

128i

�;

i.e., Z 1

0

x4px (1� x)

dx =35�

128:

143

7.8.31 2 3 P L K

Fix an integer n, positive or negative. Determine for which complexvalues of the parameter a the integralZ 1

0

xn

xa (1� x)1�adx

converges, and evaluate it.

Solution. ( Obs fel i lösningen argumentet)x 2 R+1� x 2 R+a = Re a+ i Im axa = xRe a+i Im a = xRe axi Im a

The integral is generalized in 0 and 1 " 2 R, x = 0g (x) = 1

xa�nf(x)g(x)

= 1(1�x)1�a ! 1, as x! 0R "

0g (x) dx <1, a� n < 1,

R "0f (x) dx <1, a < n+ 1.

The integral is generalized in 0 and 1 " 2 R, x = 0g (x) = 1

(1�x)1�af(x)g(x)

= xn

xa! 1, as x! 1R 1

"g (x) dx <1, 1� a < 1,

R "0f (x) dx <1, a > 0.

ThusR 10f (x) dx <1, 0 < a < n+ 1.

Show that xi Im a begränsad, set � 2 [arg x]��xi Im a�� = ��ei Im a(log x+i�)�� = ��ei Im a log x�� = 1On principal arg 0

Determination for which complex values of the parameter a the integral con-verges

144

γ4γ3

γ2γ1

VII.8.3

Set

I =

Z 1

0

xn

xa (1� x)1�adx

and integrate

f (z) =zn

za (1� z)1�a=

zn

jzjaeai arg�z j1� zj1�ae(1�a)i arg0(1�z)

along the contour in Figure VII.8.3.

145

VIII.8.3arg� z arg0 z

π π π 0 0 0

−π −π −π 0 0 0­2 ­1 1 2x

y

π π π 0 0 0

π π π 2π 2π 2π­2 ­1 1 2x

y

arg� z arg0 (1� z)

π π π 0 0 0

−π −π −π 0 0 0­2 ­1 1 2x

y

2π 2π 2π 2π π π

0 0 0 0 π π­2 ­1 1 2x

y

The argument for the function f (z) = zn

jzjaeai arg�z j1�zj1�ae(1�a)i arg0(1�z)

2π − απ 2π − απ 2π − απ 2π − 2απ π − απ π − απ

−απ −απ −απ 0 π − απ π − απ­2 ­1 1 2x

y

Residue at a simple pole at z3 =1 , where by result from Exercise VII.8.13and Rule 1,

Res [f (z) ;1] = �Res�1

w2f

�1

w

�; 0

�= �Res

"1

w2

1wn

1wa

�1� 1

w

�1�a ; 0#=

= �Res� p

w2 � 1w(1 + w2)

; 0

�= � lim

w!0

pjw2 � 1jei arg(w2�1)=2

w2 + 1= �ei�=2 = �i

Integrate along 1, and let R!1 and "! 0+. This gives

146

Z 1

f (z) dz =

=

Z 1�"

"

zn

jzjaeai arg�z j1� zj1�ae(1�a)i arg0(1�z)=

Z 1�"

"

zn

jzjaeai�0 j1� zj1�ae(1�a)i�0dx =

=

Z 1�"

"

zn

za (1� z)1�adx!

!Z 1

0

zn

za (1� z)1�adx = I:

Integrate along 2, and let "! 0+. This gives����Z 4

f (z) dz

���� � (1 + ")4

(1 + ")a (1� (1 + "))1�a� 2�" � 2�

"�a! 0:

Integrate along 3, and let R!1 and "! 0+. This gives

Z 3

f (z) dz =

=

Z "

1�"

zn

jzjaeai arg�z j1� zj1�ae(1�a)i arg0(1�z)=

Z "

1�"

zn

jzjaeai�0 j1� zj1�ae(1�a)i�2�dx =

= �e2�iaZ 1�"

"

zn

za (1� z)1�adx!

! �e2�iaZ 1

0

zn

za (1� z)1�adx = �e2�iaI:

Integrate along 4, and let "! 0+. This gives����Z 4

f (z) dz

���� � "n

"a (1� ")1�a� 2�" � 2�"n+1�a ! 0:

Using the Residue Theorem and letting R!1 and "! 0+, we obtain that

I + 0�e2�iaI + 0 = 2�i ��(�1)n (a� 1) � � � (a� n)

n!

�:

147

Solve for I, we obtain that

I =(�1)n 2�i (a�1)���(a�n)

n!

1� e2�ia =(�1)n 2�i (a�1)���(a�n)

n!

e�ia � e��ia = (�1)n �

sin (�a)

(a� 1) � � � (a� n)n!

;

i.e.,

Z 1

0

xn

xa (1� x)1�adx = (�1)n �

sin (�a)

(a� 1) � � � (a� n)n!

;0 < a < 1;n = 0; 1; 2; : : :

148

VII.8.4Show that Z 1

�1

p1� x21 + x2

dx = ��p2� 1

�:

Solution

γ4

γ2γ1

γ3

VII.8.4

Set

I =

Z 1

�1

p1� x21 + x2

dx

and integrate

f (z) =

p1� z2z2 + 1

=

pj1 + zjei arg�(1+z)=2

pj1� zjei arg0(1�z)=2

z2 + 1

along the contour in Figure VII.8.4.

149

VIII.8.4arg� z arg0 z

π π π 0 0 0

−π −π −π 0 0 0­2 ­1 1 2x

y

π π π 0 0 0

π π π 2π 2π 2π­2 ­1 1 2x

y

arg� (1 + z) arg0 (1� z)

π π 0 0 0 0

−π −π 0 0 0 0­2 ­1 1 2x

y

2π 2π 2π 2π π π

0 0 0 0 π π­2 ­1 1 2x

y

The argument for the function f (z) = z4pjzjei arg� z=2

pj1�zjei arg0(1�z)=2

3π/2 3π/2 π π π/2 π/2

−π/2 −π/2 0 0 π/2 π/2­2 ­1 1 2x

y

Residue at a simple pole at z1 = i, where by Rule 3,

Res

"pj1 + zjei arg(1+z)=2

pj1� zjei arg(1�z)=2

(z � i) (z + i) ; i

#=

=

pj1 + zjei arg(1+z)=2

pj1� zjei arg(1�z)=2

2z

�����z=i

=pj1 + ijei arg(1+i)=2

pj1� ijei arg(1�i)=2

2i=

pj1 + ije(i�=4)=2

pj1� ije(i7�=4)=2

2i

=

pp2pp

2e�i

2i=

p2(�1)2i

= �p2

2i=

ip2

150

Residue at a simple pole at z2 = �i, where by Rule 3,

Res

"pj1 + zjei arg(1+z)=2

pj1� zjei arg(1�z)=2

z2 + 1;�i#=

=

pj1 + zjei arg(1+z)=2

pj1� zjei arg(1�z)=2

2z

�����z=�i

=

=

pj1� ijei arg(1�i)=2

pj1 + ijei arg(1+i)=2

�2i =

pj1� ije(�i�=4)=2

pj1 + ije(i�=4)=2

�2i =

=

pp2pp

2

�2i =

p2

�2i =1p2i:

Residue at a simple pole at z3 =1 , where by result from Exercise VII.8.13and Rule 1,

Res [f (z) ;1] = �Res�1

w2f

�1

w

�; 0

�= �Res

24 1w2

q1� 1

w2�1+ 1

w2

� ; 035 =

= �Res� p

w2 � 1w(1 + w2)

; 0

�= � lim

w!0

pjw2 � 1jei arg(w2�1)=2

w2 + 1= �ei�=2 = �i

Integrate along 1, and let "! 0+. This gives

Z 1

f (z) dz =

=

Z 1�"

�1+"

pj1 + zjei arg�(1+z)=2

pj1� zjei arg0(1�z)=2

z2 + 1dz =

Z 1�"

�1+"

pj1 + xjei�0=2

pj1� xjei2�=2

1 + x2dx!

! �Z 1�"

�1+"

p1� x21 + x2

dx = �I:

Integrate along 2, and let "! 0+. This gives

����Z 2

f (z) dz

���� �q1� (1� ")2

1 + (1� ")2� 2�" � 2�"3=2 ! 0:

151

Integrate along 3, and let "! 0+. This gives

Z 3

f (z) dz =

=

Z �1+"

1�"

pj1 + zjei arg�(1+z)=2

pj1� zjei arg0(1�z)=2

z2 + 1dz = �

Z 1�"

�1+"

pj1 + xjei�0=2

pj1� xjei�0=2

x2 + 1dx!

! �Z 1�"

�1+"

p1� x21 + x2

dx = �I:

Integrate along 4, and let "! 0+. This gives

����Z 4

f (z) dz

���� �q1� (1� ")2

1 + (1� ")2� 2�" � 2�"3=2 ! 0:

Using the Residue Theorem and letting R!1 and "! 0+, we obtain that

�I + 0� I + 0 = 2�i ��1p2i+

1p2i� i

�i.e., Z 1

�1

p1� x21 + x2

dx = ��p2� 1

�:

152

VII.8.51 2 3 P L K

Show that the sum of the residues of a rational function on theextended complex plane is equal to zero.

Solution Ijzj=R

f (z) dz = 2�iX

Residues in �nite plane;Ijzj=R

f (z) dz = 2�iRes [f;1]

SinceH= �

H�, we get 2�i

PRes = 0.

153

VII.8.61 2 3 P L K

Find the residue of z= (z2 + 1) at each pole in the extended complexplane, and check that the sum of the residues is zero.

SolutionResidue at a simple pole at z1 = i, where by Rule 3,

Res

�z

z2 + 1; i

�=z

2z

���z=i=1

2:

Residue at a simple pole at z1 = i, where by Rule 3,

Res

�z

z2 + 1;�i�=z

2z

���z=�i

=1

2:

For residue at 1, we have that

a�1 = limz!1

zf (z) = limz!1

z2

z2 + 1= lim

z!1

1

1 + 1=z2= 1:

hence

Res

�z

z2 + 1;1�= �a�1 = �1;

and the sum of the residues is zero in the extended complex plane.

(Since zz2+1

= a�1z+O

�1z2

�as z !1).

154

VII.8.71 2 3 P L K

Find the sum of the residues of (3z4 + 2z + 1) = (8z5 + 5z2 + 2) at itspoles in the (�nite) complex plane.

SolutionFor residue at 1, we have that

a�1 = limz!1

zf (z) = limz!1

3z5 + 2z2 + z

8z5 + 5z2 + z= lim

z!1

3 + 2=z3 + 1=z4

8 + 5=z3 + 1=z4=3

8;

hence

Res [f;1] = �38:

We have that the sum of the residues in the complex �nite plane is�Res [f (z) ;1],thus the sum of residues in �nite complex plane is 3=8.

f (z) = a�1z+O

�1z2

�as z !1, so again

155

VII.8.81 2 3 P L K

Fix n � 1 and k � 0. Find the residue of zk= (zn � 1) at1 by expand-ing 1= (zn � 1) in a Laurent series. Find the residue of zk= (zn � 1)at each �nite pole, and verify that the sum of all the residues arezero.

Solution.Let f (z) = zk

zn�1 . At 1, we have

f (z) =zk

zn � 1 =zk�n

1� 1zn

= zk�n�1 + z�n + z�2n + : : :

�= zk�n+zk�2n+zk�3n+: : : :

Then Res [f;1] = �1 if k � mn = �1, thus k = mn � 1 for some integerm � 1, and otherwise Res [f;1] = 0.

Let wj be an nth root of unity, i.e wj = e2�ij=n where 0 � j � n � 1. Thenf (z) has simple poles at w0; : : : ; wn�1.Residue at a simple pole at zj = wj, where by Rule 3 and that wnj = 1

Res [f; wj] =zk

nzn�1

����z=wj

=zk+1

nzn

����z=wj

=wk+1j

nwnj=wk+1j

n:

Let S be the sum of residues, then

S =1

n

n�1Xj=0

wk+1j =1

n

�1 + w + w2 + � � �+ wn�1

�=1

n

�1� wn1� w

�= 0;

unless w = 1.If w = 1, then wk+1j = wj = 1 for every j and w = 1 , e

2�in(k+1) = 1 ,

k+1n= m, thus k = mn� 1 for some integer m � 1.

Thus we have that the sum of the residues in the complex plane are zero.

156

VII.8.91 2 3 P L K

Show that f (z) is analytic at 1, then

Res [f (z) ;1] = � limz!1

z (f (z)� f (1)) :

SolutionSince f (z) is analaytic at 1, we have

f (z) = a0 +a�1z+O

�1

z2

�:

Then

z (f (z)� f (1)) = z�a0 +

a�1z+O

�1

z2

�� a0

�= a�1 +O

�1

z

�! a�1

as z !1.From this follows that

� limz!1

z (f (z)� f (1)) = �a�1 = Res [f (z) ;1] :

157

VII.8.101 2 3 P L K

Let D be an exterior domain. Suppose that f (z) is analytic onD [ @D and at 1. Show that

1

2�i

Z@D

f (�)

� � zd� = f (�)� f (1) ; z 2 D:

Solution.Cauchy integral formula for exterior domain. If D is an exterior domaina,f (z) is analytic in D and att 1, then

1

2�i

Z@D

f (�)

� � zd� = f (z) + Res�f (�)

� � z ;1�= f (z)� f (1)

IIII

Res

�f (�)

� � z ;1�= � lim

�!1

�f (�)

� � z = �f (1) :

Or

f (�) = a0 +a�1�+a�2

�2+ : : :

f (�)

� � z =a0� � z +O

�1

�2

�

1

2�i

Z@D

f (�)

� � zd� = f (z) +1

2�i

Zjzj=R

f (�)

� � zd� =

= f (z)� 1

2�i

Zjzj=R

a0� � zd� = f (z)� a0:

158

VII.8.111 2 3 P L K

If f (z) is not integrable at 1, we de�ne the principal value

PV

Z 1

�1f (x) dx = lim

R!1PV

Z R

�Rf (x) dx:

Show that

PV

Z 1

�1

1

x� adx =

8<:i�; Im a > 0;0; Im a = 0;�i�; Im a < 0:

SolutionSet a = a1 + a2i, where a1 and a2 is �xed. We compute the principal valuefor the following 3 cases.Case 1, Im a < 0, a2 < 0. Integrate and let R!1. This gives

Z R

�R

dx

x� a =Z R

�R

dx

x� a1 � a2i= [log (x� a1 � a2i)]R�R = log

�R� a1 � a2i�R� a1 � a2i

�=

= log

���� R� a1 � a2i�R� a1 � a2i

����+ i arg (R� a1 � a2i)� i arg (�R� a1 � a2i)!! 0 + i � 0� i� = �i�:

Case 2, Im a = 0 , a2 = 0. Integrate and let R ! 1 and " ! 0+. Thisgives

Z R

�R

dx

x� a =Z R

�R

dx

x� a1=

Z a1�"

�R

dx

x� a1+

Z R

a1+"

dx

x� a1= [log jx� a1j]a1�"�R +[log jx� a1j]Ra1+" =

= log "� log jR + a1j+ log jR� a1j � log " = log����R� a1R + a1

����! 0:

Case 3, Im a > 0, a2 > 0. Integrate and let R!1. This gives

159

Z R

�R

dx

x� a =Z R

�R

dx

x� a1 � a2i= [log (x� a1 � a2i)]R�R = log

�R� a1 � a2i�R� a1 � a2i

�=

= log

���� R� a1 � a2i�R� a1 � a2i

����+ i arg (R� a1 � a2i)� i arg (�R� a1 � a2i)!! 0 + i � 2� � i� = i�:

We have that

PV

Z 1

�1

1

x� adx =

8<:i�; Im a > 0;0; Im a = 0;�i�; Im a < 0:

160

VII.8.121 2 3 P L K

Suppose that P (z) and Q (z) are polynomials such that the degreeof Q (z) is strictly greater than the degree of P (z). Suppose thatthe zeros x1; : : : ; xm of Q (z) on the real axis are all simple, and setx0 =1. Show that

PV

Z 1

�1

P (x)

Q (x)dx = 2�i

XRes

�P (x)

Q (x); zj

�+ �i

XRes

�P (x)

Q (x); zk

�;

summed over the poles zj of P (z) =Q (z) in the open upper half-plane and summed over the xk : s including1. Hint. Use the precedingexercise. See also Exercise 6.7

Solution.By VII.8.11, formula holds for 1

z�a By Exercise b.7 (p. 215.) it holds if

degQ � degP + 2;in initial case ReshP (x)Q(x)

;1i= 0. By limiticity, it hods

form sums, hence whenever degQ � degP + 1.

161

VII.8.131 2 3 P L K

Show that the analytic di¤erential f (z) dz transforms the change of variablew = 1=z to �f (1=w) dw=w2. Show that the residue of f (z) at z = 1coincides with that of �f (1=w) =w2 at w = 0.

SolutionWe have that Res [f (z) ;1] = �a1 if f (z) =

P1�1 ajz

j, jzj > R.We transform f (z) dz by the change of variable z = 1

w, then dz = � 1

w2dw,

and f (z) dz = �f�1w

�1w2dw.

Now we take the residue for �f (1=w) dw=w2 at w = 0, we have that

� f�1

w

�1

w2= � 1

w2

1X�1

ajwj=

1X�1

(�aj)w�j�2 =

= � � � � a�4w2 � a�3w � a�2 � a�1w�1 � a0w�2 � � � � ;

thus

Res

��f�1

w

�1

w2; 0

�= �a�1

We have that

Res [f (z) ;1] = �a1 = Res��f�1

w

�1

w2; 0

�:

162

VII.9.1Show using residue theory thatZ 1

0

13px2 � x3

dx =2�p3

3:

SolutionSet

I =

Z 1

0

13px2 � x3

dx

and integrate

f (z) =

Z 1

0

13pz2 � z3

dx =1

3

qjzj2e2i arg� z=3 3

pj1� zjei arg0(1�z)=3

along the contour in Figure VII.9.1.

163

VIII.9.1arg� z arg0 z

π π π 0 0 0

−π −π −π 0 0 0­2 ­1 1 2x

y

π π π 0 0 0

π π π 2π 2π 2π­2 ­1 1 2x

y

arg� z arg0 (1� z)

π π π 0 0 0

−π −π −π 0 0 0­2 ­1 1 2x

y

2π 2π 2π 2π π π

0 0 0 0 π π­2 ­1 1 2x

y

The argument for the function f (z) = 13pjzj2e2i arg� z=3 3

pj1�zjei arg0(1�z)=3

4π/3 4π/3 4π/3 2π/3 π/3 π/3

−2π/3 −2π/3 −2π/3 0 π/3 π/3­2 ­1 1 2x

y

Residue at a simple pole at z1 =1 , where by result from Exercise VII.8.13and Rule 1,

Res [f (z) ;1] = �Res�1

w2f

�1

w

�; 0

�= �Res

24 1w2

1

3

q1w2

�1� 1

w

� ; 035 =

= �Res�1

w

13pw � 1

; 0

�= � lim

w!0

13pjw � 1jei arg(w�1)=3

= �e��i=3

Integrate along 1, and let "! 0+. This gives

164

Z 1

f (z) dz =

=

Z 1�"

"

1

3

qjzj2e2i arg� z=3 3

pj1� zjei arg0(1�z)=3

dz =

=

Z 1�"

"

1

3

qjxj2e2i�0=3 3

pj1� xjei�2�=3

dx = e�2�i=3Z 1�"

"

13px2 � x3

dx!

! e�2�i=3Z 1

0

13px2 � x3

dx = e�2�i=3I:

Integrate along 2, and let "! 0+. This gives����Z 4

f (z) dz

���� � "4p" (1� ")

� 2�" � 2�"9=2 ! 0:

Integrate along 3, and let "! 0+. This gives

Z 3

f (z) dz =

=

Z "

1�"

1

3

qjzj2e2i arg� z=3 3

pj1� zjei arg0(1�z)=3

dz =

=

Z "

1�"

1

3

qjxj2e2i�0=3 3

pj1� xjei�0=3

dx = �Z 1�"

"

13px2 � x3

dx!

! �Z 1

0

13px2 � x3

dx = �I:

Integrate along 4, and let "! 0+. This gives����Z 4

f (z) dz

���� � "4p" (1� ")

� 2�" � 2�"9=2 ! 0:

Using the Residue Theorem and letting R!1 and "! 0+, we obtain that

165

e�2�i=3I + 0� I + 0 = 2�i ���e��i=3

�:

Solve for I, we obtain that

I =2�ie��i=3

1� e�2�i=3 =2�i

e�i=3 � e��i=3 =�

sin (�=3)=2�p3

3;

i.e., Z 1

0

13px2 � x3

dx =2�p3

3:

166

VII.9.2Show using residue theory thatZ 1

0

1

xpx2 � 1

dx =�

2:

Solution

­R R

γ

γ

2

6

γ8

γ4 γ1γ3

γ7γ5

VII.9.2

Set

I =

Z 1

0

1

xpx2 � 1

dx

and integrate

f (z) =1

zpz2 � 1

=1

zpjz + 1jei arg0(z+1)=2

pjz � 1jei arg�(z�1)=2

along the contour in Figure VII.9.2.

167

VIII.9.2arg0 z arg� z

π π π 0 0 0

π π π 2π 2π 2π­2 ­1 1 2x

y

π π π 0 0 0

−π −π −π 0 0 0­2 ­1 1 2x

y

arg0 (z + 1) arg0 (z � 1)

π π 0 0 0 0

π π 2π 2π 2π 2π­2 ­1 1 2x

y

π π π π 0 0

−π −π −π −π 0 0­2 ­1 1 2x

y

The argument for the function f (z) = 1

zpjz+1jei arg0(z+1)=2

pjz�1jei arg�(z�1)=2

π π π/2 π/2 0 0

0 0 π/2 π/2 π π­2 ­1 1 2x

y

Residue at a simple pole at z1 = i, where by Rule 3,

Res

"1

zpjz + 1jei arg(z+1)=2

pjz � 1jei arg(z�1)=2

; 0

#=

1pjz + 1jei arg(z+1)=2

pjz � 1jei arg(z�1)=2

�����z=0

=1

e�i=2= �i

Integrate along 1, and let R!1 and "! 0+. This gives

168

Z 1

f (z) dz =

=

Z R

1+"

1

zpjz + 1jei arg0(z+1)=2

pjz � 1jei arg�(z�1)=2

dz =

=

Z R

1+"

1

xpjx+ 1jei�0=2

pjx� 1jei�0=2

dx =

=

Z R

1+"

1

xpx2 � 1

dx!

!Z 1

1

1

xpx2 � 1

dx = I:

Integrate along 2, and let "! 0+. This gives����Z 2

f (z) dz

���� � 1

RpR2 � 1

� �R � �

R! 0:

Integrate along 3, and let R!1 and "! 0+. This gives

Z 3

f (z) dz =

=

Z �1�"

�R

1

zpjz + 1jei arg0(z+1)=2

pjz � 1jei arg�(z�1)=2

dz =

=

Z �1�"

�R

1

xpjx+ 1jei��=2

pjx� 1jei��=2

dx =Z �R

�1�"

1

xpx2 � 1

dx =

�x = �tdx = �dt

�=

Z R

1

1

tpt2 � 1

dt = I:

Integrate along 4, and let "! 0+. This gives����Z 4

f (z) dz

���� � "4p" (1� ")

� 2�" � 2�"9=2 ! 0:

Integrate along 5, and let R!1 and "! 0+. This gives

169

Z 5

f (z) dz =

=

Z �R

�1�"

1

zpjz + 1jei arg0(z+1)=2

pjz � 1jei arg�(z�1)=2

dz =

=

Z �R

�1�"

1

xpjx+ 1jei��=2

pjx� 1jei�(��)=2

dx =Z �R

�1�"

1

xpx2 � 1

dx =

�x = �tdx = �dt

�=

Z R

1

1

tpt2 � 1

dt = I:

Integrate along 6, and let "! 0+. This gives����Z 6

f (z) dz

���� � 1

RpR2 � 1

� �R � �

R! 0:

Integrate along 7, and let R!1 and "! 0+. This gives

Z 7

f (z) dz =

=

Z 1+"

R

1

zpjz + 1jei arg0(z+1)=2

pjz � 1jei arg�(z�1)=2

dz =

=

Z 1+"

R

1

xpjx+ 1jei�2�=2

pjx� 1jei�0=2

dx =

=

Z R

1+"

1

xpx2 � 1

dx!

!Z 1

1

1

xpx2 � 1

dx = I:

Integrate along 8, and let "! 0+. This gives����Z 4

f (z) dz

���� � "4p" (1� ")

� 2�" � 2�"9=2 ! 0:

Using the Residue Theorem and letting R!1 and "! 0+, we obtain that

170

I + 0 + I + 0 + I + 0 + I + 0 = 2�i � (�i) ;i.e., Z 1

0

1

xpx2 � 1

dx =�

2:

171

VII.9.3Show using residue theory thatZ 1

�1

1

(x2 + 1)p1� x2

dx =�p2

Solution

­R R

γ2

γ8

γ4

γ6

γ1

γ3

γ7

γ5

VII.9.3

Set

I =

Z 1

�1

1

(x2 + 1)p1� x2

dx

and integrate

f (z) =1

(z2 + 1)p1� x2

=1

(z � i) (z + i)pj1 + zjei arg�(1+z)=2

pj1� zjei arg0(1�z)=2

along the contour in Figure VII.9.3.

172

VIII.9.3arg� z arg0 z

π π π 0 0 0

−π −π −π 0 0 0­2 ­1 1 2x

y

π π π 0 0 0

π π π 2π 2π 2π­2 ­1 1 2x

y

arg� (1 + z) arg0 (1� z)

π π 0 0 0 0

−π −π 0 0 0 0­2 ­1 1 2x

y

2π 2π 2π 2π π π

0 0 0 0 π π­2 ­1 1 2x

y

The argument for the function f (z) = 1

(z�i)(z+i)pj1+zjei arg�(1+z)=2

pj1�zjei arg0(1�z)=2

3π/2 3π/2 π π π/2 π/2

−π/2 −π/2 0 0 π/2 π/2­2 ­1 1 2x

y

Residue at a simple pole at z1 = i, where by Rule 3,

Res

"1

(z � i) (z + i)pj1 + zjei arg(1+z)=2

pj1� zjei arg(1�z)=2

; i

#=

1

(z + i)pj1 + zjei arg(1+z)=2

pj1� zjei arg(1�z)=2

�����z=i

=

1

(i+ i)pj1 + zjei arg(1+i)=2

pj1� zjei arg(1�i)=2

=1

2ipj1 + zje(i�=4)=2

pj1� zje(i7�=4)=2

=1

2ipp

2pp

2e�i=

1

2ip2(�1)

= � 1

2p2i=

1

2p2i

173

Residue at a simple pole at z2 = �i, where by Rule 3,

Res

"1

(z � i)pj1 + zjei arg(1+z)=2

pj1� zjei arg(1�z)=2

;�i#=

=1

(z � i)pj1 + zjei arg(1+z)=2

pj1� zjei arg(1�z)=2

�����z=�i

=

1

(�i� i)pj1 + zjei arg(1�i)=2

pj1� zjei arg(1+i)=2

=1

�2ipj1 + zje(�i�=4)=2

pj1� zje(i�=4)=2

=

=1

�2ipp

2pp

2e0i=

1

�2ip2= � 1

2p2i=

1

2p2i

Integrate along 1, and let "! 0+. This gives

Z 1

f (z) dz =

=

Z 1�"

�1+"

1

(z2 + 1)pj1 + zjei arg�(1+z)=2

pj1� zjei arg0(1�z)=2

dz =

=

Z 1�"

�1+"

1

(z2 + 1)pj1 + xjei0=2

pj1� xjei2�=2

dx =�Z 1�"

�1+"

1

(x2 + 1)p1� x2

dx!

�Z 1

�1

1

(x2 + 1)p1� x2

dx = �I:

Integrate along 2, and let "! 0+. This gives����Z 4

f (z) dz

���� � "4p" (1� ")

� 2�" � 2�"9=2 ! 0:

Integrate along 3, and let "! 0+. This gives

174

Z 1

f (z) dz =

=

Z �1+"

1�"

1

(z2 + 1)pj1 + zjei arg�(1+z)=2

pj1� zjei arg0(1�z)=2

dz =

=

Z �1+"

1�"

1

(z2 + 1)pj1 + xjei�0=2

pj1� xjei�0=2

dx =�Z 1�"

�1+"

1

(x2 + 1)p1� x2

dx!

! �Z 1

�1

1

(x2 + 1)p1� x2

dx = �I:

Integrate along 4, and let "! 0+. This givesZ 4

f (z) dz ! 0:

Integrate along 5, and let R!1 and "! 0+. This gives

Z 5

f (z) dz =

=

Z �R

�1�"

1

(z2 + 1)pj1 + zjei arg(1+z)=2

pj1� zjei arg(1�z)=2

dz =

=

Z �R

�1�"

1

(z2 + 1)pj1 + xjei(��)=2

pj1� xjei0=2

dx = �Z �R

�1�"

1

(x2 + 1)p1� x2 (�i)

dx!

! �Z �1

�1

1

(x2 + 1)p1� x2 (�i)

dx = �J:

Integrate along 6, and let R!1. This gives����Z 6

f (z) dz

���� � 1

(R2 � 1)pR2 � 1

� 2�R � 2�

R3! 0:

Integrate along 7, and let R!1 and "! 0+. This gives

175

Z 7

f (z) dz =

=

Z �1�"

�R

1

(z2 + 1)pj1 + zjei arg(1+z)=2

pj1� zjei arg(1�z)=2

dz =

=

Z �1�"

�R

1

(z2 + 1)pj1 + xjei�=2

pj1� xjei2�=2

dx = �Z �R

�1�"

1

(x2 + 1)p1� x2 (�i)

dx!

!Z �1

�1

1

(x2 + 1)p1� x2 (�i)

dx = �J:

Integrate along 8, and let "! 0+. This gives����Z 4

f (z) dz

���� � "4p" (1� ")

� 2�" � 2�"9=2 ! 0:

Using the Residue Theorem and letting R!1 and "! 0+, we obtain that

�I + 0� I � J + 0 + J + 0 = 2�i ��1

2p2i+

1

2p2i

�i.e., Z 1

�1

1

(x2 + 1)p1� x2

dx =�p2:

176

VII.9.4Show using residue theory thatZ 1

�1

1

3

q(1� x) (1 + x)2

dx =2�p3:

SolutionSet

I =

Z 1

�1

1

3

q(1� x) (1 + x)2

dx

and integrate

f (z) =1

3

q(1� z) (1 + z)2

=1

3pj1� zjei arg0(1�z)=3

qj1 + zj2e2i arg�(1+z)=3

along the contour in Figure VII.9.4.

177

VIII.9.4arg0 z arg� z

π π π 0 0 0

π π π 2π 2π 2π­2 ­1 1 2x

y

π π π 0 0 0

−π −π −π 0 0 0­2 ­1 1 2x

y

arg0 (1� z) arg� (1 + z)

2π 2π 2π 2π π π

0 0 0 0 π π­2 ­1 1 2x

y

π π 0 0 0 0

−π −π 0 0 0 0­2 ­1 1 2x

y

The argument for the function f (z) = 13pj1�zjei arg0(1�z)=3

pj1+zj2e2i arg�(1+z)=3

4π/3 4π/3 2π/3 2π/3 π/3 π/3

−2π/3 −2π/3 0 0 π/3 π/3­2 ­1 1 2x

y

Residue at a simple pole at z1 =1 , where by result from Exercise VII.8.13and Rule 1,

178

Res [f (z) ;1] = �Res�1

w2f

�1

w

�; 0

�= �Res

24 1w2

1

3

q�1� 1

w

� �1 + 1

w

�2 ; 035 =

= �Res

24 1w

1

3

q(w � 1) (w + 1)2

; 0

35 =� lim

w!0

24 1

3pjw � 1jei arg(w�1)=3

qjw + 1j2e2i arg(w+1)=3

35 = �e��i=3Integrate along 1, and let "! 0+. This gives

Z 1

f (z) dz =

=

Z 1�"

�1+"

1

3pj1� zjei arg0(1�z)=3

qj1 + zj2e2i arg�(1+z)=3

dz =

=

Z 1�"

�1+"

1

3pj1� xjei2�=3

qj1 + xj2e2i�0=3

dx = e�2�i=3Z 1�"

�1+"

1

3

q(1� x) (1 + x)2

dx!

! e�2�i=3Z 1

�1

1

3

q(1� x) (1 + x)2

dx = e�2�i=3I:

Integrate along 2, and let "! 0+. This gives����Z 4

f (z) dz

���� � "4p" (1� ")

� 2�" � 2�"9=2 ! 0:

Integrate along 3, and let "! 0+. This gives

179

Z 3

f (z) dz =

=

Z �1+"

1�"

1

3pj1� zjei arg0(1�z)=3

qj1 + zj2e2i arg�(1+z)=3

dz =

=

Z �1+"

1�"

1

3pj1� xjei�0=3

qj1 + xj2e2i�0=3

dx = �Z 1�"

�1+"

1

3

q(1� x) (1 + x)2

dx!

!Z 1

�1

1

3

q(1� x) (1 + x)2

dx = �I:

Integrate along 4, and let "! 0+. This gives����Z 4

f (z) dz

���� � "4p" (1� ")

� 2�" � 2�"9=2 ! 0:

Using the Residue Theorem and letting R!1 and "! 0+, we obtain that

e�2�i=3I + 0� I + 0 = 2�i ���e��i=3

�:

Solve for I, we obtain that

I =�2�ie��i=3e�2�i=3 � 1 =

2�i

e�i=3 � e��i=3 =�

sin (�=3)=2�p3;

i.e., Z 1

�1

1

3

q(1� x) (1 + x)2

dx =2�p3:

180

VII.9.5Show using residue theory thatZ 1

0

1

(x+ 1) 4

qx (1� x)3

dx =�4p2:

SolutionSet

I =

Z 1

0

1

(x+ 1) 4

qx (1� x)3

dx

and integrate

f (z) =1

(z + 1) 4

qz (1� z)3

=1

(z + 1) 4pjzjei arg�z=4 4

qj1� zj3e3i arg0(1�z)=4

along the contour in Figure VII.9.5.

181

VIII.9.5arg� z arg0 z

π π π 0 0 0

−π −π −π 0 0 0­2 ­1 1 2x

y

π π π 0 0 0

π π π 2π 2π 2π­2 ­1 1 2x

y

arg� z arg0 (1� z)

π π π 0 0 0

−π −π −π 0 0 0­2 ­1 1 2x

y

2π 2π 2π 2π π π

0 0 0 0 π π­2 ­1 1 2x

y

The argument for the function f (z) = 1

(z+1) 4pjzjei arg�z=4 4

pj1�zj3e3i arg0(1�z)=4

7π/4 7π/4 7π/4 3π/2 3π/4 3π/4

−π/4 −π/4 −π/4 3π/4 3π/4 3π/4­2 ­1 1 2x

y

Residue at a simple pole at z1 = �1, where by Rule 3,

182

Res

24 1

(z + 1) 4pjzjei arg(z)=4 4

qj1� zj3e3i arg(1�z)=4

;�1

35 =1

4pjzjei arg(z)=4 4

qj1� zj3e3i arg(1�z)=4

������z=�1

=

=1

4pjzjei arg(z)=4 4

qj1� zj3e3i arg(1�z)=4

=

=1

4p8e��i=4

=14p8

�1p2+ i

1p2

�=

1

2 4p2+ i

1

2 4p2

Residue at a simple pole at z2 =1 , where by result from Exercise VII.8.13and Rule 1,

Res [f (z) ;1] = �Res�1

w2f

�1

w

�; 0

�= �Res

24 1w2

1�1w+ 1�

4

q1w

�1� 1

w

�3 ; 035 =

= � limw!0

24 1

(1 + w) 4

q(w � 1)3

; 0

35 = 0Residue at a simple pole at z2 = �i, where by Rule 3,Integrate along 1, and let R!1 and "! 0+. This gives

Z 1

f (z) dz =

=

Z 1�"

"

1

(z + 1) 4pjzjei arg�z=4 4

qj1� zj3e3i arg0(1�z)=4

dz =

=

Z 1�"

"

1

(x+ 1) 4pjxjei�0=4 4

qj1� xj3e3i�2�=4

dx = i

Z 1�"

"

1

(x+ 1) 4

qx (1� x)3

dx!

! i

Z 1

0

1

(x+ 1) 4

qx (1� x)3

dx = iI:

183

Integrate along 2, and let "! 0+. This gives����Z 4

f (z) dz

���� � "4p" (1� ")

� 2�" � 2�"9=2 ! 0:

Integrate along 3, and let R!1 and "! 0+. This gives

Z 3

f (z) dz =

=

Z "

1�"

1

(z + 1) 4pjzjei arg�z=4 4

qj1� zj3e3i arg0(1�z)=4

dz =

=

Z "

1�"

1

(x+ 1) 4pjxjei�0=4 4

qj1� xj3e3i�0=4

dx = �Z 1�"

"

1

(x+ 1) 4

qx (1� x)3

dx!

! �Z 1

0

1

(x+ 1) 4

qx (1� x)3

dx = �I:

Integrate along 4, and let "! 0+. This gives����Z 4

f (z) dz

���� � "4p" (1� ")

� 2�" � 2�"9=2 ! 0:

Using the Residue Theorem and letting R!1 and "! 0+, we obtain that

iI + 0� I + 0 = 2�i ��

1

2 4p2+ i

1

2 4p2+ 0

�;

i.e., Z 1

0

1

(x+ 1) 4

qx (1� x)3

dx =�4p2:

184

VIII 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 1912345678

1

VIII.1.11 2 3 P L K

LLLShow that z4 + 2z2 � z + 1 has exactly one root in each quadrant.

Solution

VIII.1.1

Set p (z) = z4+2z2�z+1, and compute4 arg p (z) along the three segmentsin the sector path in �gure VIII.1.1. First of all p (z) have no real zeros.

The positive real axis 1 from 0 toR we parametizize as x : z = t; 0 � t � R,and our function p (z) becomes p (t) = t4 + 2t2 � t + 1 = (t2 + 1)2 � t > 0.Since p (x) > 0 for t � 0, their is no change of argument on this line segmenton the real axis, thus 4 arg p (z) = 0.

The arc 2 from R to iR we parametizize as 2 : z = Reit, 0 � t � �=2,and our function p (z) becomes p (t) = R4e4it + 2R2e2it � Reit + 1. Sincethe variation in argument is determined by the dominating term R4e4it for Rlarge the change of argument on this arc is 4 times the variation in argumentfor the arc, and thus 4 arg p (z) � 4 � �

2= 2�.

The positive imaginary axis 3 from iR to 0 we parametizize as 3 : z =it; 0 � t � R, and our function p (z) becomes f (t) = (it)4 + 2 (it)2 � it + 1= (t2 � 1)2 � it . We �nd that the real part have a zero at t = 1, and theimaginary part have a zero for t = 0. We make the following table and dothe sketches,

2

t Re z Im z arg z

1 + � 01 0 � ��

2

0 1 0 0

VIII.1.1 (R = 4)

­2 ­1 1 2 3 4

­2

­1

1

2

3

4

VIII.1.1 (R = 4)

Thus when we move from iR to 0, the argument for p (z) remain in the 4-thquadrant except for touching the imaginary axis at �i and terminating at1, and their is no change of argument on this line segment on the imaginaryaxis, thus 4 arg p (z) � 0.

Now we have that the total change of argument for p (z) is � 2�, so we haveexactly one zero in the �rst quadrat. Because that the roots come in complexconjugate pairs it is plain that p (z) have a root in the fourth quadrant too.Becauce that p (x) = (x2 + 1)2�x > 0 it is clear that p (z) have no real rootsand p (z) must have on has one zero in each quadrant.

3

VIII.1.21 2 3 P L K

Find the number of zeros of the polynomial p (z) = z4+z3+4z2+3z+2in each quadrant.

Solution

VIII.1.2

Set p (z) = z4 + z3 + 4z2 + 3z + 2, and compute 4 arg p (z) along the threesegments in the sector path in �gure VIII.1.2. First of all p (z) have no realzeros.

The positive real axis 1 from 0 toR we parametizize as x : z = t; 0 � t � R,and our function p (z) becomes p (t) = t4+t3+4t2+3t+2 > 0. Since p (x) > 0for t � 0, their is no change of argument on this line segment on the realaxis, thus 4 arg p (z) = 0.

The arc 2 from R to iR we parametizize as 2 : z = Reit, 0 � t � �=2, and

our function p (z) becomes p (t) = R4e4it+R3e3it+4R2e2it+3Reit+2. Sincethe variation in argument is determined by the dominating term R4e4it for Rlarge the change of argument on this arc is 4 times the variation in argumentfor the arc, and thus 4 arg p (z) � 4 � �

2= 2�.

The positive imaginary axis 3 from iR to 0 we parametizize as 3 : z =it; 0 � t � R, and our function p (z) becomes f (t) = (it)4 + (it)3 + 4 (it)2 +3it+ 2 = t4 � 4t2 + 2 + i (�t3 + 3t) . We �nd that the real part have a zero

4

at t =p2�

p2 and t =

p2 +

p2, and the imaginary part have a zero for

t = 0 and t =p3. We make the following table and do the sketches,

t Re z Im z arg z

1 + � 0p2 +

p2 0 � ��

2p3 � 0 ��p2�

p2 0 + �3�

2

0 + 0 �2�

VIII.1.2 (R = 10)

­4 ­2 2 4

­4

­2

2

4

VIII.1.2 (R = 10)

Thus when we move from iR to 0, the argument for p (z) on this line segmenton the imaginary axis changes so that, 4 arg p (z) � �2�.

Now we have that the total change of argument for p (z) is � 0, so we haveno zero in the 1st quadrant.Since p (z) has real coe¢ cients, the roots come in conjugate pairs (since thereare nor real roots), therefore there are no roots in the 4th quadras as well.By symmetry, there are 2 roots each in the 2nd and 3rd quadrants.

5

VIII.1.31 2 3 P L K

Find the number of zeros of the polynomial p (z) = z6 + 4z4 + z3 +2z2 + z + 5 in the �rst quadrant fRe z > 0; Im z > 0g.

Solution

VIII.1.3

Set p (z) = z6 + 4z4 + z3 + 2z2 + z + 5, and compute 4 arg p (z) along thethree segments in the sector path in �gure VIII.1.3. First of all p (z) have noreal zeros.

The positive real axis 1 from 0 toR we parametizize as x : z = t; 0 � t � R,and our function p (z) becomes p (t) = t6 + 4t4 + t3 + 2t2 + t+ 5 > 0. Sincep (t) > 0 for t � 0, their is no change of argument on this line segment onthe real axis, thus 4 arg p (z) = 0.

The arc 2 from R to iR we parametizize as 2 : z = Reit, 0 � t � �=2, and

our function p (z) becomes p (t) = R6e6it+4R4e4it+3R3e3it+2R2e2it+Reit+5.Since the variation in argument is determined by the dominating term R6e6it

for R large the change of argument on this arc is 6 times the variation inargument for the arc, and thus 4 arg p (z) � 6 � �

2= 3�.

The positive imaginary axis 3 from iR to 0 we parametizize as 3 : z =it; 0 � t � R, and our function p (z) becomes f (t) = (it)6 + 4 (it)4 + (it)3 +2 (it)2+ it+5 = �t6+4t4� 2t2+5+ i (�t3 + t) . We �nd that the real parthave a zero at t � 1; 95, and the imaginary part have a zero for t = 0 andt = 1. We make the following table and do the sketches,

6

t Re z Im z arg z

1 � � ��1; 95 0 � ��

2

1 + 0 00 + 0 0

­1000 ­500 500 1000

­500

500

1000

VIII.1.3  (R = 3)

­2 2 4 6 8 10

­6

­4

­2

VIII.1.3  (R = 3)

Thus when we move from iR to 0, the argument for p (z) on this line segmenton the imaginary axis changes so that, 4 arg p (z) � �.

Now we have that the total change of argument for p (z) is � 4�, so wehave that p (z) has exactly two zeros in the �rst quadrat. (None on real orimaginary axis)

7

VIII.1.41 2 3 P L K

LLLFind the number of zeros of the polynomial p (z) = z9+2z5�2z4+z+3in the right half-plane.

Solution

VIII.1.4

Set p (z) = z9 + 2z5 � 2z4 + z + 3, and compute 4 arg p (z) along the threesegments in the sector path in �gure VIII.1.4. First of all p (z) have no realzeros.

The positive real axis 1 from 0 toR we parametizize as x : z = t; 0 � t � R,and our function p (z) becomes p (t) = t9 + 2t5 � 2t4 + t+ 3. Since p (x) > 0for t � 0, their is no change of argument on this line segment on the realaxis, thus 4 arg p (z) = 0.

The arc 2 from R to iR we parametizize as 2 : z = Reit, 0 � t � �=2, and

our function p (z) becomes p (t) = R9e9it+2R5e5it�2R4e4it+Reit+3. Sincethe variation in argument is determined by the dominating term R9e9it for Rlarge the change of argument on this arc is 9 times the variation in argumentfor the arc, and thus 4 arg p (z) � 9 � �

2= 9�

2.

The positive imaginary axis 3 from iR to 0 we parametizize as 3 : z =it; 0 � t � R, and our function p (z) becomes f (t) = (it)9+2 (it)5�2 (it)4+it + 3 = �2t4 + 3 + iy (y4 + 1)2. We �nd that the real part have a zero att = 4

q32, and the imaginary part have only a zero for t = 0. We make the

following table and do the sketches,

8

t Re z Im z arg z

1 � + �2

4

q32

0 + �2

0 + 0 0­600­400­200 200 400 600

­600

­400

­200

200

400

600

VIII.1.4  (R = 2)

­20 ­10 10 20

10

20

30

VIII.1.4  (R = 2)

Thus when we move from iR to 0, the argument is changed from �=2 to 0, sothe charge of argument on this on the imaginary axis is �, thus 4 arg p (z) =��=2.

Now we have that the total change of argument for p (z) is � 4�, so wehave exactly two zeros in the �rst quadrat. Because that the roots comein complex conjugate pairs it is plain that p (z) has four zeros in the righthalf-plane.

9

VIII.1.51 2 3 P L K

For a �xed real number �, �nd the number of zeros z4+z3+4z2+�z+3satisfying Re z < 0. (Your answer depends on �.)

Solution

VIII.1.5

Set p (z) = z4 + z3 + 4z2 + �z + 3, and compute 4 arg p (z) along the threesegments in the contuour in �gure VIII.1.5.

The positive real axis 1 from 0 toR we parametizize as x : z = t; 0 � t � R,and our function p (z) becomes p (t) = t4 + t3 + 4t2 + at + 3 = (t2 + 1)2 � t> 0. Since p (x) > 0 for t � 0, their is no change of argument on this linesegment on the real axis, thus 4 arg p (z) = 0.

The arc 2 from R to iR we parametizize as 2 : z = Reit, 0 � t � �=2, and

our function p (z) becomes p (t) = R4e4it+R3e3it+4R2e2it+aReit+3. Sincethe variation in argument is determined by the dominating term R4e4it for Rlarge (independent of the value of a) the change of argument on this arc is 4times the variation in argument for the arc, and thus4 arg p (z) � 4 � �

2= 2�.

The positive imaginary axis 3 from iR to 0 we parametizize as 3 : z =it; 0 � t � R, and our function p (z) becomes f (t) = (it)4 + (it)3 + 4 (it)2 +ait + 3 = (t2 � 1) (t2 � 3) + i (�t3 + at). We �nd that the real part have azero at t = 1 and t =

p3 and the imaginary part have a zero for t = 0 and

and a second root t =pa if a � 0. We make the following table and do the

sketches,

10

a < 0

t Re z Im z arg z

1 + � 0p3 0 � ��

2

1 0 � ��2

0 + 0 0

VIII.1.5 (a = ­4, R = 10) VIII.1.5 (a = ­4, R = 10)

a = 0

t Re z Im z arg z

1 + � 0p3 0 � ��

2

1 0 � ��2

0 + 0 0

VIII.1.5 (a = 0, R = 10) VIII.1.5 (a = 0, R = 10)

0 < a < 1

t Re z Im z arg z

1 + � 0p3 0 � ��

2

1 0 � ��2p

a + 0 00 + 0 0

VIII.1.5 (a = 1/2, R = 10) VIII.1.5 (a = 1/2, R = 10)

11

a = 1

t Re z Im z arg z

1 + � 0p3 0 � ��

2

1 0 0 �0 + 0 0

VIII.1.5 (a = 1, R = 10) VIII.1.5 (a = 1, R = 10)

1 < a < 3

t Re z Im z arg z

1 + � 0p3 0 � ��

2pa � 0 ��

1 0 + �3�2

0 + 0 �2�

VIII.1.5 (a = 2, R = 10) VIII.1.5 (a = 2, R = 10)

a = 3

t Re z Im z arg z

1 + � 0p3 0 0 �

1 0 + �2

0 + 0 0

VIII.1.5 (a = 3, R = 10) VIII.1.5 (a = 3, R = 10)

12

a > 3

t Re z Im z arg z

1 + � 0pa + 0 0p3 0 + �

2

1 0 + �2

0 + 0 0

VIII.1.5 (a = 9, R = 10) VIII.1.5 (a = 9, R = 10)

Thus when we move from iR to 0, the argument for p (z) only change if1 < a < 3 and we have that 4 arg p (z) � �2�, for all other value on a wehave that 4 arg p (z) � 0. Remark that we can se that for a = 1 and a = 3we have zeros on the imaginary axis.

Now we have that the total change of argument for p (z) is � 2�, so we haveexactly one zero in the �rst quadrat. Because that the roots come in complexconjugate pairs it is plain that p (z) has one zero in each quadrant.

If 1 � � � 3, the total change of argument for p (z) is � 0 and no zeros onthe imaginary axis, thus we must have four zeros in the open left half-plane.If � < 1 and � > 3, the total charge of argument for p (z) is � 2� and nozeros of the imaginary axis, thus we must have two zeros in the open lefthalf-plane.If � = 1 and � = 3, the total charge of argument for p (z) is � 2� and wehave two zeros on the imaginary axis, thus we must have remaining two zerosin the open left half-plane.

13

VIII.1.61 2 3 P L K

For a �xed real number �, �nd the number of solutions of z5+2z3�z2+z = �satisfying Re z > 0.

Solution

VIII.1.6

Set p (z) = z5 + 2z3 � z2 + z � �, and compute 4 arg p (z) along the twosegments in the contuour in �gure VIII.1.6.

The arc 1 from�iR to iR we parametizize as 2 : z = Reit, ��=2 � t � �=2,and our function p (z) becomes p (t) = R5e5it+2R3e3it�R2e2it+Reit��. Sincethe variation in argument is determined by the dominating term R5e5it for Rlarge (independent of the value of �) the change of argument on this arc is 5times the variation in argument for the arc, and thus4 arg p (z) � 5 � �

2= 5�

2.

The positive imaginary axis 2 from iR to �iR we parametizize as 2 : z =it; �R � t � R, and our function p (z) becomes f (t) = (it)5 + 2 (it)3 �(it)2+ it�� = t2��+ it (t2 � 1)2. We �nd that the real part have a zero att = �

p� and t =

p� and the imaginary part have a zero for t = �1, t = 0

and t = 1. We make the following table and do the sketch,

14

t Re z Im z arg z

1

VIII.1.6 (a = 4, R = 4) VIII.1.6 (a = 4, R = 4)

t Re Im arg

1 + 0 0p3 0 � ��

2pa � 0 ��

1 0 + �3�2

0 3 0 �2�

VIII.1.5 (a = 2, R = 10) VIII.1.5 (a = 2, R = 10)

t Re z Im z arg z

1

VIII.1.5 (a = 0, R = 10) VIII.1.5 (a = 0, R = 10)

15

Thus when we move from iR to 0, the argument for p (z) remain in the 4-thquadrant except for touching the imaginary axis at �i and terminating at1, and their is no change of argument on this line segment on the imaginaryaxis, thus 4 arg p (z) = 0.

Now we have that the total change of argument for p (z) is � 2�, so we haveexactly one zero in the �rst quadrat. Because that the roots come in complexconjugate pairs it is plain that p (z) has one zero in each quadrant.

16

VIII.1.71 2 3 P L K

For a �xed complex number �, show that if m and n are large integers, thenthe equation ez = z + � has exactly m + n solutions in the horizontal stripf�2�im < Im z < 2�ing.

Solution

VIII.1.7

Set p (z) = z4+2z2�z+1, and compute4 arg p (z) along the three segmentsin the contuour in �gure VIII.1.7.The positive real axis 1 from 0 toR we parametizize as x : z = t; 0 � t � R,and our function p (z) becomes p (t) = t4 + 2t2 � t + 1 = (t2 + 1)2 � t > 0.Since p (x) > 0 for t � 0, their is no change of argument on this line segmenton the real axis, thus 4 arg p (z) = 0.

The arc 2 from R to iR we parametizize as 2 : z = Reit, 0 � t � �=2,and our function p (z) becomes p (t) = R4e4it + 2R2e2it � Reit + 1. Sincethe variation in argument is determined by the dominating term R4e4it for Rlarge the change of argument on this arc is 4 times the variation in argumentfor the arc, and thus 4 arg p (z) � 4 � �

2= 2�.

The positive imaginary axis 3 from iR to 0 we parametizize as 3 : z =it; 0 � t � R, and our function p (z) becomes f (t) = (it)4 + 2 (it)2 � it + 1= (t2 � 1)2 � it . We �nd that the real part have a zero at t = 1, and theimaginary part have a zero for t = 0 on the positive imaginary axis. Wemake the following table and do the sketch,

17

t Re Im arg

1 + � 01 0 � ��

2

0 + 0 0 ­20 ­10 10 20

­20

­10

10

20

VIII.1.7

Thus when we move from iR to 0, the argument for p (z) remain in the 4-thquadrant except for touching the imaginary axis at �i and terminating at1, and their is no change of argument on this line segment on the imaginaryaxis, thus 4 arg p (z) = 0.

Now we have that the total change of argument for p (z) is � 2�, so we haveexactly one zero in the �rst quadrat. Because that the roots come in complexconjugate pairs it is plain that p (z) has one zero in each quadrant.

18

VIII.1.81 2 3 P L K

LLLShow that if Re� > 1, then the equation ez = z + � has exactly onesolution in the left half-plane.

Solution

VIII.1.8

Set p (z) = z4+2z2�z+1, and compute4 arg p (z) along the three segmentsin the contuour in �gure VIII.1.8.The positive real axis 1 from 0 toR we parametizize as x : z = t; 0 � t � R,and our function p (z) becomes p (t) = t4 + 2t2 � t + 1 = (t2 + 1)2 � t > 0.Since p (x) > 0 for t � 0, their is no change of argument on this line segmenton the real axis, thus 4 arg p (z) = 0.

The arc 2 from R to iR we parametizize as 2 : z = Reit, 0 � t � �=2,and our function p (z) becomes p (t) = R4e4it + 2R2e2it � Reit + 1. Sincethe variation in argument is determined by the dominating term R4e4it for Rlarge the change of argument on this arc is 4 times the variation in argumentfor the arc, and thus 4 arg p (z) � 4 � �

2= 2�.

The positive imaginary axis 3 from iR to 0 we parametizize as 3 : z =it; 0 � t � R, and our function p (z) becomes f (t) = (it)4 + 2 (it)2 � it + 1= (t2 � 1)2 � it . We �nd that the real part have a zero at t = 1, and theimaginary part have a zero for t = 0 on the positive imaginary axis. Wemake the following table and do the sketch,

19

t Re Im arg

1 + � 01 0 � ��

2

0 + 0 0 ­20 ­10 10 20

­20

­10

10

20

VIII.1.8

Thus when we move from iR to 0, the argument for p (z) remain in the 4-thquadrant except for touching the imaginary axis at �i and terminating at1, and their is no change of argument on this line segment on the imaginaryaxis, thus 4 arg p (z) = 0.

Now we have that the total change of argument for p (z) is � 2�, so we haveexactly one zero in the �rst quadrat. Because that the roots come in complexconjugate pairs it is plain that p (z) has one zero in each quadrant.

Set f (z) = ez�z��, Re� > 1, and get f (iy) = cos y�Re�+ i (sin y � y)�i Im�. Thus when we move from f (�iR) to f (iR), f (iy) remains in the lefthalf-plane and the change in arg f is � �.When we move from f (iR) to f (�iR) along f

�Rei�

�, �2< � < 3�

2, we cross

the real axis from the 4-th to the 1 st quadrant, thus the change in arg f isagain � �. (May also use Rouche�s Theorem.)

t Re Im arg

1 � + �2

4

q32

0 + �2

0 + + 0

­100 ­50 50 100

­100

­50

50

100

VIII.1.4 (R = 2)

­20 ­10 10 20

10

20

30

VIII.1.4  (R = 2)

20

t Re Im arg

1 � + �2

4

q32

0 + �2

0 + + 0

­400 ­200 200 400

­400

­200

200

400

VIII.1.4 (R = 2)

­20 ­10 10 20

10

20

30

VIII.1.4  (R = 2)

(it)4 � (it)3 + 13 (it)2 � it +36 = t4 + it3 � 13t2 � it+ 36

21

VIII.1.91 2 3 P L K

Show that if f (z) is analytic in a domain D, and if is a closed curve in Dsuch that the values of f (z) on lie in the slit plane Cn (�1; 0] then theincrease in the argument of f (z) around is zero.

Solution

22

VIII.2.11 2 3 P L K

LLLShow that 2z5 + 6z � 1 has one root in the interval 0 < x < 1 andfour roots in the annulus f1 < jzj < 2g.

SolutionSet �rst p (z) = f1 (z) + h1 (z), where f1 (z) = 2z5 and h1 (z) = 6z � 1. Onthe circle jzj = 2 we have,jf1 (z)j = j2z5j = 2 � 25 = 64 and jh1 (z)j = j6z � 1j � 13, then jf1 (z)j >jh1 (z)j on jzj = 2 so by Rouche�s Theorem p (z) has all 5 roots in the diskjzj = 2.Set now p (z) = f2 (z) + h2 (z), where f2 (z) = 6z and h2 (z) = 2z5 � 1. Onthe circle jzj = 1 we have, jf2 (z)j = j6zj = 6 and jh2 (z)j = j2z5 � 1j � 3,then jf2 (z)j > jh2 (z)j on jzj = 1 so by Rouche�s Theorem p (z) has 1 rootsin the disk jzj = 1.It follows that p (z) = 2z5 + 6z � 1 has 5 � 1 = 4 roots in the annulus1 < jzj < 2. As any complex roots come in conjugate pairs that have thesame magnitude, the root in the disk jzj = 1 must be real, we have thatp (0) = �1 and p (1) = 7, thus this root must be in the interval 0 < x < 1.

23

VIII.2.21 2 3 P L K

LLLHow many roots does z9 + z5� 8z3 + 2z + 1 have between the circlesfjzj = 1g and fjzj = 2g?

SolutionSet �rst p (z) = f1 (z)+h1 (z), where f1 (z) = z9 and h1 (z) = z5�8z3+2z+1.On the circle jzj = 2 we have,jf1 (z)j = jz9j = jzj9 = 29 = 512 and jh1 (z)j = jz5 � 8z3 + 2z + 1j � 101,then jf1 (z)j > jh1 (z)j on jzj = 2 so by Rouche�s Theorem p (z) has 9 rootsin the disk jzj = 2.Set now p (z) = f2 (z) + h2 (z), where f2 (z) = z9 + z5 + 2z + 1 and h2 (z) =�8z3. On the circle jzj = 1 we have, jf2 (z)j = j�8z3j = 8 and jh2 (z)j =jz9 + z5 + 2z + 1j � 5, then jf2 (z)j > jh2 (z)j on jzj = 2 so by Rouche�sTheorem p (z) has 3 roots in the disk jzj = 1.Thus it follows that p (z) = z9 + z5 � 8z3 + 2z + 1 has 9� 3 = 7 roots in theannulus 1 < jzj < 2.

24

VIII.2.31 2 3 P L K

LLLShow that if m and n are positive integers, the the polynomial

p (z) = 1 + z +z2

2!+ � � �+ z

m

m!+ 3zn

has exactly n zeros in the unit disk.

SolutionSet p (z) = f (z)+h (z), where f (z) = 3zn and h (z) = 1+ z+ z2

2!+ � � �+ zm

m!.

On the circle jzj = 1 we have,jf (z)j = j3znj = 3 jzjn = 3 � 1n = 3 and jh (z)j =

���1 + z + z2

2!+ � � �+ zm

m!

��� �P1k=0

1k!= e, then jf (z)j > jh (z)j on jzj = 1 so by Rouche�s Theorem

p (z) = 1+z+ z2

2!+ � � �+ zm

m!+3zn has exactly n roots in the unit disk jzj = 1.

25

VIII.2.41 2 3 P L K

LLLFix a complex number � such that j�j < 1. For n � 1, show that(z � 1)n ez �� has n zeros satisfying jz � 1j < 1 and no other zeros inthe right half-plane. Determine the multiplicity of the zeros.

SolutionSet p (z) = f (z) + h (z), where f (z) = (z � 1)n ez and h (z) = ��. On thecircle jz � 1j = 1 which we parametrize by z = 1 + ei�, 0 � � � 2� we have,jf (z)j = j(z � 1)n ezj =

���1 + ei� � 1���n ��e1+i��� = e1+cos � � 1, then 0 � � �2� and jh (z)j = j��j = j�j < 1 from the text, then jf (z)j > jh (z)j onjz � 1j = 1 so by Rouche�s Theorem p (z) = (z � 1)n ez �� has n roots inthe disk jz � 1j = 1. We have that f 0 (z) = n (z � 1)n�1 ez + (z � 1)n ez =(n+ z � 1) (z � 1)n�1 ez, thus the zeros of of f 0 (z) are at z = 1 and atz = 1 � n, where the only zeros in the right half-plane are at z = 1. Sincef (1) = ��, the zeros of f (z) must be simple unless � = 0, in which case wege a zero of order n at z = 1.

***********Här är lite konstigt den derivarande satsen, hur funkar den med nollställen***********

26

VIII.2.51 2 3 P L K

For a �xed � satisfying j�j < 1, show that (z � 1)n ez + � (z + 1)n hasn zeros in the right half-plane, which are all simple if � 6= 0.

Solution

27

VIII.2.61 2 3 P L K

Let p (z) = z6 + 9z4 + z3 + 2z + 4 be the polynomial treated in the examplein this section.(a)Determine which quadrants contain the four zeros of p (z) that lie inside theunit circle.(b)Determine which quadrants contain the two zeros of p (z) that lie outside theunit circle.(c)Show that the two zeros of p (z) that lie outside the unit circle satisfyfjz � 3ij < 1=10g.

Solution

28

VIII.2.71 2 3 P L K

LLLLet f (z) and g (z) be analytic functions on the bounded domain D thatextend continuously to @D and satisfy jf (z) + g (z)j < jf (z)j + jg (z)j on@D. Show that f (z) and g (z) have the same number of zeros in D, countingmultiplicity. Remark. This is a variant of Rouche�s theorem, in which thehypotheses are symmetric in f (z) and g (z). Rouche�s theorem is obtainedby setting h (z) = �f (z)�g (z). For the solution of the exercise, see Exercise9 in the preceding section.

Solution

29

VIII.2.81 2 3 P L K

LetD be a bounded domain, and let f (z) and h (z) be meromorphic functionson D that extend to be analytic on @D. Suppose that jh (z)j < jf (z)j on@D. Show by example that f (z) and f (z)+h (z) can have di¤erent numbersof zeros on D. What can be said about f (z) and f (z) + g (z)? Prove yourassertion.

Solution

30

VIII.2.91 2 3 P L K

Let f (z) be a continuously di¤erentiable function on a domain D. Supposethat for all complex constants a and b, the increase in the argument of f (z)+az+b around any small circle in D on which f (z)+az+b 6= 0 is nonnegative.Show that f (z) is analytic.

Solution

31

VIII.3.11 2 3 P L K

LLLLet ffk (z)g be a sequence of analytic functions onD that converges normallyto f (z), and suppose that f (z) has a zero of orderN at z0 2D. Use Rouche�stheorem to show that there exists � > 0 such that for k large, fk (z) hasexactly N zeros counting multiplicity on the disk fjz � z0j < �g.

Solution

32

VIII.3.21 2 3 P L K

LLLLet S be the family of univalent functions f (z) de�ned on the open unit diskfjzj < 1g that satisfy f (0) = 0 and f 0 (0) = 1. Show that S is closed undernormal convergence, that is, if a sequence in S converges normally to f (z),then f 2 S. Remark. It is also true, but more di¢ cult to prove, that S isa compact family of analytic functions, that is, every sequence in S has anormally convergent subsequence.

Solution

33

VIII.4.11 2 3 P L K

LLLSuppose D is a bounded domain with piecewise smooth boundary. Let f (z)be meromorphic and g (z) analytic on D. Suppose that both f (z) and g (z)extend analytically across the boundary of D, and that f (z) 6= 0 on @D.Show that

1

2�i

I@D

g (z)f 0 (z)

f (z)dz =

nXj=1

mjg (zj) ;

where z1; : : : ; zn are the zeros and poles of f (z), and mj is the order of f (z)at zj.

Solution

34

VIII.4.21 2 3 P L K

Let f (z) be a meromorphic function on the complex plane that is doublyperiodic. Suppose that the zeros and the poles of f (z) are at the pointsz1; : : : ; zn and at their translates by periods of f (z), and suppose no zj is atranslate by a period of another zk. Let mj be the order of f (z) at zj. Showthat

Pmjzj is a period of f (z). Hint. Integrate zf 0 (z) =f (z) around the

boundary of the fundamental parallelogram P constructed in Section VI.5.

Solution

35

VIII.4.31 2 3 P L K

LLLLet ffk (z)g be a sequence of analytic functions on a domainD that convergesnormally to f (z). Suppose that fk (z) attains each value w at most m times(counting multiplicity) in D. Show that either f (z) is constant, or f (z)attains each value w at most m times in D.

Solution

36

VIII.4.41 2 3 P L K

Let f (z) be an analytic function on the open unit disk D = fjzj < 1g.Suppose there is an annulus U = fr < jzj < 1g such that the restriction off (z) to U is one-to-one. Show that f (z) is one-to-one on D.

Solution

37

VIII.4.51 2 3 P L K

Let f (z) = p (z) =q (z) be a rational function, where p (z) and q (z) arepolynomials that are relatively prime (no common zeros). We de�ne thedegree of f (z) to be the larger of the degrees of p (z) and q (z). Denote thedegree of f (z) by d.(a)Show that each value w 2 C, w 6= f (1), is assumed d times by f (z) on C.(b)Show that f (z) attains each value w 2 C� d times on C� (as always, countingmultiplicity).

Solution

38

VIII.4.61 2 3 P L K

Let f (z) be a meromorphic function on the complex plane, and suppose thereis an integer m such that f�1 (w) has at most m points for all w 2 C. Showthat f (z) is a rational function.

39

VIII.4.71 2 3 P L K

Let F (z; w) be a continuous function of z and w that depends analyticallyon z for each �xed w, and let F1 (z; w) denote the derivative of F (z; w) withrespect to z. Suppose F (z0; w0) = 0, and F1 (z0; w0) 6= 0. Choose � suchthat F (x;w0) 6= 0 for 0 < jz � z0j � �.(a)Show that there exists � > 0 such that if jw � w0j < �, there is a uniquez = g (w) satisfying jz � z0j < � and F (z; w) = 0.(b)Show that

g (w) =1

2�i

Zj��z0j=�

�F1 (�; w)

F (�; w)d�; jw � w0j < �:

(c)Suppose further that F (z; w) is analytic in w for each �xed z, and let F2 (z; w)denote the derivative of F (z; w) with respect to w. Show that g (w) is ana-lytic, and

g0 (w) = �F2 (g (w) ; w) =F1 (g (w) ; w) :(d)Derive the inverse function theorem given in this section, together with theformula for the derivative of the inverse function, as a corollary of (a), (b),and (c).Remark. this is the implicit function theorem for analytic functions. Notethat a speci�c formula is given for the function g (w) de�ned implicitely byF (g (w) ; w) = 0.

40

VIII.4.81 2 3 P L K

LetD be a bounded domain, and let f (z) be a continuous function onD[@Dthat is analytic on D. Show that @ (f (D)) � f (@D), that is, the boundaryof the open set f (D) is contained in the image under f (z) of the boundaryof D.

Solution

41

VIII.5.11 2 3 P L K

Find the critical points and critical values of f (z) = z + 1=z. Sketch thecurves where f (z) is real. Sketch the regions where Im f (z) > 0 and whereIm f (z) < 0.

Solution.

42

VIII.5.21 2 3 P L K

Suppose g (z) is analytic at z = 0, with power series g (z) = 2+ iz4+O (z5).Sketch and label the curves passing through z = 0 where Re g (z) = 2 andIm g (z) = 0.

Solution

43

VIII.5.31 2 3 P L K

Find the critical points and critical values of f (z) = z2 + 1. Sketch the setof points z such that jf (z)j � 1, and locate the critical points of f (z) on thesketch.

Solution

44

VIII.5.41 2 3 P L K

Suppose that f (z) is analytic at z0. Show that if the set of z such thatRe f (z) = Re f (z0) consists of just one curve passing through z0, thenf 0 (z0) 6= 0. Show also that if the set of z such that jf (z)j = jf (z0)j consistsof just one curve passing through z0, then f 0 (z0) 6= 0.

Solution

45

VIII.5.51 2 3 P L K

How many critical points, counting multiplicity, does a polynomial of degreem have in the complex plane? Justify your answer.

Solution

46

VIII.5.61 2 3 P L K

Find and plot the critical points and critical values of f (z) = z2 + 1 and ofits iterates f (f (z)) = (z2 + 1)

2+ 1 and f (f (f (z))). Suggestion. Use the

chain rule.

Solution

47

VIII.5.71 2 3 P L K

Let f (z) be a polynomial of degreem. How many (�nite) critical points doesthe N � fold iterate f � � � � � f (N times) have? Describe them in terms ofthe critical points of f (z).

Solution

48

VIII.5.81 2 3 P L K

We de�ne a pole of f (z) to be a critical point of f (z) of order k if z0 is acritical point of 1=f (z) of order k. We de�ne z =1 to be a critical point off (z) of order k if w = 0 is a critical point of g (w) = f (1=w) of order k. Showthat with this de�nition, a point z0 2 C� is a critical point of order k for ameromorphic function f (z) if and only if there are open sets U containingz0 and V containing w0 = f (z0) such that each w 2 v, w 6= w0, has exactlyk + 1 preimages in U . Remark. We say that f (z) is a (k + 1) � sheetedcovering of f�1 (V n fw0g) \ U over V n fw0g.

Solution

49

VIII.5.91 2 3 P L K

Show that a polynomial of degree m, regarded as a meromorphic function onC�, has a critical point of order m� 1 at z0 =1.

Solution

50

VIII.5.101 2 3 P L K

Locate the critical points ana critical values in the extended complex plane ofthe polynomial f (z) = z4 � 2z2. Determine the order of each critical point.Sketch the set of points z such that Im z � 0.

Solution

51

VIII.5.111 2 3 P L K

Show that if f is a rational function, and if g is a fractional linear trans-formation, then f and g � f have the some critical points in the extendedcomplex plane C�. What can be said about the critical values of g �f? Whatcan be said about the critical points and critical values f � g?

Solution

52

VIII.5.121 2 3 P L K

Let f (z) = p (z) =q (z) be a rational function of degree d, so that p (z) andq (z) are reactively prime, and d is the larger of the degrees of p (z) andq (z). (See Exercise 4.5.) Show that f (z) has 2d� 2 critical points, countingmultiplicity, in the extended complex plane C�. Hint if deg p 6= deg q, thenthe number of critical points of f (z) in the �nite plane C is deg (qp0 � q0p) =deg p+deg q� 1, while the order of the critical points at1 is jdeg p� deg qj�1.

Solution

53

VIII.5.131 2 3 P L K

Show that the set of solution points (w; z) of the equation z2�2 (cosw)+1 = 0consists of the graphs of two entire functions z1 (w) and z2 (w) of w. Specifythe entire functions, and determine where their graphs meet. Remark. Thesolutions set forms a reducible one-dimensional analytic variety in C2.

Solution

54

VIII.5.141 2 3 P L K

Let a0 (w) ; : : : ; am�1 (w) be analytic in a neighborhood of w = 0 and vanishat w = 0. Consider the monic polynomial in z whose coe¢ cients are analyticfunctions in w,

P (z; w) = zm + am�1 (w) zm�1 + � � �+ a0 (w) ; jwj < �:

Suppose that for each �xed w, 0 < jwj < �, there are m distinct solutions ofP (z; w) = 0.(a)Show that the m roots of the equation P (z; w) = 0 determine analyticfunctions z1 (w) ; : : : ; zm (w) in the slit disk fjwj < �n (��; 0]g. Hint. Usethe implicit function theorem (Exercise 4.7).(b)Glue together branch cuts to form an m � sheeted (possibly disconnected)surface over the punctured disk f0 < jwj < �g on which the branches zj (w)determine a continuous function.(c)Suppose that the indices are arranged so that for some �xed k, 1 � k � n, thecontinuation of zj (w) once around w = 0 is zj+1 (w) for 1 � j � k� 1, whilethe continuation of zk (w) once around w = 0 is z1 (w). Show that Q (z; w) =(z � z1 (w)) � � � (z � zk (w)) determines a polynomial in z whose coe¢ cientsare analytic functions of w for jwj < �. Show further that the polynomialQ (z; w) is an irreducible factor of P (z; w), and that all irreducible factors ofP (z; w) arise from subsets of the zj (w)�s in this way.(d)Show that if f (z) is an analytic function that has a zero of order m atz = 0, and z1 (w) ; : : : ; zm (w) are solutions of the equation w = f (z), thenthe polynomial P (z; w) = (z � z1 (w)) � � � (z � zm (w)) is irreducible.

Solution

55

VIII.5.151 2 3 P L K

Consider monic polynomial in z of the form

P (z; w) = zm + am�1 (w) + � � �+ a0 (w) ;where the functions a0 (w) ; : : : ; am�1 (w) are de�ned and meromorphic insome disk centered at w = 0. Let P0 (z; w) and P1 (z; w) be two such poly-nomials, and consider the following algorithm. Using the division algorithm,�nd polynomialsA2 (z; w) and P2 (z; w) such that P0 (z; w) = A2 (z; w)P1 (z; w)+P2 (z; w) and the degree of P2 (z; w) is less then the degree of P1 (z; w). Con-tinue in this fashion, �nding polynomials Aj+1 (z; w) and Pj+1 (z; w) suchthat

Pj�1 (z; w) = Aj+1 (z; w)Pj (z; w) + Pj+1 (z; w)

and degPj+1 (z; w) < degPj (z; w), until eventually we reach

Pl (z; w) = 0; Pl�1 (z; w) = Al (z; w)Pl (z; w) :

Let D (z; w) be the monic polynomials in z obtained by dividing Pl (z; w) bythe coe¢ cient of the highest power of z.(a)Show that D (z; w) is the greatest common divisor of P0 (z; w) and P1 (z; w),in the sense that D (z; w) divides both P0 (z; w) and P1 (z; w), and eachpolynomial that divides both P0 (z; w) and P1 (z; w) also divides D (z; w).(b)Show that there are polynomials A (z; w) and B (z; w) such that D = AP0+BP1.(c)Show, that if P0 (z; w) and P1 (z; w) are relatively prime (that is D (z; w) =1), then there is " > 0 such that for each �xed w, 0 < jwj < ", the polynomialsP0 (z; w) and P1 (z; w) have no common zeros.(d)Show that any polynomial P (z; w) as above can be factored as a product ofirreducible polynomials, and the factorization is unique up to the order ofthe factors and multiplication of a factor by a meromorphic function in w.(e)

56

Show that if the coe¢ cients of P (z; w) are analytic at w = 0, then theirreducible factors of P (z; w) can be chosen so that their coe¢ cients areanalytic at w = 0.(f)Show that if P (z; w) is irreducible, then there is " > 0 such that for each�xed w, 0 < jwj < ", the roots of P (z; w) are distinct.(g)Show that the results of Exercise 14(a)-(c) hold without the supposition thatthe solutions of P (z; w) = 0 are distinct.

Solution

57

VIII.6.11 2 3 P L K

Sketch the closed path (t) = eit sin (2t), 0 � t � 2t, and determine thewinding number W ( ; �) for each point � not on the path.

Solution

58

VIII.6.21 2 3 P L K

Sketch the closed path (t) = e�2it cos t, 0 � t � 2�, and determine thewinding number W ( ; �) for each point � not on the path

Solution

59

VIII.6.31 2 3 P L K

Let f (z) be analytic on an open set containing a closed path , and supposef (z) 6= 0 on . Show that the increase in arg f (z) around is 2�W (f � ; 0).

Solution

60

VIII.6.41 2 3 P L K

Let D be a domain, and suppose z0 and z1 lie in the same connected com-ponent of CnD.(a)Show that the increase in the argument of f (z) = (z � z0) (z � z1) aroundany closed curve in D is an even multiple of 2�.(b)Show that (z � z0) (z � z1) has an analytic square in D.(c)Show by example that (z � z0) (z � z1) does not unnecessarily have an ana-lytic cube in D.

Solution

61

VIII.6.51 2 3 P L K

Show that if is a piecewise smooth closed curve in the complex plane, withtrace �, and if z0 62 �, thenZ

1

(z � z0)ndz = 0; n � 2:

Solution

62

VIII.6.61 2 3 P L K

Let be a closed path in a domain D such that W ( ; �) = 0 for all � 62 D.Suppose that f (z) is analytic on D except possibly at a �nite number ofisolated singularities z1; : : : ; zm 2 Dn�. Show thatZ

f (z) dz = 2�iX

W ( ; zk) Res [f; zk] :

Hint. Consider the Laurent decomposition at each zk, and use Exercise 5.

Solution

63

VIII.6.71 2 3 P L K

Evaluate

1

2�i

Z

dz

z (z2 � 1) ;

where is the closed path indicated in the �gure. Hint. Either use Exercise6, or proceed directly with partial fractions.

64

VIII.6.81 2 3 P L K

Let (t) and � (t), a � t � b, be closed paths.(a)Show that if � 2 C does not lie on the straight line segment between (t)and � (t), for a � t � b, then W (�; �) =W ( ; �).(b)Show that if j� (t)� (t)j < j� � (t)j for a a � t � b, then W (�; �) =W ( ; �).

Solution

65

VIII.6.91 2 3 P L K

Let f (z) be a continuous complex-valued function on the complex planesuch that f (z) is analytic for jzj < 1, f (z) 6= 0 for jzj � 1, and f (z)! 1 asz !1. Show that f (z) 6= 0 for jzj < 1.

Solution

66

VIII.6.101 2 3 P L K

Let K be a nonempty closed bounded subset of the complex plane, and letf (z) be a continuous complex-valued function on the complex plane thatis analytic on CnK and at 1. Show that every value attained by f (z) onC� = C[ f1g is attained by f (z) somewhere on K, that is f (C�) = f (K).

Solution

67

VIII.7.11 2 3 P L K

Let f (z) be an entire function, and suppose g (�) is analytic for � in the openupper hand lower half-planes and across the interval (�1; 1) on the real line.Suppose that Z 1

�1

f (x)

z � � dx = g (�)

for � in the upper half-plane. What is the value of the integral when � is inthe lower half-plane? Justify your answer carefully.

Solution

68

VIII.7.21 2 3 P L K

Show that Z 1

�1

dx

x� � = Log�� � 1� + 1

�; � 2 Cn [�1;+1] :

(Note that we use the principal branch of the logarithm here.) Reconcile thisresult with your solution to Exercise 1.

Solution

69

VIII.7.31 2 3 P L K

Find the Cauchy integrals of the following functions around the unit circle� = fjzj =g, positively oriented.(a) z (b) 1

z(c) x = Re z (d) y = Im (z)

Solution

70

VIII.7.41 2 3 P L K

Suppose f (z) is analytic on an annulus f� < jzj < �g, and let f (z) = f0 (z)+f1 (z) be the Laurent decomposition of f (z). (See Section VI.1.) Fix rbetween � and �, and let F (�) be the Cauchy integral of f (z) around thecircle jzj = r. Show that f0 (�) = F (�) for j�j < r, and f1 (�) = �F (�) forj�j > r. Show further that f0 (�) = F� (�) and f1 (�) = �F+ (�). RemarkThe formula f (z) = f0 (z) + f1 (z) re�ects the jump theorem for the Cauchyintegral of f (z) around circles jzj = r.

Solution

71

VIII.7.51 2 3 P L K

Let be a piecewise smooth curve, and let F (�) be the Cauchy integral (7.1)of a continuous function f (z) on . Show that if g (z) is a smooth functionon the complex plane that is zero o¤ some bounded set, thenZ

g (z) f (z) dz = 2i

ZZC

@g

@�zF (z) dx dy:

Hint. Recall Pompeiu�s formula (Section IV.8).

72

VIII.7.61 2 3 P L K

Determine whether the point z lie inside or outside. Explain.

Solution

73

VIII.7.71 2 3 P L K

A simple arc � in C is the image of a continuous one-to-one function (t)from a closed interval [a; b] to the complex plane. Show that a simple arc� in C has a connected complement, that is Cn� is connected. You mayuse the Tietze extension theorem, that a continuous real-valued function ona closed subset of the complex plane can be extended to a continuous real-valued function on the entire complex plane. Hint. Suppose z0 belongs toa bounded complement of Cn�. Find at continuous determination h (z) oflog (z � z0) on �, extend h (z) to a continuous function on C�, and de�nef (z) = z� z0 on the component Cn� containing z0, and f (z) = eh(z) on theremainder of C�n�. Consider the increase in the argument of f (z) aroundcircles centered at z0.

Prove the Jordan curve theorem for a simple closed curve by �lling in thefollowing proof outline.(a)Show that each component Cn� has boundary �. Hint. For z0 = (t0) 2 �,apply the preceding exercise to the simple arc �n (I), where I is a smallopen parameter interval containing t0.(b)Prove the Jordan curve theorem in the case where contains a straight linesegment.(c)Show that for any z0 = (t0) 2 �, any small disk D0 containing z0, andcomponent U of Cn�, there are points z1 = (t1) and z2 = (t2) such thatthe image of the parameter segment between t1 and t2 is contained in D0 andsuch that z1 and z2 can be joined by a broken line segment in U \D0.(d)With notation as in (b), let �be the simple curve obtained by replacing thesegment of in D0 between z0 and z1 by the broken line segment in U \D0

between them, and let � be the simple closed curve in D0 obtained by thefollowing the segment of in D0 from z0 to z1 and returning to z0 along thebroken line segment. Show that W (� ; �) = 0 and W ( ; �) = W (�; �) for� 2 Cn�, � 6= D0.(e)

74

Using (b) and (d), show that Cn� has at least two components and thatW ( ; �) = �1 for � in each bounded component of Cn�.(f)By taking U in (c) to be a bounded component of Cn�, show thatW ( ; �) =0 for � in any other component of Cn�.

Solution

75

VIII.8.11 2 3 P L K

Which of the following domains in C are simply connected? Justify youranswers. (a) D = fIm z > 0g n [0; i], the upper half-plane with a verticalslit from 0 to i. (b) D = fIm z > 0g n [i; 2i], the upper half-plane with avertical slit from i to 2i. (c) D = Cn [0;+1], the complex plane slit alongthe positive real axis. (d) D = Cn [�1; 1], the complex plane with an intervaldeleted.

Solution

76

VIII.8.21 2 3 P L K

Show that a domain D in the extended complex plane C� = C [ f1g issimply connected if and only if its complement C�nD is connected. Hint. IfD 6= C�, move a point in the complement of D to1. If D = C�, �rst deforma given closed path to one that does not cover the sphere, then deform it toa point by pulling along arcs of great circles.

Solution

77

VIII.8.31 2 3 P L K

Which of the following domains in C� are simply connected? Justify youranswers. (a) D = C�n [�1; 1], the extended plane with an interval deleted,(b) D = C�n f�1; 0; 1g, the thrice-punctured sphere.

Solution

78

VIII.8.41 2 3 P L K

Show that a domain D in the complex plane is simply connected if and onlyif any analytic function f (z) on D does not vanish at any point of D has ananalytic logarithm on D. Hint if f (z) 6= 0 on D, consider the function

G (z) =

Z z

z0

f 0 (z)

f (w)dw:

Solution

79

VIII.8.51 2 3 P L K

Show that a domainD is simply connected if and only if any analytic functionf (z) on D that does not vanish at any point of D has an analytic squareroot on D. Show that this occurs if and only if for any point z0 62 D thefunction z � z0 has an analytic square root on D.

Solution

80

VIII.8.61 2 3 P L K

Show that a domain D is simply connected if and only if each continuousfunction f (z) on D that does not vanish at any point D has continuouslogarithm on D.

Solution

81

VIII.8.71 2 3 P L K

Let E be a closed connected subset of the extended complex plane C�. Showthat each connected component C�nE is simply connected.

Solution

82

VIII.8.81 2 3 P L K

Show that simple connectivity is a "topological property" that is, if U andV are domains, and ' is a continuous map of U onto V such that '�1 is alsocontinuous, then U is simply connected if and only if V is simply connected.

Solution

83

VIII.8.91 2 3 P L K

Suppose that f (z) is analytic on a domain D, and f 0 (z) has no zeros on D.Suppose also that f (D) is simply connected, and that there is a branch g (w)of f�1 that is analytic on w0 = f (z0) and that can be continued analyticallyalong any path in f (D) starting at w0. Show that f (z) is one-to-one on D.

Solution

84

VIII.8.101 2 3 P L K

We de�ne an integral 1� cycle in D to be an expression of the form � =Pkj j, where 1; : : : ; m are closed paths in D and k1; : : : ; km are integers.

We de�ne the winding number of � about � to be W (�; �) =PkjW

� j; �

�,

� 2 CnD. Show that if h (�) is a continuous integer-valued function onC�nD such that h (1) = 0, then there is an integral 1-cycle � on D suchthat W (�; �) = h (�) for all � 2 CnD.

Solution

85

VIII.8.111 2 3 P L K

An integral 1-cycle � is homologous to zero i D if W (�; �) = 0 whenever� 62 D. Let U be a bounded domain whose boundary consists of a �nitenumber of piecewise smooth closed curves 1; : : : ; m, oriented positively withrespect to U , such that U together with its boundary is contained inD. Showthat the 1-cycle @U =

P j is homologous to zero in D.

Solution

86

VIII.8.121 2 3 P L K

Let D be a domain in C such that C�nD consists of m+1 disjoint closed con-nected sets. Show that there arem piecewise smooth closed curves 1; : : : ; msuch that every integral 1-cycle � can be expressed uniquely in the form� = �0+

Pkj j, where the kj�s are integers and �0 is homologous to zero in

D. Remark. The j�s form a homology basis for D.

Solution

87