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Math 325: Complex variables and applications

Xiaokui Yang

Department of Mathematics, Northwestern University,

[email protected]

Contents

1 Complex numbers, 1-4 21.1 Basic algebraic properties of complex numbers . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2 Exponential forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Elementary functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Functions and limits, 5-7 92.1 Complex functions and limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.2 Differentiations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

3 Analytic functions, 7-9 143.1 Cauchy-Riemann equations and analytic functions . . . . . . . . . . . . . . . . . . . . . . . 143.2 Harmonic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

4 Complex functions and integrations, 10-12 184.1 Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184.2 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184.3 Contour integrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

5 Cauchy’s integral formula, 12-17 235.1 Cauchy-Goursat Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235.2 Simply and multiply connected domains . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255.3 Cauchy integral formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275.4 Cauchy’s inequality and Liouville’s theorem. . . . . . . . . . . . . . . . . . . . . . . . . . 295.5 Maximum modulus principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315.6 Schwarz’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

6 Power series, 18-20 336.1 Taylor’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336.2 Laurent’s series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 356.3 Uniform properties of series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 376.4 Review for Midterm II. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

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7 Residue, 21-23 407.1 Cauchy’s residue theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407.2 Three types of singularities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 417.3 Applications of Residue. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437.4 Zeros of an analytic function are isolated . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

8 Conformal mapping and Linear fractional transformations, 23-26 468.1 Elementary functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 468.2 Linear fractional transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 478.3 f : D →D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 508.4 f : H →D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 518.5 Schwarz-Pick. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

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1 Complex numbers, 1-4

1.1 Basic algebraic properties of complex numbers

1.1.1 Sums and products

Let a, b be two real numbers. Let i be the imaginary unit, i.e.

i2 = i ·i = −1. (1.1)The set of complex numbers is dened as

C := {(a, b) | a, b R } (1.2)with the correspondence

(a, b) ←→a + ib. (1.3)If z = a + ib with a, b R , a is called the real part of z and it is denoted by a = Re(z). b is called the

imaginary part of z, and denoted by Im (z). That is z = Re(z) + iIm (z).Let z = a + ib and w = c + id .

(1) −w = −c −id ;(2) w− 1 = 1w =

c− idc2 + d2 if w = 0 ;

(3) z + w = ( a + c) + i(b + d);

(4) z −w = z + ( −w) = ( a −c) + i(b−d);(5) z ·w = ( a + ib) ·(c + id) = ( ac −bd) + i(ad + bc);(6) if w = 0 ,

zw = z ·

1w =

(ac + bd)+ i(bc− ad )c2 + d2 .

Example 1.1.(3 + 5 i)(2 + i) = 6 + 10 i + 3 i + 5 i2 = 6 + 13 i −5 = 1 + 13 i

(2 + i)− 1 = 12 + i

= 2−i22 + 1 2

= 2−i

53 + 5 i2 + i

= (3 + 5 i)(2 −i)

5 =

11 + 7i5

1.1.2 Conjugates and norms

Let z = a + ib with a, bR

, z = a −ib (1.4)is called the complex conjugate of z.

|z| = a2 + b2 (1.5)is called the absolute value (norm) of z.

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Example 1.2. Let a R , a = a . i = −i . Let z = 3 + 4 i , thenz = 3 −4i

and

|z

|=

32 + 4 2 = 5

Theorem 1.3 (Basic properties) . Let z, w C . Then

(1) Re(z) = z+ z2 ; Re(z) = Re(z);

(2) Im (z) = z− z2i ; Im (z) = −Im (z);(3) Re(z + w) = Re(z) + Re(w);

(4) Im (z + w) = Im (z) + Im (w);

(5) z R Im (z) = 0 z = z;(6) z = z;

(7) z + w = z + w; z −w = z −w;(8) z ·w = z ·w;(9) zw =

zw if w = 0 .

(10) |z|2 = z ·z;(11) |z| = 0 if and only if z = 0 ;(12) |z ·w| = |z| · |w|;

(13)zw =

|z ||w| ;

(14) |z + w| ≤ |z|+ |w|.Proof. The proofs of these facts are straightforward and we omit the details. As examples, we will prove(12) and (14). Let z = a + ib, w = c + id . Then

z ·w = ( ac −bd) + i(ad + bc)and so

|z ·w|2 = ( ac −bd)2 + ( ad + bc)2 = ( a2 + b2)(c2 + d2) = |z|2 · |w|2.For (14), we use the trick |z|2 = z ·z. That is

|z + w|2 = ( z + w) ·(z + w) = |z|2 + |w|2 + zw + zw = |z|2 + |w|2 + 2 Re(z ·w)On the other hand,

Re(z ·w) ≤ |z ·w| = |z||w| = |z||w|Hence,

|z + w|2 = |z|2 + |w|2 + 2 Re(z ·w) ≤ |z|2 + |w|2 + 2 |z||w| = ( |z|+ |w|)2.

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1.2 Exponential forms

1.2.1 Exponential functions

Denition 1.4. For z = x + iy where x, y R , we dene the exponential function ez by

ez

= ex

(cos y + i sin y). (1.6)Theorem 1.5. The exponential function has the following properties

(1) eiθ = cos θ + i sin θ for θ R (Euler formula);

(2) ez+ w = ez ·ew for z, w C ;(3) |ez | = ex if z = x + iy with x, y R ;(4) ez+2 πi = ez for z C (the exponential function is 2πi -periodic);(5) ez = 1 if and only if z = 2πi ·n with n Z ;(6) e

w= e

zif and only if w = z + 2 πi ·n with n Z .

Proof. (2). Let z = x + iy and w = u + iv with a, b, u, v R . Recall that the magic formulascos(y + v) = cos y cos v −sin y sin v, sin(y + v) = sin y cos v + cos y sin v.

Hence

ez ·ew = ex (cos y + i sin y) ·eu (cos v + i sin v)= ex+ u ((cos y cos v −sin y sin v) + i (sin y cos v + cos y sin v))= ex+ u (cos(y + v) + i sin(y + v))= e(x+ u)+ i(y+ v)

= ez+ w

(5). Let z = x + iy , then ez = ex cos y + iex sin y = 1 . That is

ex cos y = 1 and ex sin y = 0

It implies sin y = 0 and cos y = ±1. Hence ex = 1 , i.e. x = 0 . Then cos y = 1 .sin y = 0 and cosy = 1

implies y = 2πi ·n with n Z .

1.2.2 Polar formsLet (r, θ ) be the polar coordinates of a point (x, y ) R 2 , i.e.

x = r cos θ, y = r sin θ.

Let z = x + iy , then z can be written in polar form as

z = r (cos θ + i sin θ) (1.7)

If z = r (cos θ + i sin θ),

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(1) r = |z| is the norm/length/radius/magnitude of z;(2) θ is called an argument of z and denoted by arg (z). Since

z = r (cos θ + i sin θ) = r (cos(θ + 2 nπ ) + i sin(θ + 2 nπ )) ,

such angles are not unique. The principle value of arg (z), denoted by Arg (z), is that unique value Θsuch that −π < θ 0 ≤π . Hencearg (z) = Arg (z) + 2 nπ, n Z .

It is obvious thattan θ0 =

yx

(3) We can rewrite z asz = x + iy = |z| (cos θ0 + i sin θ0)

Example 1.6. (1) z = i, arg (z) = π/ 2. z = eπi/ 2;

(2) z = −1, arg (z) = π , z = eπi ;(3) Let z = 1 + i . It is easy to see that tan θ0 = 1 and so Θ = π/ 4.

z = 1 + i = |z|(cos(θ0) + i sin θ0) = √ 2(cos(π/ 4) + i sin(π/ 4));(4) z = −2i , arg (z) = −π2 , and

z = 2e−πi

2 .

Theorem 1.7. If z = r (cos θ + i sin θ) = re iθ , then

zn = r n (cos(nθ) + i sin(nθ))

Example 1.8. Compute(1 + √ 3i)300 .

Let z = 1 + √ 3i . It is easy to see thatz = 2(cos( π/ 3) + i sin(π/ 3)) = 2 e

πi

3

Hencez300 = 2 300e300π/ 3 = 2 300 .

1.2.3 Roots of complex numbers

Example 1.9. Let z = cos( π/ 7) + i sin(π/ 7) = eπi

7 . Then we have

z7 = e7πi/ 7 = eπi = −1, z14 = 1Here z is called 7-th root of −1, and 14-th root of 1.

Question 1.10. (1) How to nd an n-th root of a given number z0 , i.e.

zn = z0.

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(2) In general, how many distinct n-th roots for a given nonzero number z0.

At rst, we writez0 = r 0eiθ 0

where θ0 is the principle argument of z0 . Suppose z = re iθ , then zn = r n einθ . Now we get the equation

zn = z0 r n einθ = r0eiθ0That is

r n = r 0 and nθ = θ0 + 2 kπ, k = 0 , ±1, ±2, · · ·or equivalently,

z = n√ r 0ei“θ 0n +

2kπn ”, k = 0 , ±1, ±2, · · ·

Note that all of the distinct roots are obtained when k = 0 , 1, 2 · · · , n −1.Let ck denote these distinct roots and write

ck = n

√ r 0ei“θ 0n + 2kπn ”

, k = 0 , 1, 2, · · · , n −1. (1.8)when k = 0 ,c0 = n√ r 0e iθ 0n (1.9)

is called the principle root.

Example 1.11. Find all values of 11n .

Since 1 = ei0 , we getck = e

2kπin , k = 0 , 1, 2, · · · , n −1.

Let

ω = c1 = e2πi

n

, (1.10)then all roots are

c0 = 1 ,ω ,ω2, · · · , ωn − 1.Note that ωn = 1 . See the distribution of those points on the unit circle.

Example 1.12. Find all values of (−8i)1/ 3 .Note that −8i = 8e−

πi

2 . Hence

ck = 3√ 8ei(− π3 · 2 + 2kπ3 ), k = 0 , 1, 2.

Hence

c0 = 2 e− πi

6 = √ 3 −ic1 = 2eπi/ 2 = 2 ic2 = ei

7π6 = −√ 3 −i

Show the geometry of these points (roots).

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Example 1.13. Find solutions toz2 = −7 + 24i

in rectangular coordinates.Here we assume z = a + ib with a, b R .

z2

= a2

−b2

+ 2 abi = −7 + 24iand so

a2 −b2 = −7, 2ab = 24It is also equivalent to

a2 −(12/a )2 = −7 = a4 + 7 a2 −144 = 0 = (a2 + 16)( a2 −9) = 0Therefore a = 3 , b = 4 or a = −3, b = −4. We get two solutions

z = 3 + 4 i, z = −3 −4i1.3 Elementary functions

In the real casexa , sin x, log x, ex

1.3.1 Trigonometric functions

Recall that, for real number x

eix = cos x + i sin x, e− ix = cos x −i sin xThat is

sin x = eix

−e− ix

2i , cosx = e

ix+ e

− ix

2For any complex number z C , we dene

sin z = eiz −e− iz

2i , cosz =

eiz + e− iz

2 . (1.11)

Henceeiz = cos z + i sin z.

Example 1.14. (1) sin(π/ 2) = 1 ;

(2) sin(πi/ 2) = e−

π2 − e

π2

2i .(3) sin(z) = 0 if and only if z = πn for some n Z . In fact, eiz −e− iz = 0 implies e2iz = 1 , i.e.2iz = 2iπn for some n .(4) sin and cos have the same properties as the real case.

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1.3.2 Logarithms.

Recall in the real case a = eb, then b = log a .Let z C . Every solution w C to the equation

z = ew (1.12)

is called a logarithm of z. A number z C has a logarithm if and only if z = 0 .Let z = re iθ with r > 0 and θ R . Then

z = elog r + iθ = ew

Hence w = log r + iθ + i2πn for any n Z . So every z C \{0}has innitely many logarithms. We willdenote by log z. Note that it is multi-valued.

Principle value of log z is denoted by Logz

Logz = log r + iθ0 (1.13)

where z = re iθ0 and θ0 is the principle argument of z. Therefore

log z = Logz + 2 nπi, n Z (1.14)

Example 1.15. If z = −1. Then z = eπi . Hencelog(−1) = πi + 2 nπi, n = 0 , ±1, ±2, · · ·

Log(−1) = πiProposition 1.16. For any complex numbers z1 and z2 ,

log(z1z2) = log z1 ·log z2 (1.15)and

log z1z2

= log z1 −log z2. (1.16)

1.3.3 Complex exponents

Recall, xa , x R , a R . Let c be a complex number. The complex function zc is dened as

zc := ( elog z )c = ec log z

Note that, it is a multi-valued function. The principle value of zc is

P.V. (zc) = ecLogz (1.17)

and sozc = P.V. (zc) ·ec2nπi , n Z

Example 1.17. (−i)i .(1). z = −i = e−

πi

2 , θ0 = −π2 .(2). Logz = −πi2 .(3). P.V. (−i)i = eiLogz = eπ/ 2.(4). (−i) i = e

π

2 ·ei·2nπi = e(1 − 4n ) π

2 , n = 0 , ±1, ±2, · · · ,

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2 Functions and limits, 5-7

2.1 Complex functions and limits

2.1.1 Complex functions

Let S

C be a subset of complex numbers. f : S →

C is called a complex function. For example,f (z) = z2 : C →C . However, f (z) = 1z2 +1 is not well-dened if z2 + 1 = 0 , i.e.

f (z) = 1z2 + 1

: C \ {i, −i} →C .f (z) = z2 and f (z) = z2 are all functions dened on C .On the other hand, we know √ z has two values. If z = re iθ0 with θ0 9 −π, π ], then

c1 = √ re iθ 02 and c2 = −√ reiθ 0

2

Now we can dene a function

f (z) = √ reiθ 0

2

, z = 00, z = 0It is a single-valued function on the whole C and satises

f 2(z) = z

for any z C .2.1.2 Limits of a complex function

Denition 2.1. Let f be a complex function. Suppose the function f is dened in some deleted neighbor-hood of z0 . For any ε > 0, there exists some δ > 0 such that for any z

S with

|z

−z0

|< δ ,

|f (z) −w0| < ε (2.1)then we say that the limit of f at point z0 is w0 , and write it as

limz→z0

f (z0) = w0

• By triangle inequality, we see that if a limit of a function f (z) exists at z0, it is unique.Let f (z) be a complex function. Since we can write z = x + iy and f (z) C , we can write

f (z) = u(x, y) + iv(x, y).

Example 2.2. (1) f (z) = z2 = ( x2 −y2) + i2xy . Henceu(x, y ) = x2 −y2, v(x, y) = 2 xy.

(2) f (z) = ez = ex+ iy = ex eiy = ex cos y + iex sin y.

u(x, y) = ex cos y, v(x, y) = ex sin y.

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Conversely, if a complex function f (z) is written in terms of (x, y ) R 2 , we can rewrite it in terms of zand z by the relation

z = x + iy,z = x −iy.

x = z+ z2 ,y = z− z2i .

Example 2.3. Let f (z) = ( x2

−y2

+ x) + i(y −2xy). We can see that f (z) = z2

+ z.Example 2.4. (1) Let f (z) = iz2 , then

limz→1

f (z) = i2

.

(2) then function f (z) = zz is dened is a neighborhood of 0 C . But it is undened at 0.

limz→0

f (z) =? .

We say that f (z) = (x2 − y2 )+ i2xy

x2 + y2 . We know the real part of f (z) is

u(x, y ) = x2

−y2

x2 + y2 .

The limit of u(x, y) at point 0 does not exist!.

Example 2.5. For any z S := C \ {Re(z) ≤0, Im (z) = 0}, we dene a function

F (z) =√ z, z C \ {0},0, z = 0

(1) F is not a continuous function on C ;

(2) F is not continuous at points of {Re(z) ≤0, Im (z) = 0};(3) F is a continuous function on S .

Theorem 2.6 (Basic properties of limits) . (1) Let f (z) = u(x, y) + iv(x, y) and z = x + iy . If

limz→z0

f (z) = w0

and in the real coordinates, z0 = x0 + iy0 , w0 = u0 + iv0 , then

lim(x,y )→(x0 ,y0 )

u(x, y) = u0, lim(x,y )→(x0 ,y0 )

v(x, y ) = v0 (2.2)

(2) Suppose that

limz→z0 f 1(z) = w1, limz→z0 f 2(z) = w2then

limz→z0

(f 1(z) + f 2(z)) = limz→z0f 1(z) + limz→z0

f 2(z) = w1 + w2

limz→z0

(f 1(z) ·f 2(z)) = limz→z0 f 1(z) · limz→z0 f 2(z) = w1 ·w2if w2 = 0 , we have

limz→z0

f 1(z)f 2(z)

= limz→z0 f 1(z)limz→z0 f 2(z)

= w1w2

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Denition 2.7. Let f : S → C be a complex function. Then f (z) is call continuous at point z = z0 if limz→z0 f (z) exists andlim

z→z0f (z) = f (z0). (2.3)

Or equivalently, for any ε > 0, there exists some δ > 0 such that for any z S with |z −z0| < δ ,

|f (z) −f (z0)| < ε. (2.4)Theorem 2.8 (basic properties of continuous functions) . (1) A composition of continuous function is itself

continuous.

(2) If a function f (z) is continuous and nonzero at point z0 , then there exists a small neighborhood U of z0 , such that f (z) = 0 over U .

(3) Let f (z) = u(x, y) + iv(x, y ). Then f (z) is continuous if and only if both u(x, y) and v(x, y ) arecontinuous.

(4) Let R be a closed and bounded subset of C . If f is continuous on R , then f is bounded, i.e. there existsa nonnegative real number M such that

|f (z)| ≤M for all z R. (2.5) In particular, we can take

M = supz R

f (z) = maxz R

f (z)

Recall that, a continuous function attains its maximum and minimum on a closed and bounded R .

2.2 Differentiations

2.2.1 Denition

Recall the real case.

Let U C be an open subset and f : U →C be a complex function.(1) f is called differentiable at point z0 U if

limz→z0

f (z) −f (z0)z −z0

= lim∆ z→0

f (z0 + ∆ z) −f (z0)∆ z

exists. This limit is called the derivative of f at z0 and denoted by f ′ (z0);

(2) f is call differentiable on U if f is differentiable at every point of U .

Example 2.9. Let f (z) = z2 . f ′ (i) =? .

f ′ (i) = lim∆ z→0

f (i + ∆ z) −f (i)∆ z = lim∆ z→0(∆ z)2 + 2 i∆ z

∆ z = 2i

In general, it is easy to see that f ′(z) = 2 z

Example 2.10. f (z) = z. f ′ (0) =? .

f (∆ z) −f (0)∆ z

= ∆ z∆ z

= ∆ x −i∆ y∆ x + i∆ y

Does not exist!!!

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2.2.2 Derivatives of elementary functions

(1) (zn ) ′ = nz n − 1 .

(2) (ez ) ′ = ez , and (eiz ) ′ = ie iz

(3) (sin z) ′ = cos z and (cos z) ′ =

−sin z. In fact, by denition


2i , cosz =

eiz + e− iz

2 .

Example 2.11. f (z) = √ z, z S = C \{Re(z) ≤0, Im (z) = 0}is differentiable and satises f 2(z) = zfor every z S !Proof. We show it by denition. Let z0 S , and let z be close to z0 , i.e. √ z + √ z0 = 0 . Hence

f (z) −f (z0)z −z0

= f (z) −f (z0)

z −z0 · f (z) + f (z0)f (z) + f (z0)

= 1

f (z) + f (z0)

since f (z) = √ z. Since f (z) is continuous if z S , we havelim

z→z0

f (z) −f (z0)z −z0

= 1

2f (z0) =

12√ z0 (2.6)

Hence the derivative of √ z is(√ z) ′ = 1

2√ z , z C \ {Re(z) ≤0, Im (z) = 0}. (2.7)

Question 2.12. Does there exist a differentiable function f : C

→C such that

f 2(z) = z, z C . (2.8)No. If so

2f ′ (z)f (z) = 1

but it is obvious f (0) = 0 .

Example 2.13. The derivative of Log(z) when z C \ {Re(z) ≤0, Im (z) = 0 }is(Log(z)) ′ =

1z

(2.9)

Proof. When z C \ {Re(z) ≤0, Im (z) = 0}, we haveef (z) = z

By chain rulef ′ (z)ef (z) = 1

that is f ′ (z) = 1z .

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2.2.3 Basic properties of differential functions

Theorem 2.14. (1) If f : U →C is differentiable, then f is continuous on U ;(2) Let a, b C be two constants and f, g : U C be differentiable. Then

(1) af + bg is differentiable and, (af + bg) ′ = af ′ + bg′ ;(2) f ·g is differentiable on U and (f ·g) ′ = f ′ ·g + f ·g′ ;(3) If g = 0 on U , then f g is differentiable on U and

f g

′

= f ′ ·g −f ·g′

g2 . (2.10)

(4) If f, g : C →C are differentiable, then g ◦ f is also differentiable and (g ◦ f ) ′ = ( g′ ◦ f ) ·f ′

Proof.

g(f (z0 + ∆ z)) −g(f (z0))∆ z = g(f (z0 + ∆ z)) −g(f (z0))f (z0 + ∆ z) −f (f 0) · f (z0 + ∆ z) −f (f 0)∆ z

Let w0 = f (z0), then ∆ w = f (z0 + ∆ z) −f (z0). It is easy to see that ∆ w →0 when ∆ z →0 since f iscontinuous at point z0 .

g(f (z0 + ∆ z)) −g(f (z0))∆ z

= g(w0 + ∆ w) −g(w0)

∆ w · f (z0 + ∆ z) −f (f 0)

∆ z

Example 2.15. Find the derivative of esin( z) . It is cos(z)esin z .

Theorem 2.16. Let ΩC

be a region, and f : Ω→C

is differentiable. If f ′

(z) ≡0 , then f is a constant function. Explain region.Proof. Let z0 Ω. We will prove f (z) ≡f (z0) over Ω. In fact, if sufcient to prove it is constant on eachline from z0. For any z Ω, we assume there is a line segment connecting z0 and z, i.e.

γ (t) = tz + (1 −t)z0Then we have a real function h(t) = f (γ (t)) : [0, 1] →C such that

h ′(t) = 0

hence h(t) = h(0) = f (z0).

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3 Analytic functions, 7-9

3.1 Cauchy-Riemann equations and analytic functions

Theorem 3.1. Let U be an open subset of C and

f (z) = u(x, y ) + iv(x, y)then f is differentiable if and only if u, v are in C 1(U ) and

ux = vy ,uy = −vx

They are called Cauchy-Riemann equations for the differentiable function f .

Proof. Since f is differentiable, we can compute

f ′ (z) = lim∆ z→0

f (z + ∆ z) −f (z)∆ z

along two different paths and the limits must be the same.(1) Path One: ∆ z = ∆ x →0. Hence

f (z + ∆ z) −f (z)∆ z

= u(x + ∆ x, y ) + iv(x + ∆ x, y) −u(x, y) −iv(x, y)

∆ x →ux (x, y) + ivx (x, y )(2) Path two: ∆ z = i∆ y →0

f (z + ∆ z) −f (z)∆ z

= u(x, y + ∆ y) + iv(x, y + ∆ y) −u(x, y) −iv(x, y)

i∆ y → −iu y(x, y ) + vy(x, y)Hence the CR equations hold.

If u and v satisfy the CR equation, then

u(x + h, y + k) = u(x, y) + ux (x, y)h + vx (x, y )k + E x

v(x + h, y + k) = v(x, y) + vy(x, y)h + vy(x, y)k + E yLet c = ux (x, y) + ivx (x, y), now we want to show that

f ′ (z) = ux (x, y) + ivx (x, y )

Let w = h + ik . Now we can show

limw→0

f (z + w) −f (z)w −c = 0 .

In fact,

f (z + w) −f (z)w −c

=f (z + w) −f (z) −c(h + ik)

w

= |u(x + h, y + k) −u(x, y) −hu x (x, y) −ku y(x, y ) + i (v(x + h, y + k) −v(x, y) −hvx (x, y) −kvy(x, y|w|

→0

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Corollary 3.2. If f (z) is differentiable, f ′(z) is differentiable and so f (n ) (z) is differentiable for any n .

Proof. If f (z) is differentiable, we have

f ′ (z) = ux (x, y) + ivx (x, y ) =

u(x, y) + iv(x, y )

It is easy to see thatux (x, y) = uxx (x, y) = vxy (x, y) = vy(x, y )

and similarlyuy(x, y) = −vx (x, y)

hence f ′(z) = u + iv satises CR equations, and so f ′ (z) is differentiable.

Example 3.3. f (z) = z2 = ( x2 −y2) + 2 xyi . u(x, y) = x2 −y2 and v(x, y) = 2 xy .ux = vy = 2x,uy = −vx = −2y.

Example 3.4. Let f (z) = ez = ex (cos y + i sin y). Then

f ′(z) = ux + ivx = ex (cos y + i sin y) = f (z)

Example 3.5. Let f (z) = |z|2 = x2 + y2 . Then u = x2 + y2 and v = 0 . They do not satisfy the C-Requation except z = 0 .

Denition 3.6. A function f of the complex variable z is called analytic at a point z0 if it has a derivative ateach point in some open neighborhood of z0 .

Remark 3.7. (1) If f is analytic at z0 , it must be analytic at each point in some neighborhood of z0 .

(2) f is analytic in an open set U if it has a derivative everywhere in that set.

(3) f is analytic in an arbitrary set S means, it is analytic in an open set containing S .Example 3.8. f (z) = 1 /z is analytic at each point except z = 0 . However, f (z) = |z|2 is only differen-tiable at point z = 0 . Hence f is not analytic at any point of z C .Denition 3.9. A complex function f is called an entire function if it is analytic at all points of C . Hence,polynomial functions are entire functions.

Let H (U ) be the set of complex functions which are holomorphic over U

Proposition 3.10. Let f H (U ).(1) If Im (f ) = c1 or Re(f ) = c2 , then f is constant.

(2) If f H (U ) , then f is constant.(3) If |f | ≡c , then f is a constant.(4) Suppose f (z) = 0 throughout U . If Arg (f (z)) is a constant, then f is a constant.

Proof. (1) Suppose Im (f ) = c2 , that is f (z) = u(x, y ) + ic2. Hence, by C-R equations, we have

ux = 0 , uy = 0

Therefore u must be a constant. Similarly, we can show if Re(f ) = 0 , f must be a constant.

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Theorem 3.17. Suppose u(x, y) is the real part of some f H (C ) , thenf (z) = 2 u

z2

, z2i −u(0, 0) + iC (3.1)

where C is a real constant.

Proof.

u(x, y) = f (z) + f (z)

2 =

f (x + iy) + f (x −iy)2

Replace x = z2 and y = z2i , we have

uz2

, z2i

= f (z) + f (0)

2

If we assume f (0) = u(0, 0) −iC , we obtainf (z) = 2 u

z2

, z2i −u(0, 0) + iC

Example 3.18. u(x, y ) = y3 −3x2y and f (2) = 9 i .

f (z) = 2 z3

(2i)3 −3z2

22z2i

= iz 3 + iC.

and sof (z) = iz 3 + i.

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4 Complex functions and integrations, 10-12

4.1 Derivatives

Let u(t), v(t) be two real functions, i.e. u, v : R →R . We can make a complex valued function

w(t) = u(t) + iv(t) : R →CThen the derivative of w(t) with respect to real variable t is dened as

w′(t) = u ′ (t) + iv ′(t)

Proposition 4.1. Let z0 = x0 + iy0 be a constant. Then

ddt

(z0w(t)) = z0w′ (t) (4.1)

and d

dtez0 t = z0ez0 t (4.2)

Proof. We show it by denition. For example

ez0 t = etx 0 + ity 0 = etx 0 (cos(ty0) + i sin(ty0)) = etx 0 cos(ty0) + ie tx 0 sin(ty0)

Hence by denition,

ddt

ez0 t = etx 0 cos(ty0)′ + i etx 0 sin(ty0)

′ = ( x0 + iy0)(etx 0 cos(ty0) + ie tx 0 sin(ty0)) = z0etz 0

Example 4.2. (1) (e− tz20 + it ) ′ = (

−z20 + i)e− tz

20 + it .

4.2 Integration

Let w(t) = u(t) + iv(t) be a complex-valued function of a real variable t [a, b]. The integration of w(t)is dened as

ba w(t)dt := ba u(t)dt + i ba v(t)dt (4.3)Corollary 4.3.

Re

b

aw(t)dt =

b

aRe (w(t)) dt (4.4)

Im ba w(t)dt = ba Im (w(t)) dt (4.5)Example 4.4.

10 (1 + it )2dt = 10 (1 −t2) + i2t dt = 23 + i.

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Fundamental theorem of Calculus: Real version:If F (t) is a function dened on [a, b], such that

F ′ (t) = f (t)

then

b

af (t)dt = F (b) −F (a).

Complex version: Let w(t) = u(t) + iv(t) be a complex valued function with a real variable t [a, b]. If there exists a complex valued function

W (t) = U (t) + iV (t)

such that W ′ (t) = w(t), i.e.U ′ (t) = u(t), V ′(t) = v(t)

then

b

aw(t)dt = W (b)

−W (a).

Example 4.5. It is easy to see that(eit /i ) ′ = eit

So

π0 eit dt = eiti |t= πt=0 = eiπi − 1i = 2 iExample 4.6. Compute

π0 et cos(2t)dtConsider the function w(t) = et+2 it = et (cos(2t) + i sin(2t)) . Hence, we have

π0 w(t)dt = 12i + 1 e(1+2 i)t |πt=0 = e0 −e(1+2 i)π1 + i = 1−eπ1 + i = 1−eπ2 −i 1 −eπ2On the other hand, it is easy to see

π0 et cos(2t)dt = Re π0 w(t)dt = 1 −eπ24.3 Contour integrations

Let x(t), y(t) be real functions with t [a, b].

z(t) = x(t) + iy(t) : [a, b] →C is a curve C in the complex plane C .

(1) C is called a simple curve, if z(t1) = z(t2) if t1, t2 (a, b) and t1 = t2;(2) C is a called a simple closed curve if it is a simple curve and z(a) = z(b).

(3) C is called a simple closed curve with positive direction if the curve is in the counterclockwise directionwhen t [a, b] is increasing.

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Example 4.7. (1) If z(t) = eit , t [0, π ]. It is a simple curve, but it is not closed.

(2) If z(t) = eit , t [0, 2π]. It is a simple closed curve with positive direction.

(3) If z(t) = e− it , t [0, 2π]. It is a simple closed curve with negative direction.

(4) If z(t) = e2it , t [0, 2π]. It is not simple curve.

Let C be a curve with a given direction. −C is the same curve with opposite direction.Example 4.8. Let C be the unite circle with positive direction, then −C is the unite circle with negativedirection. By using parametrizations, we have

C : z(t) = eit , t [0, 2π]

and

−C : z(t) = e− it , t [−2π, 0]•. Arc-length of a complex curve z(t) = x(t) + iy(t), t [a, b].

L = b

a |z ′ (t)|2dt = n

a [x′(t)]2 + [y′(t)]2dt

4.3.1 Integrations

Let f (z) be a complex function, we want to integrate it along a curve C :

C f (z)dzSuppose z = z(t) for t [a, b] is a parametrization for the curve C , we dene

C

f (z)dz :=

b

a

f (z(t))z ′(t)dt.

Example 4.9. Let C be the directed line segment from 0 to 1 + i . Compute C zdz.We choose a parametrization for C :z(t) = t + it, t [0, 1]

Then

10 (t + it )(1 + i)dt = (1 + i)22 t2|t=1t=0 = iExample 4.10. (1) Compute | z |=1 zdz .z = eit with t [0, 2π]. Hence

|z |=1 zdz = 2π

0eit ie it dt = 0

(2) Compute |z |=1 1z dz .z = eit with t [0, 2π]. Hence | z |=1 1z dz = 2π0 e− it ie it dt = 2πi

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Theorem 4.11 (Basic properties) . Let z0 C and C , C 1 and C 2 be three curves, then(1) C z0f (z)dz = z0 C f (z)dz ;(2) − C f (z)dz = − C f (z)dz;(3)

C 1 + C 2 f (z)dz =

C 1 f (z)dz + C 2 f (z)dz.

4.3.2 Basic estimates

Lemma 4.12. Let w(t) be a continuous complex valued function dened on an interval a ≤ t ≤b , then

ba w(t)dt ≤ ba |w(t)|dt (4.6)Proof. Let

re iθ = ba w(t)dt.That is

LHS = b

aw(t)dt = r.

We obtain

r = e− iθ ba w(t)dt= ba e− iθ w(t)dt= Re ba e− iθ w(t)dt=

b

aRe e− iθ w(t) dt

≤ ba |e− iθ w(t)|dt= RHS

Theorem 4.13. Let C be a contour of length L , and suppose that a function f (z) is continuous on C . If M is a nonnegative constant such that

|f (z)| ≤M for all z C , then C f (z)dz ≤ML. (4.7)Proof. Let z(t) t [a, b] be a parametrization of C . By denition,

C f (z)dz = ba f (z(t))z ′(t)dt21

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Hence, by previous lemma, we have

C f (z)dz = ba f (z(t))z ′ (t)dt≤

b

a |f (z(t))z ′(t)

|dt

≤ M ba |z ′(t)|dt= ML.

Note that the arc-length of the contour C is

L = ba |z ′ (t)|dt.Example 4.14. Let C be the arc of the circle

|z

| = 2 from z = 2 to z = 2 i that lies in the rst quadrant.

Show that

C dzz2 −1 ≤ π3Proof. We know that

1

|z2 −1| ≤ 1

|z|2 −1 =

13

and L = π .

4.3.3 Anti-derivatives

Theorem 4.15. Suppose that a function f (z) is continuous on a domain D . If any one of the followingstatements is true, then so are the others.

(1) f (z) has an antiderivative F (z) throughout D , i.e. F ′(z) = f (z)

(2) Let C be any contour in D connection z1 and z2 , then

C f (z)dz = F (z)|z2z1 = F (z2) −F (z1) (4.8)that means the integral does not depend on the contour C but only on the endpoints z1 and z2 .

(3) the integrals of f (z) around closed contours lying entirely in D all have value zero, i.e.

C f (z)dz = 0 for any closed curve in D .

Example 4.16. (1) Let f (z) = z2 , it has an anti-derivative F (z) = z3/ 3. Hence

1+ i0 z2dz = (1 + i)23 = 2i −23 .22

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(2) Let C be any contour connecting z1 and z2 , then

C zn dz = zn +12 −zn +11n + 1 (4.9)where n Z + .

(3) Let C be a closed curve lying on the right half plane D = {z C | Re(z) > 0}. Then

C 1z dz = 0 . (4.10)Since F (z) = Logz is a well-dened holomorphic function on D with

F ′ (z) = 1z

By part (3) , we get the result.

5 Cauchy’s integral formula, 12-17

5.1 Cauchy-Goursat Theorem

Recall line integral

C xdyTheorem 5.1. Let C be a positively oriented, piecewise smooth, simple closed curve in a plane, and let Rbe the region bounded by C . If P, Q are functions with continuous derivatives, then

C P dx + Qdy = R (Qx −P y) dxdy (5.1)

Theorem 5.2 (Cauchy-Goursat) . Let f (z) be analytic at all points interior to and on a simple closed contour C , then

C f (z)dz = 0 . (5.2)Proof. In the following, we show it with extra condition f ′ (z) is continuous.

Let z = z(t), t [a, b] be a parametrization of contour C with positive direction. Then

C f (z)dz = ba f (z(t))z ′(t)dtLet f (z) = u(x, y) + iv(x, y ) and z(t) = x(t) + iy(t), then the integrand

f (z(t))z ′ (t) = ux ′ −vy′ + i(vx ′ + uy ′ )

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Hence

C f (z)dz = ba f (z(t))z ′(t)dt=

b

a

(ux ′ −vy′)dt + i

b

a

(vx ′ + uy ′)dt

= ba udx −vdy + i C vdx + udy= R (−vx −uy) dxdy + i R (ux −vy)dxdy= 0

We do not assume the continuity of f ′ (z). Let C be a rectangular with edges parallel to x, y axes.Let I 0 = C f (z)dz , we cut the domain enclosed by C evenly, then C f (z)dz =

eC 1

+ + ·+ eC 4

dz

There exists a C i := C 1 such that C 1 f (z)dz ≥ 14 C f (z)dz . By induction, we have C n f (z)dz ≥ 14n |f (z)dz| (5.3)

The rectangles C n converges to a point z0 . For any ε > 0, there exists δ > 0 such that if |z −z0| < δ ,f (z) −f (z0)

z −z0 −f ′ (z0) < ε (5.4)

That is

|f (z)

−f (z0)

−(z

−z0)f ′(z0)

|< ε

|z

−z0

|. (5.5)

On the other hand, if n is large enough, the neighborhood

|z −z0| < δ is contained in C n .

By anti-derivative theorem, we have

C n f (z)dz = C n (f (z) −f (z0) −(z −z0)f (z0))dztherefore,

C n f (z)dz = C n (f (z) −f (z0) −(z −z0)f (z0))dz < C n ε|z −z

0||dz|Let L be the perimeter of C and d the diagonal of C , then dn = z− n d, Ln = 2− n L, hence

C n f (z)dz = C n (f (z) −f (z0) −(z −z0)f (z0))dz < C n ε|z −z0||dz| ≤4− n dLεSo

C n f (z)dz ≤dLε24

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which is arbitrary small. Now we conclude that

C n f (z)dz = 0 .Example 5.3. Let C be a simple closed curve containing the point z = 0 inside of C . Then

C 1z dz = 2πiwhere the integration is in the positive direction.

Proof. Let C 1 be the circle |z| = r inside of C . Then

C 1 1z dz = 2πiby standard parametrization. Now we consider the domain enclosed by C and C 1 , we can cut it into twopieces with simple closed boundaries (pictures), and so by Cauchy-Gaursat, we have

C f (z)dz = C − C 1 f (z)dz + C 1 f (z)dz = 2πi.5.2 Simply and multiply connected domains

5.2.1 Simply connected domains

Denition 5.4. A simply connected domain U is a domain such that every simple closed contour within itencloses only points of D .

Example 5.5. D = {z C | |z| < 1}is a simply connected domain. However, R = {z C | 1 < |z| < 2}is not simply connected.

Theorem 5.6. Let U be a simply connected domain, and f H (U ). Then for any closed contour (not necessarily simple closed contour) in U ,

C f (z)dz = 0 . (5.6)Remark 5.7. If the contour C has nite self intersection, we can apply Cauchy-Goursat theorem to proveit.

Example 5.8. If C is a closed contour lying in the open disk |z| < 2, then

C zez(z2 + 5) 2013 dz = 0 . (5.7)Corollary 5.9. Let U be a simply connected domain. A function f H (U ) must have an anti-derivativeeverywhere in U .

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5.2.2 Multi-connected domains

Denition 5.10. A domain that is not simply connected is said to be multi-connected.

Example 5.11. Let U = {1 < |z| < 2}. It is multi-connected.Theorem 5.12. Suppose that

(1) C is a simple closed contour in the positive direction;

(2) C k (k = 1 , · · · , n) are simple closed contours interior to C and all of them are in the negative direction.Suppose they are disjoint and their interiors are also disjoint.

If f is analytic throughout this multi-connected domain, then

C f (z)dz + nk=1 C k f (z)dz = 0 (5.8)Proof. Show it by a simple picture and then use Cauchy-Goursat.

Corollary 5.13. Let C 1 and C 2 be two positive oriented simple closed contours where C 1 is interior to C 2 . If f is analytic in the closed region enclosed by these two contours, then

C 1 f (z)dz = C 2 f (z)dz. (5.9)Example 5.14. Let C be a simple closed contour with positive direction. 0 is an interior point of C , then

C 1z dz = 2πi. (5.10)Example 5.15. f (z) = 1

z2

− 6iz − 8 compute

(1) |z |=1 f (z)dz = 0 .(2) |z |=3 f (z)dz = −π(3) |z |=3 f (z)dz = 0Proof. Since

f (z) = 12i

1z −4i −

1z −2i

Hence, f (z) is analytic in |z| ≤1. (1) is obvious by Cauchy-Gaursat.For part (2) |z |=3 f (z)dz = −12i |z |=3 1z −2i dz = | z− 2i|= 12 1z −2i dz = −π.For (3) ,

|z |=5 f (z)dz = 12i |z |=5 1z −4i dz − 12i |z |=5 1z −2i dz = 0

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5.3 Cauchy integral formula

Recall

|z− z0 |= r 1z −z0 dz = 2πiThe following is a fundamental result of complex analysis

Theorem 5.16 (Cauchy integral formula) . Let C be a simple closed contour in the positive direction and Dbe the domain enclosed by C . Suppose f H (D C ) and z0 D , then

f (z0) = 12πi C f (z)z −z0 dz (5.11)

or equivalently

C f (z)z −z0 dz = 2πi ·f (z0). (5.12)Proof. Let r be every small, and the circle |z −z0| = r is contained in D . It is easy to see that

C f (z)z −z0 dz = | z− z0 |= r f (z)z −z0 dz.Hence,

C f (z)z −z0 dz −2πi ·f (z0) = C f (z)z −z0 dz − |z− z0 |= r f (z0)z −z0 dz= | z− z0 |= r f (z) −f (z0)z −z0 dz

Since, f (z) is analytic and so it is continuous. Let ε be any number, then there exists δ > 0, such that if

|z

−z0

|< δ,

|f (z) −f (z0)| < ε.Hence, when r < δ,

C f (z)z −z0 dz −2πi ·f (z0) = |z− z0 |= r f (z) −f (z0)z −z0 dz≤ | z− z0 |= r f (z) −f (z0)z −z0 |dz|≤ ε |z− z0 |= r 1z −z0 |dz| = ε 1r ·2πr = 2πε.

Hence the left hand side is smaller than any number ε > 0, and it must be zero.Example 5.17.

| z |=2 z(9 −z2)(z + i) dz = |z |=2 z/ 9 −z2z + i dz = 2πi −i9 −(−i)2 = π5Example 5.18.

|z |=3 1z2 −6iz −8dz = |z |=31

z− 4iz −2i

dz = 2πi · 1

2i −4i = −π.

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Theorem 5.19. Let C be a simple closed contour in the positive direction and D be the domain enclosed by C . Suppose f H (D C ) and z0 D , then

f (n ) (z0) = n!2πi C f (z)(z −z0)n +1 dz (5.13)

or equivalently

C f (z)z −z0 dz = 2πin! ·f (n )(z0). (5.14)

Proof. We sketch the proof.

f ′(z0) = lim∆ z0 →0

f (z0 + ∆ z0) −f (z0)∆ z0

= lim∆ z0 →0

12πi

1∆ z0 C f (z)z −z0 −∆ z0 − f (z)z −z0 dz

= lim∆ z0 →0

12πi C f (z)(z −z0 −∆ z0)(z −z0) dz

= 12πi C f (z)(z −z0)2 dz

Similarly, we have

f ′′ (z0) = lim∆ z0 →0

f ′ (z0 + ∆ z0) −f ′ (z0)∆ z0

= lim∆ z0 →0

12πi

1∆ z0 C f (z)(z −z0 −∆ z0)2 − f (z)(z −z0)2 dz

= lim∆ z0 →0

12πi C 2(z −z0)f (z)(z −z0 −∆ z0)2(z −z0)2 dz + ∆ z0

= 22πi C f (z)(z −z0)3 dz.

Example 5.20. (1)

|z |=1e2z

z4 dz = 2πi

3! (e2z

)′′′

|z=0 = 8πi

3 .

(2) Let z0 be any point interior to a positively oriented simple closed contour C . Then

C 1z −z0 dz = 2πi. (5.15)and C 1(z −z0)n +1 dz = 0 (5.16)for any n ≥1.

Theorem 5.21. If f is analytic at a given point, then f (n ) are all analytic at that point.

Recall real and imaginary parts and derivatives.

Corollary 5.22. If f = u + iv is analytic at a point z = x + iy , then u and v have continuous partialderivatives of all orders at that point.

Theorem 5.23 (Morera) . Let f be a continuous function on a domain D . Suppose

C f (z)dz = 0 for every closed contour C in D , then f is analytic throughout of D .

Proof. This follows by Theorem 4.15 .

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5.4 Cauchy’s inequality and Liouville’s theorem.

Recall Cauchy’s integral formula for derivatives:

f (n ) (z0) = n!2πi C f (z)(z −z0)n +1 dz (5.17)

Theorem 5.24 (Cauchy’s inequality) . Let C be the contour |z −z0| = R with positive direction. Suppose f is analytic inside of |z −z0| ≤R and

M R := M ax | z− z0 |= R |f (z)| (5.18)then

|f (n ) (z0)| ≤ n!M R

Rn . (5.19)

Proof. By Cauchy’s integral formula for derivatives:

|f (n ) (z0)

| =

n!

2πi |z− z0 |= R

f (z)

(z −z0)n +1dz

≤ 12π | z− z0 |= R f (z)z −z0 |dz|

≤ 12π

M RRn +1 |z− z0 |= R |dz|

= n!M R

Rn

Theorem 5.25 (Liouville) . Bounded entire functions must be constant.

Proof. Suppose |f (z)| ≤M for any z C . For any xed z0 C , we apply Cauchy’s inequality that

f ′(z0) = 12π |z− z0 |= R f (z)(z −z0)2 dz

and

|f ′ (z0)| ≤ M R

when R → ∞, we have f ′ (z0) = 0for any z0 .

Proposition 5.26. Let u(x, y) : R 2 →R be a harmonic function. If it is bounded, then it must be a constant.Proof. Let u be the real part of some entire function f : C →C , i.e. f = u + iv . Then

g = ef = eu ·eiv .So g is a bounded entire function, and it is a constant. Finally, we know u is a constant.

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Theorem 5.27. (1) Let f (z) be an entire function such that

|f (z)| < |z|n for some positive integer n and all sufcient large |z|. Show that f (z) is a polynomial with degree at most n .

(2) Let f (z) be an entire function such that

|f (z)| < |z|12

for all sufcient large |z|. Show that f is a constant. That means, entire functions with “slow growth”must be constant. It is a generalization of Liouville’s theorem that: any bounded entire function must be constant.

Proof. It follows by the formula

f (n ) (z0) = n!2πi C f (z)(z −z0)n +1 dz.

Theorem 5.28 (Fundamental theorem of algebra) . Any polynomial

P (z) = an zn + an − 1zn − 1 · · ·+ a0, an = 0of degree n ≥1 has at least one zero. That is there exists at least one point z0 such that P (z0) = 0 .Proof. If P (z) is not zero for any z C , then

h(z) := 1P (z)

(5.20)

is clearly an entire function. Now we want to show that h(z) is also bounded. Hence by Liouville’s theorem,h(z) is constant which is absurd.

Next we show h(z) is bounded. We set

w = a0zn

+ a1zn − 1 · · ·+

an − 1z

and sof (z) = ( w + an )zn

Now we choose large R such that

|a0|Rn ≤ |

an |2n

, |a1|Rn − 2 ≤ |

an |2n

, · · · , |an − 1|

R ≤ |an |2n

Hence

|w| ≤ |a0zn |+ |

a1zn − 1 |+ · · ·+ |

an − 1z |

and when |z| ≥R , we have|w| ≤n × |

an |2n

= |an |2

.

Therefore, when |z| ≥R ,

|f (z)| = |w + an ||zn |≥ | |w| − |an ||Rn ≥ |an |Rn

2

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Hence when |z| ≥R ,|h| ≤

2

|an |Rn := M 1

On the other hand, since h(z) is analytic over |z| ≤R , hence it is bounded, i.e.

|h| ≤

M 2

when

|z| ≤

R

In summary, we obtain

|h| ≤max{M 1, M 2}

Theorem 5.29. Any polynomial

P (z) = an zn + an − 1zn − 1 + · · ·+ a0, an = 0of degree n ≥1 can be written as

P (z) = an (z

−zn )(z

−zn − 1)

· · ·(z

−z0)

5.5 Maximum modulus principle

Lemma 5.30. Suppose that for any z with |z −z0| < ε , we have

|f (z)| ≤ |f (z0)|then f (z) has the constant value f (z0) throughout that neighborhood.

Proof. For any 0 < r < ε we have

f (z0) = 1

2πi |z− z0 |= r

f (z)

z −z0dz =

1

2π 2π

0

f (z0 + re iθ )dθ Mean value property

hence,

|f (z0) ≤ 12π 2π0 |f (z0 + re iθ )|dθ ≤ 12π 2π0 |f (z0)|dθ = |f (z0)|

Therefore, for any θ [0, 2π],

|f (z0 + re iθ )| = |f (z0)| (5.21)Since r is arbitrary, we know for any z with |z −z0| < ε , we have

|f (z)| = |f (z0)| = cHence, f (z) must be a constant if

|z

−z0

|< ε .

Theorem 5.31 (Maximum modulus principle) . If f H (U ) and f is not a constant. Then |f (z)| has nomaximum value in U . That is, there is no point z0 U such that

|f (z)| ≤f (z0) for all z U .Proof. It follows from the above lemma and connecting trick.

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Example 5.32. f (z) = 1z− 1 which is analytic on |z| < 1. It is easy to see thatsup|z |< 1 |f (z)| = ∞

Corollary 5.33. If f is continuous on a closed and bounded region R . If f is analytic and not constant inthe interior of R. Then

maxz R |f (z)| = max∂R |f (z)| (5.22)That is, the maximum value of |f (z)| in R which is always attained on the boundary of R , and never in theinterior of R .

Corollary 5.34. Let u be a non-constant harmonic function over Ω R 2 = C . Then u attains both itsmaximum and minimum value on the boundary of Ω.

Proof. Suppose u attains its maximum value at some point (x0, y0) Ω, then we can pick a small disk centered at z0 = x0 + iy0 . In that small disk, we have

f = u + iv, g = ef

Hence,

|g

|attains its maximum value inside of that small disk, and so it is a constant. It is absurd.

If we consider −u , it is also harmonic, and so it attains its maximum value on the boundary of Ω. That isu attains its minimum value on the boundary.Example 5.35. f (z) = ez on |z| ≤1.5.6 Schwarz’s Lemma

Theorem 5.36 (Schwarz’s Lemma) . Let D be the unit disk in C and f : D →C be a holomorphic functionsatisfying f (0) = 0 and |f (z)| ≤1 for any z D .(1)

|f (z)| ≤ |z| for all z D (5.23)and

|f ′ (0)| ≤1. (5.24)(2) If there exists a point z0 D \ {0}such that we have equality in ( 8.3 ), i.e. |f (z0)| = |z0| , then there

exists θ0 R such that f (z) = eiθ0 z for all z D . (5.25)Proof. (1) Let f (z) = zh(z), then h(z) is analytic over |z| < 1. We shall show |h(z)| ≤ 1. For anyz0 D , we choose r such that |z0| < r < 1, then h(z) is analytic over D r and by maximum modulus

principle

|h(z0)| ≤max∂D r |h(z)| = max∂D r |f (z)||z| ≤

1r

Let r →1, we have |h(z0)| ≤1. Therefore|f (z)| = |z||h(z)| ≤ |z|

and

|f ′ (0)| = |h(0)| ≤1.(2) If |f (z0)| = |z0|, we have |h(z0)| = 1 , by maximum modulus principle h must be a constant h(z0). We

set h(z0) −eiθ 0 and so f (z) = eiθ0 .

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6 Power series, 18-20

6.1 Taylor’s Theorem

Denition 6.1. A sequence {zn }converges to w0 if for any ε > 0, there exists some N such that is n > N

|zn −w| < ε.Example 6.2. {zn = 1n 2 + i} →i.Denition 6.3. An innite series

∞

n =1zn = z1 + z2 + · · ·

converges to the sum S , if the partial sums

S N =N

n =1zn

converges to S as a sequence.Example 6.4. Consider

∞

n =0zn

That is zn = zn .

S N = 1 + z + z2 + zN − 1 = 1−zN

1 −zlim

N →∞S N =

11 −z

if

|z

|< 1. If

|z

|> 1, then it does not converge.

Example 6.5.1

1 −x = 1 + x + x2 + · · ·=

∞

n =0xn , |x| < 1

Hence1

1 −x = 1 + x + · · ·+ xN − 1 +

xN

1 −xCorollary 6.6. If a series of complex numbers an converges, lim |an | = 0 .Theorem 6.7. Suppose that a function f is analytic throughout a disk

|z −z0| < R 0centered at z0 with radius R0 . Then f (z) has the power series representation

f (z) =∞

n =0an (z −z0)n , |z −z0| < R 0 (6.1)

where

an = f (n ) (z0)

n! =

12πi | z− z0 |= r f (z)(z −z0)n +1 dz (6.2)

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Proof. We use Cauchy’s integral formula.For a given z, we set |z −z0| = r . Now we choose a contour |s −z0| = r 0 with r < r 0 < R 0 . Recall that

1s −z

= 1

s −z0 −(z −z0)

=

1

s −z0 · 1

1 − z− z0s− z0

= 1s −z0

1 +z −z0s −z0

+ · · ·+z −z0s −z0

N − 1

+z− z0s− z0

N

1 − z− z0s− z0=

N − 1

n =0

(z −z0)n(s −z0)n +1

+ (z −z0)N

(s −z)(s −z0)N Hence, we have

f (z) = 12πi

|s− z |= ε

f (s)s

−z

ds = 12πi

|s− z

0|= r

0

f (s)s

−z

ds

= 12πi |s− z0 |= r 0 f (s) N − 1n =0 (z −z0)n(s −z0)n +1 + (z −z0)N (s −z)(s −z0)N ds

=N 1

n =0

f (n ) (z0)n!

(z −z0)n + E N where

E N = (z −z0)N

2πi |s− z0 |= r 0 f (s)(s −z)(s −z0)N ds (6.3)It is easy to see that

|s

−z

|≥ | |s

−z0

| − |z

−z0

||= r 0

−r

Let M be the maximum value of |f (s)| on C = {|z −z0| = r 0}, we haveE N ≤

rN

2πM

(r 0 −r )r N 0 2πr 0=

Mr 0r 0 −r

rr 0

N

since r/r 0 < 1, when N → ∞, E N →0.Example 6.8.

ez =∞

n =0

zn

n!, |z| < ∞

Example 6.9.

z2e2z = z2∞

n =0

3n znn!

=∞

n =0

3nn!

zn +2 , |z| < ∞Example 6.10.

sin z =∞

n =0(−1)n

z2n +1

(2n + 1)!, |z| < ∞

since


2i

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For example, we have

b1 = 12πi |z |= 12 f (z)dz = 1 , a0 = 12πi | z |= 12 f (z)z dz = 0

bn =

1

2πi |z |=12

f (z)

z− n +1 dz = 0 , if n ≥0.a1 =

12πi |z |= 12 f (z)z1+1 dz = −1

(2) Domain 1 < |z| < ∞.1

1 + z2 =

1z2 ·

11 + 1z2

= 1z2 ·

∞

n =0(−1)n

1z2

n

=∞

n =0(−1)n z− 2n − 2

f (z) = z− 1 +∞

n =0

(−1)n +1 z− 2n − 1 =∞

n =1

(−1)n +1 z− 2n − 1.For example, we have

b1 = 12πi |z |=3 f (z)dz = 0

an = 0

In particular,1

2πi | z |=3 1z(1 + z2) dz = 0Example 6.16. Let f (z) = e

1z

(1) For the domain 0 < |z| < ∞. e 1z =∞

n =0

z− n

n!

Hence,a0 = 1 , an = 0 when n ≥1.

b1 = 12πi |z |=1 e 1z dz = 1

bn = 12πi |z |=1 e 1z zn − 1dz = 1n!

(2) Although f (z) = e1/z

is analytic at point z = 1 , we can formulate the Laurent series for 1 < |z −1|

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e1z =

∞

n =0an (z −1)n +

∞

n =1bn (z −1)− n

Hence, we getan = 0 when n ≥1

a0 = 12πi |z− 1|=2

f (z)z −1dz = 1 .

b1 = 12πi |z− 1|=2 f (z)dz = 1 .

6.3 Uniform properties of series

Example 6.17. For which values of z, the following series converge

∞

n =0

z + 12z + 3

n

Since we know n zn converges when |z| < 1. Hence, we know when

z + 12z + 3

< 1

that series converges.

|z + 1 | < |2z + 3 |Let z = x + iy , we get

(x + 1) 2 + y2 < (2x + 3) 2 + y2

and so x = Re(z) > −43 or x = Re(z) < −2.Denition 6.18. Convergence radius R of a series n an (z −z0)n . For any point z, with |z −z0| < R , theseries converges, and for any z with |z −z0| > R , the series does not converge. In general, we do not knowthe case |z −z0| = R .Example 6.19. (1) n z

n , the converges radius is R = 1 .

(2) nz nn ! , the converges radius is R = ∞.

Theorem 6.20. Let n an zn be a series. If

limn

anan +1

exists, then

R = limn

anan +1

is the convergence radius of n an zn .

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Example 6.21.∞

n =0(−1)n 2− n z2n

Here, we set w = −z2

2 , then it is

n =0w

k=

11 −w =

11 + z22

The convergence radius is | −z2/ 2| < 1, that is |z| < √ 2.Denition 6.22. A series n an z

n is absolutely convergent, if

n|an zn |

converges.

Theorem 6.23. If a power series∞

n =0

an (z

−z0)n

converges when z = z1 (z1 = z0) , then it is absolutely convergent at each point z in the open disk |z −z0| 0. Then the function

f (z) =∞

n =0

an zn , |z| < R

is analytic. The derivatives of f (z) are obtained by differentiating the series term by term,

f ′ (z) =∞

n =1na n zn − 1, f ′′ (z) =

∞

n =2n(n −1)an zn − 2, |z| < R

and similarly for the higher-order derivatives. The coefcients of the series are given by

an = f (n )(0)

n! , n = 0 , 1, · · ·

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Example 6.25.1

1 −z =

∞

n =0zn , |z| < 1

1

(1 −z)2 =

∞

n =1

nz n − 1,

|z

|< 1

2(1 −z)3

=∞

n =2n(n −1)zn − 2, |z| < 1

6.4 Review for Midterm II.

Example 6.26. Compute

|z |=2 ezz2(z + 1)( z2 + 3) dz.

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7 Residue, 21-23

7.1 Cauchy’s residue theorem

Question. How to compute

|z |=1

sin1

zdz.

We say that z0 is an isolated singular point of f if f is analytic throughout some deleted neighborhood

0 < |z −z0| < εExample 7.1.

z + 1z3(z2 + 1)

has three isolated singular points z = 0 , z = ±i .Let z0 be an isolated singular point of a function f , there exists a positive number R such that f is analytic

throughout 0 < |z −z0| < R . f has Laurent seriesf (z) =

∞

n =0an (z −z0)n +

∞

n =1

bn(z −z0)n

(7.1)

where

an = 12πi C f (z)(z −z0)n +1 dz, n = 0 , 1, 2, · · · , (7.2)

bn = 12πi C f (z)(z −z0)− n +1 dz, n = 0 , 1, 2, · · · , (7.3)

Here,

b1 = 12πi |z− z0 |= R 1 f (z)dz

the complex number b1 is called the residue of f at the singular point z0 and we write it asb1 = Res z= z0 f (z) (7.4)

Or equivalently,

C f (z)dz = 2πiRes z= z0 f (z).Example 7.2. Consider the integral

|z |=1 z2 sin 1z dzSince

sin z = z

− z

3! +

z5

5! +

· · ·we havesin(1/z ) =

1z −

13!z3

+ · · ·and so

z2 sin(1/z ) = z − 13! ·

1z

+ · · ·That is

|z |=1 z2 sin(1/z )dz = 2πi ·(−1/ 3!) = −π/ 3.40

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Theorem 7.3 (Cauchy’s Residue Theorem) . Let C be a simple closed contour, describe in the positivedirection. If a function f is analytic insider and on C except for a nite number of singular points zk(k = 1 , · · · , n) inside of C . Then

C

f (z)dz = 2πin

k=1

Res z= zk f (z) (7.5)

Proof. It follows by Cauchy-Goursat theorem.

Remark 7.4. That means we can compute an integral by using Laurent series.


|z |=2 5z −2z(z −1) dz.Solution. We have two methods to compute it.

(1) By Laurent series. Residue at point z = 0 . Since

11 −z

= zn , |z| < 1and so

5z −2z(z −1)

= 2−5z

z (1 + z + z2 + · · ·) =

2z −5 + · · ·= Res z=0 f (z) = 2

Similar, the Residue at point z = 1 is

5z −2z(z −1)

= 5(z −1) + 3

z −1 · 1 + ( z −1) = 5 +

3z −1

1 −(z −1) + ( z −1)2 + · · ·Hence Res z=1 f (z) = 3

(2) By Cauchy’s integral formula. it is easy to see that

|z |=2 5z −2z(z −1) dz = |z |= ε 5z −2z(z −1) dz+ | z− 1|= ε 5z −2z(z −1) dz = 5 ·0 −20 −1 ·2πi + 5 ·1 −21 2πi = 10πi7.2 Three types of singularities

7.2.1 Poles of order m .

Let z0 be a singular point of f and f be analytic over 0 < |z −z0| < R . The Laurent series is

f (z) =∞

n =0an (z −z0)n +

∞

n =1bn(z −z0)n

If bm = 0 and bm +1 = bm +2 = · · ·= 0 , i.e.

f (z) =∞

n =0an (z −z0)n +

b1(z −z0)n

+ b1

(z −z0)n + · · ·+

bm(z −z0)n

Then z0 is called a pole of order m.

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7.2.3 Essential singular points.

If a singular point is neither a pole nor a removable singular point, it is called an essential singular point.

Example 7.11.

e1/z =∞

n =0

z− n

n! , 0 <

|z

|<

∞suppose f is analytic at point z0 . If f (z0) = 0 and, there exists some positive integer m such that

f (m ) (z0) = 0 , f (z0) = f ′

(z0) = · · ·= f (m − 1) (z0) = 0 .Then z0 is called a zero of order m .

Example 7.12. f (z) = ( z −1)3 + ( z −1)4 . Then z0 = 1 is a zero of order 3.Proposition 7.13. Let z0 be a zero point of f . z0 is a zero of order m if and only if f (z) can be written inthe form

f (z) = g(z)(z

−z0)m

where g(z) is analytic and g(z0) = 0 .

Proposition 7.14. If f (z) and h(z) are analytic at z0 , and if h(z) has a simple zero at z0 , then

Res z= z0f (z)h(z)

= f (z0)h ′ (z0)

Proof. It follows by Cauchy’s integral formula.

Example 7.15.

Res z= iz3

z2 + 1 = −

12

7.3 Applications of Residue.


∞0 x2x6 + 1 dx = π6Proof. Consider f (z) = z

6

z6 +1 around the simple closed contour C R consisting of two components

C R = C 1 + C 2

where C 1 is the line segment on the real x-axis connection −R and R and C 2 = {|z| = R, Im (z) > 0}.Three singular points inside of the contour C R are

c0 = eiπ/ 6, c1 = i, c2 = ei5π/ 6.

Hence, by Cauchy’s residue theorem,

C R f (z)dz = 2πi Resf (z) = 2 πi z26z5 |c0 ,c1 ,c2 = π343

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where we have used the fact that (Proposition 7.14).

Res z= iz2

z6 + 1 =

z2

(z6 + 1) ′ |z= i = z2

6z5 |z= iOn the other hand,

B R f (z)dz = C 1 z2

z6 + 1 dz + C 2 z2

z6 + 1 dz

It is obvious that

C 1 z2z6 + 1 dz = R− R x2x6 + 1 dx →2 + ∞0 x2x6 + 1 dx as x → ∞.On the other hand,

C 2 z2z6 + 1 dz ≤ C 2 |z|2|z|6 −1 |dz| ≤ R2R6 −1πR →0 as R → ∞Finally, we obtain,

2 + ∞

0

x2

x6 + 1dx = lim

R →∞ C R z2

z6 + 1dz − C 2 z

2

z6 + 1dz =

π6

.

Example 7.17.

∞−∞ cos(3x)(x2 + 1) 2 = 2πe3Proof. Consider the function

f (z) = ei3z

(z2 + 1) 2

around around the simple closed contour C R consisting of two components

C R = C 1 + C 2

where C 1 is the line segment on the real x-axis connection −R and R and C 2 = {|z| = R, Im (z) > 0}.The singular point inside of the contour C R is z = i. It is a pole of order 2. Hence

C R f (z)dz = 2πi e3iz(z + i)2 ′z= i = 2 πi 3ie3iz(z + i)2 z= i − 2e3iz(z + i)3 |z= i= 2 πi

3ie− 3

4i − 2e− 3

8i3=

2πe3

On the other hand, C 1 f (z)dz = R− R ei3x(x2 + 1) 2 dx.As similar as the previous example, we have

+ ∞−∞ cos(3x)(x2 + 1) 2 dx = limR →∞ Re C R ei3z(z2 + 1) 2 dz − C 2 ei3z(z2 + 1) 2 dz = 2πe3 .

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7.4 Zeros of an analytic function are isolated

Theorem 7.18. If Ω is a domain, and f (z) is an analytic function on Ω that is not identically zero, then the zeros of f (z) are isolated.

Proof. Let z0 Ω be a zero point of f . Let |z −z0| < r be a small disk contained in Ω. We have

f (z) =∞

n =0

f (n ) (z0)n!

(z −z0)n

for any z with |z −z0| < r . Suppose f (z) is not identically zero in the small disk |z −z0| < r , we see thatthere exists some m0 such that if m ≥m0 , then f (m ) (z0) = 0 and f (k) (z0) = 0 if k < m 0 . That is, z0 is azero of zero m0 , so

f (z) = ( z −z0)m h(z)with h(z) not zero at z0 . Hence z0 is an isolated zero point.

If f (z) is identically zero at that small disk |z −z0| < r , we can use connecting trick to show f (z) isconstantly zero.

Example 7.19. In the real variable case, the above result is not true. For example,

f (x)x2 sin 1x , x = 0 .0, x = 0

This function is differentiable over R . It has zeros

± 1nπ

They are not isolated since 0 is a limit point of them.

Theorem 7.20 (Uniqueness Principle) . If f (z) and g(z) are analytic on a domain Ω , and if f (z) = g(z) for z belonging to a set that has a nonisolated point, then f (z) = g(z) for all z Ω. In particular, if twoholomorphic functions coincide on a small open subset of Ω , then they are the same.

Proof. We can consider the function F (z) := f (z) −g(z)

Corollary 7.21. Let f (z), g(z) be analytic over Ω. If f (z)g(z) = 0 for all z Ω , then either f (z) ≡0 or g(z) ≡0.Example 7.22. Prove that sin2 z + cos2 z = 1 .

Proof. Since F (z) = sin 2 z + cos 2 z −1 is zero when z R . Hence the entire function F (z) must be zerofor all z C .Example 7.23. sin(z) has innitely many zero points z = nπ .

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8 Conformal mapping and Linear fractional transformations, 23-26

Denition 8.1. A function f is called conformal at a point z0 if it preserves the angle at that point. Moreprecisely, if there are two curves pass through z0 , with an angle θ0 , then the image curves under f have thesame angle θ0 . (picture here.)

Theorem 8.2. . If f (z) is analytic at z0 and f ′ (z0) = 0 , then f (z) is conformal at z0 .

Proof. Let γ : [−1, 1] →C be a smooth curve such that γ (0) = z0 . It is easy to see that(f ◦γ ) ′ (0) = f ′(z0) ·γ ′ (0)

If we have two such curves, γ 1 and γ 02, then the angle between them is preserved under f , since the imagetangent vectors are a nonzero constant scalar multiple of the original tangent vectors.

Denition 8.3. Let Ω1 and Ω2 be two domains of C . A conformal mapping f : Ω1 →Ω2 is a continuouslydifferentiable function that

(1) f is conformal at each point of Ω1 , and

(2) f maps Ω1 one-to-one onto Ω2 , i.e. f is bijective.

Proposition 8.4. Let f : Ω1 →Ω2 be a conformal mapping, then the inverse map f − 1 : Ω2 →Ω1 is also aconformal mapping. Moreover, if w0 = f (z0) , then, f ′ (z0) = 0 and

[f − 1(w0)]′ = 1f ′ (z0)

.

8.1 Elementary functions

Example 8.5.f (z) = z2 :

{Re(z) > 0

} →C

\(

−∞, 0]

Example 8.6. Fix θ0 , 0 < θ0 < π . If 0 < θ < π/a ,

f (z) = za : {|argz | < θ 0}→{|argz | < aθ 0}In particular, the function

f (z) = zπ

2θ 0 : {|argz | < θ 0}→{|argz | < π/ 2}Example 8.7. f (z) = ez is conformal at every point z C . But it is NOT a conformal mapping, since it isnot injective (it is periodic). However,

f (z) = ez :

{|Im (z)

|< π

} →C

\(

−∞, 0]

is a conformal mapping.

Example 8.8. The principle Log(z) is well-dened and holomorphic over C \ (−∞, 0]. In fact,f (z) : C \ (−∞, 0] → {Im (z) < π }

In fact, it is the inverse mapping of

f (z) = ez : {|Im (z)| < π } →C \ (−∞, 0].

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8.2 Linear fractional transformations

8.2.1 Stereographic projection.

Consider the unit sphere S 2 in R 3

S 2 =

{(x1, x2, x3) R 3

|x2

1 + x2

2 + x2

3 = 1

}.

Let N = (0 , 0, 1) be the North pole on S 2 . We can do the projection from the sphere to the complex planeC = {(x,y, 0) R 2}as a subspace of R 3 . More precisely, for any point (x1, x2, x3) (other than (0, 0, 1))on the sphere, there is a line through (x1, x2, x3) and (0, 0, 1). The intersection point with C is denoted by

z = x + iy.

In fact, that line isX = tx 1,Y = tx 2,Z = 1 + t(x3 −1).

We see at the intersection point (x,y, 0), we have Z = 0 , i.e.

t = 1

1 −x3.

Hence,

x = x11 −x3

, y = x21 −x3

, z = x + iy = x1 + ix 2

1 −x3Or equivalently,

|z|2 = x21 + x

22

(1− x3 )2 = 1+ x31− x3 ;

x3 = |z |2 − 1

|z|2 +1 ;

x1 = z+ z|z|2 +1 ;

x2 = z− zi( |z |2 +1)Now we set

Ψ : S 2 \ {N } →Cis bijective. Note also that the point N is mapped into the point ∞. We also get the map

Ψ : S 2 →C {∞}where C {∞}is called the extended complex plane, and denoted by C .A fractional linear transformation is a function of the form

f (z) = az + bcz + d (8.1)

where a,b, c, d are complex constants satisfying

ad −bc = 0 . (8.2)Fractional linear transformations are also called M öbius transformations. Since

f ′ (z) = ad −bc(cz + d)2

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the condition ad −bc = 0 simply guarantees that f (z) is not constant.Inverse map. Let

w = az + bcz + d

, −→z = b−dwcw −a

Example 8.9. (1) f (z) = z + b, a = 0 . It is called a translation.

(2) f (z) = az with a = 0 is called a dilation.

(3) f (z) = 1z is called an inversion

It is convenient to regard a fractional linear transformation as a map from the extended complex plane

C = C ∞to itself. If f (z) is afne, we dene f (∞) = ∞ Otherwise, f (z) has standard form with c = 0 , and wedene

f (−d/c ) = ∞and

f (∞) = limz→∞ f (z) = limz→∞a + b/zc + d/z

= ac

.

Proposition 8.10. The fractional linear transformation satises

f (z0) = 0 , f (z1) = 1 , f (z2) = ∞is given by

f (z) = z −z0z −z2

z1 −z2z1 −z0

for z0, z1, z2 C . If z0 = ∞ , then we get f (z) = z1 −z2z −z2

Theorem 8.11. . Given any three distinct points z0, z1, z2 in the extended complex plane, and given anythree distinct values w0, w1, w2 in the extended complex plane, there is a unique fractional linear transfor-mation w = F (z) such that

F (z0) = w0, F (z1) = w1, F (z2) = w2.

Proof. Letf (z0) = 0 , f (z1) = 1 , f (z2) = ∞

and

g(w0) = 0 , g(w1) = 1 , g(w2) = ∞then F = g− 1 ◦ f satises F (z0) = w0, F (z1) = w1, F (z2) = w2.

Theorem 8.12. All linear fractional transformations are compositions of these three transformations.

Theorem 8.13. A fractional linear transformation maps circles in the extended complex plane to circles.

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Example 8.14. . Find the fractional linear transformation mapping −1 to 0, ∞to 1, and i to ∞.z + 1z −i

Example 8.15. Find the fractional linear transformation mapping 0 to −1, i to 0, and ∞to 1.Since f maps i to 0, it is of the form

f (z) = a(z −i)

cz + dSince f maps ∞to 1, we have a = c.

f (z) = a(z −i)

az + dSince f maps 0 to −1, we have d = ia

f (z) = z −iz + i

Example 8.16. Determine the images of each of the following sets under the above fractional linear trans-formation:

(1) the imaginary axis,

(2) the right half-plane,

(3) the real axis,

(4) the upper half-plane,

(5) the unit disk.

Sketch the images of horizontal lines and of vertical lines under the transformation.Solution.

(1) Pick three points to gure out the circle. 0 → −1, i →0, ∞ →1. The image is the real axis;(2) Right half plane to the lower half plane.

(3) The unit circlea −ia + i

= 1 .

(4) The upper half plane to the unit disk since the boundary of upper half plane is corresponding to theboundary of the unit disk. f (i) = 0 .

(5) f (−1) = i , f (1) = −i , f (i) = 0 . Hence f maps unit circle to the imaginary line. since f (0) = −1, f maps unit circle to left half plane. We also see that

f (eiθ ) = − cosθ1 + sin θ

i.

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8.3 f : D →DLemma 8.17. If g : D →D is a conformal mapping with

g(0) = 0 ,

then g(z) must be a rotation, i.e. g(z) = eiθ z, z D for some xed θ [0, 2π].

Proof. We apply Schwarz’ Lemma to g(z), we obtain

|g(z)| ≤ |z|When we apply Schwarz’s Lemma to g− 1, we have

|g− 1(w)| ≤ |w|If we choose w = g(z), that is |z| ≤ |g(z)|, therefore

|g(z)| = |z|Hence g(z)/z is a holomorphic function with constant modulus, and it must be a constant. Since |g(z)| ≤|z|, we see g(z) = eiθ z.Theorem 8.18. If f : D →D is a conformal mapping, then

f (z) = eiθ z −a1 −az

for

|a

|< 1.

Proof. We set

g(z) = z −a1 −az

.

It is easy to see that g(z) maps the unit circle to unit circle. Since a D , we see g(z) maps the unit disk tounit disk.

Let f : D →D be an arbitrary conformal mapping. Let a = f − 1(0) . Thenh(z) = f ◦g− 1(z) : D →D , h(0) = 0

Hence h(z) = f ◦g− 1(z) = eiθ z, and so f (z) = eiθ z− a1− az .

Example 8.19. Find a conformal mapping f fromD

toD

such that

f (12

) = i4

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8.4 f : H →DIf f : H D is a linear fractional transformation, then

f (z) = eiθz −az −a

with a H .Example 8.20. Find a conformal mapping of the part of the upper half-plane outside the unit circle onto theentire upper half-plane mapping −1 to −1, i to 0, and +1 to +1 . (picture.)(1) ξ = z− 1z+1 maps the upper half-plane outside the unit circle onto the rst quadrant.

ξ (−1) = −∞, ξ (0) = −1, ξ (1) = 0(2) η = ξ 2 maps the rst quadrant onto the upper half plane;

η(−∞) = + ∞, η(−1) = 1 , η(0) = 0(3) w = 1+ η1− η maps the upper half plane to the upper half plane with the points description.

w(+ ∞) = −1, w(1) = 0 , w(0) = 1(4) Finally, it is

w = 1 + ( z −1/z + 1) 21 −(z −1/z + 1) 2

= 12

(z + z− 1)

8.4.1 Summary

Let H be the upper half plane {Im (z) > 0}and D be the unit disk {|z| < 1}.(1) ez maps strip onto C \ (−∞, 0]:

f (z) = ez : {|Im (z)| < π } →C \ (−∞, 0]and Log(z) is the inverse map of ez ,

g(z) = Log(z) : C \ (−∞, 0] → {|Im (z)| < π }(2) zα maps circular sector to circular sector.

f (z) = z2 : {Re(z) > 0} →C \ (−∞, 0]and

f (z) = √ z : C \ (−∞, 0] → {Re(z) > 0}(3) The fractional linear transformation satises

f (z0) = 0 , f (z1) = 1 , f (z2) = ∞is given by

f (z) = z −z0z −z2

z1 −z2z1 −z0

for z0, z1, z2 C . If z0 = ∞, then we getf (z) =

z1 −z2z −z2

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(4) If f : D →D is a conformal mapping, then f is of the formf (z) = eiθ

z −a1 −az

for |a| < 1.(5) If f : H D is a conformal mapping, then f is of the form

f (z) = eiθz −az −a

with a H .

8.5 Schwarz-Pick.

Theorem 8.21 (Schwarz’s Lemma) . Let D be the unit disk in C and f : D →C be a holomorphic functionsatisfying f (0) = 0 and |f (z)| ≤1 for any z D .(1)

|f (z)| ≤ |z| for all z D (8.3)and

|f ′ (0)| ≤1. (8.4)(2) If there exists a point z0 D \ {0}such that we have equality in ( 8.3 ), i.e. |f (z0)| = |z0| , then there

exists θ0 R such that f (z) = eiθ0 z for all z D . (8.5)The Schwarz-Pick lemma observes that there is no need to restrict to f (0) = 0 .

Lemma 8.22 (Schwarz-Pick) . Let f : D →D be holomorphic, then(1) For any z1, z2 D , |f (z1) −f (z2)|

|1 −f (z1)f (z2)| ≤ |z1 −z2||1 −z1z2|

(2) For any z D ,|f ′ (z)| ≤

1− |f (z)|21 − |z|2

Proof. We x z2 D . Firstg(z) =

z −z21 −z2z

: D →D , g(z2) = 0

h(z) = z −f (z2)1 −f (z2)z: D →D , g(f (z2)) = 0

HenceF (z) = h(f (g− 1(z))) : D →D , F (0) = 0

By Schwarz Lemma, we have

|F (z)| ≤ |z|In particular

|F (g(z1))| ≤ |g(z1)|, = | h(f (z1)) | ≤ |g(z1)|.52

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That is (1) . For (2) , since we have

|f (z1) −f (z2)||1 −f (z1)f (z2)| ≤

|z1 −z2||1 −z1z2|

= |f (z1) −f (z2)||z1 −z2| ≤

|1 −f (z1)f (z2)||1 −z1z2|

when z1

→z2 , we get

|f ′ (z2)| ≤ 1− |f (z2)|21 − |z2|2

Example 8.23. Suppose that f (z) is analytic on D and f (0) = 0 . If Ref (z) < 1, then we get

(1) |f (z)| ≤ 2|z|1−| z | ;(2) |f ′ (0)| ≤2.Proof. We have

g = z

z −2 :

{Rez < 1

} →D

Now we get

g ◦ f = f (z)f (z) −2

: D →Dthen we have

|g ◦ f (z)| ≤ |z| = | f (z)| ≤ 2|z|1 − |z|

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