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The University of Sydney

School of Mathematics and Statistics

Solutions to Tutorial 11: Complex integrals

MATH3068 Analysis Semester 2, 2009

Web Page: http://www.maths.usyd.edu.au:8000/u/UG/SM/MATH3068/Lecturer: Donald Cartwright

1.   Evaluate the integral

 C 

|z| dz  along the curve  C , where C   is

(a) the line segment from 0 to 1− i;

Solution:   The curve C   is parametrized by  z (t) = (1 − i)t, 0 ≤ t ≤ 1. On this segment  C ,|z| =

√ t2 + t2 =

√ 2 t  (since  t ≥ 0), and z ′(t) = (1 − i). So

 C 

|z| dz  =

   10

|z(t)|z′(t) dt  =

   10

√ 2 t(1− i) dt  =

√ 2(1− i)

t2

2

10

=

√ 2

2  (1− i).

(b) the circle C 1(0) = {z ∈ C   :   |z| = 1}.

Solution:   On C , |z| = 1 and so C |z| dz  =

 C 

 1  dz  = 0 by Cauchy’s Theorem, because theconstant function 1 is differentiable. Alternatively,  C  is parametrized by z  =  eit, 0 ≤ t ≤ 2π,and  z ′(t) =  ieit, so

 C 

|z| dz  =

   2π0

|z(t)|z′(t) dt  =

   2π0

ieit dt =  eit2π0

= 0.

2.   Evaluate

 C 

ez dz, where C   is

(a) the line segment from 1 to i;

Solution:   Because ez

= F ′

(z) for all z, where F (z) =  ez

, the integral of  ez

along a curve C is F (β ) − F (α) = eβ − eα, where  α  and  β  are the start and finish points of  C .

In both parts (a) and (b),   C   is a curve from 1 to   i. The integral therefore has the samevalue in parts (a) and (b):

 C 

ez dz  =  ei − e = cos 1− e + i sin1.

(b) the circular arc from 1 to i  along the circle |z| = 1;

Solution:  See the solution of the previous part.

(c) the circle C 2(0) = {z ∈ C   :   |z| = 2}, taken once anti-clockwise.

Solution:   The integral is around a closed curve, starting at  α = 2 and ending at  β  = 2.So it equals eβ − eα = 0.

3.   Evaluate

 C 

(2z + i) dz  where C   is

(a) the circular arc from 3 + 4i to −5 along the circle |z| = 5;

Solution:   Since 2z + i =  F ′(z) for F (z) =  z2 + iz, the integral of 2z + i along a curve C   isF (β )−F (α), where α  and  β  are the start and finish points of  C . So the value of 

 C 

 2z + i dzdepends only on the end-points of  C .

 C 

2z + i dz  =  F (−5) − F (3 + 4i) =

z2 + iz

−5

3+4i

= 36− 32i.

This method would work with 2z + i   replaced by any polynomial.

Copyright   c 2009 The University of Sydney   1

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(b) the straight line segment from −5 to 3 + 4i;

Solution:   The curve has the same end-points as in part (a) but is directed from −5 to3 + 4i, so that   α   and   β   are the reverse of what they were in part (a). The value of theintegral is therefore −36 + 32i.

(c) the circle C 5(0) = {z ∈ C   :   |z| = 5}, taken once anti-clockwise.

Solution:   The curve is closed, i.e.,  α =  β , and so the integral is zero.

4.   Evaluate C 

 f (z) dz , where f (z) =   ez

z  and C  is the circle C 2(1) = {z ∈ C   :   |z− 1| = 2}, traversed

in the anti-clockwise direction.

Solution:   Since ez differentiable on and inside C , we can put a  = 0 in Cauchy’s Integral Formula,obtaining

 C 

ez

z  dz = 2πie0 = 2πi.

5.   Evaluate the following integrals C 

 f (z) dz . All the circles are traversed once in the anticlockwisedirection.

(a)   f (z) =   2zz2+1

,  C  is the circle C 3(0) = {z ∈ C   :   |z| = 3}.

Solution:

1z + i

 +   1z − i

 =   z − i + x + i(z + i)(z − i)

 =   2zz2 + 1

.

So C 

2z

z2 + 1  dz  =

 C 

  1

z + i +

  1

z − i

  dz  =

 C 

1

z + i dz +

 C 

1

z − i  dz = 2πi + 2πi  = 4πi

since    C 

1

z − a dz  = 2πi

whenever  C  winds once around  a, as shown in lectures.

(b)   f (z) =   1z2+5z+6

,  C  is the unit circle  C 1(0) = {z ∈ C   :   |z| = 1}.

Solution:   Because z 2 + 5z + 6 = (z + 2)(z + 3), the only problem points are

 −2 and

 −3,

which are outside   C . So the integrand is differentiable on and inside   C , and therefore C 

dzz2+5z+6

 = 0 by Cauchy’s theorem.

(c)   f (z) =   cos zz(z2+1)

,  C  is the circle  C 13

(0) = {z ∈ C   :  |z| =   13}.

Solution:  The only problem points for this function are at 0 and ±i. Of these, only  z  = 0lies inside   C . So we can apply the Cauchy Integral Formula to   g(z) =   cos z

z2+1  and   a   = 0,

getting  C 

cos z dz

z(z2 + 1)  dz =

 C 

g(z)

z  dz  = 2πig(0) = 2πi.

(d)   f (z) =   z3+2z(z−1)3

,  C  is the circle  C 12

(0) = {z ∈ C   :   |z| =   12}.

Solution:   The integral is 0 by Cauchy’s Theorem, since   f (z) is differentiable on andinside C .

6.   Evaluate the integral    ∞

−∞

1

x2 + 2x + 2  dx

by integrating   1z2+2z+2

 around the closed curve consisting of the line segment from −R to R (where

R >√ 

2), followed by the semicircle above the  x-axis going from  R  around to −R.

Solution:

1+i

R

2

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The denominator  z2 + 2z + 2 factors into (z − α)(z − β ), where  α = −1 + i  and  β  = −1 − i. So

we can think of the integrand as   f (z)z−α

, where f (z) =   1z−β

, which is differentiable on and inside  C .So the integral around  C   is

2πif (α) = 2πi  1

α − β   = π.

The horizontal part of the curve is parametrized by  z (t) = t, −R ≤ t ≤ R. So the integral alongthat part is simply    R−R

1

t2 + 2t + 2 dt.

On the semicircular part  C ′ of  C ,  z2 + 2z + 2 is dominated by the  z2 term when  R   is large. Oneway to express this is by writing

|z2 + 2z + 2| = R21 +

 2

z  +

  2

z2

≥ R2

1−   2

R −   2

R2

≥ R2

1 −  2

3 − 2

9

 =

  R2

9  once R ≥ 3.

Hence the integrand is in modulus at most   9R2   on  C ′. Now C ′ has length  π R, and so

 C ′

1

z2 + 2z + 2   dz ≤

  9

R2 · πR  =

  9π

R ,

which tends to 0 as  R →∞. Therefore the second term on the right in the equation

π =

 C 

1

z2 + 2z + 2 dz  =

   R−R

1

t2 + 2t + 2  dt +

 C ′

1

z2 + 2z + 2  dz

is thought of as an error term  E (R), and has modulus at most 9π/R. Hence

   R−R

1

t2 + 2t + 2 dt =  π − E (R) → π   as  R →∞.

That is,    ∞−∞

1t2 + 2t + 2

  dt =  π.

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