Lecture Notes Complex Analysis MATH 435 - New …iavramid/notes/complanal.pdfLecture Notes Complex...

Lecture NotesComplex Analysis

MATH 435Instructor: Ivan Avramidi

Textbook: J. E. Marsden and M. J. Hoffman, (Freeman, 1999)

New Mexico Institute of Mining and Technology

Socorro, NM 87801

July 10, 2009

Author: Ivan Avramidi; File: complanal.tex; Date: December 8, 2009; Time: 12:32

Contents

1 Analytic Functions 11.1 Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1.1 Complex Number System . . . . . . . . . . . . . . . . . 11.1.2 Algebraic Properties . . . . . . . . . . . . . . . . . . . . 21.1.3 Properties of Complex Numbers . . . . . . . . . . . . . . 3

1.2 Elementary Functions . . . . . . . . . . . . . . . . . . . . . . . . 71.2.1 Exponential Function . . . . . . . . . . . . . . . . . . . . 71.2.2 Trigonometric Functions . . . . . . . . . . . . . . . . . . 71.2.3 Logarithm Function . . . . . . . . . . . . . . . . . . . . 81.2.4 Complex Powers . . . . . . . . . . . . . . . . . . . . . . 91.2.5 Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.3 Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . . 111.3.1 Open Sets . . . . . . . . . . . . . . . . . . . . . . . . . . 111.3.2 Mappings, Limits and Continuity . . . . . . . . . . . . . 121.3.3 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . 131.3.4 Closed Sets . . . . . . . . . . . . . . . . . . . . . . . . . 141.3.5 Connected Sets . . . . . . . . . . . . . . . . . . . . . . . 161.3.6 Compact Sets . . . . . . . . . . . . . . . . . . . . . . . . 171.3.7 Uniform Continuity . . . . . . . . . . . . . . . . . . . . . 181.3.8 Path Covering Lemma . . . . . . . . . . . . . . . . . . . 191.3.9 Riemann Sphere and Point at Infinity . . . . . . . . . . . 19

1.4 Analytic Functions . . . . . . . . . . . . . . . . . . . . . . . . . 201.4.1 Conformal Maps. . . . . . . . . . . . . . . . . . . . . . . 221.4.2 Cauchy-Riemann Equations . . . . . . . . . . . . . . . . 231.4.3 Inverse Functions . . . . . . . . . . . . . . . . . . . . . . 24

1.5 Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271.5.1 Exponential Function and Logarithm . . . . . . . . . . . 27

I

II CONTENTS

1.5.2 Trigonometric Functions . . . . . . . . . . . . . . . . . . 271.5.3 Power Function . . . . . . . . . . . . . . . . . . . . . . . 281.5.4 The n-th Root Function . . . . . . . . . . . . . . . . . . . 28

2 Cauchy Theorem 292.1 Contour Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . 292.2 Cauchy Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 342.3 Cauchy Integral Formula . . . . . . . . . . . . . . . . . . . . . . 372.4 Maximum Modulus Theorem and Harmonmic Functions . . . . . 41

2.4.1 Maximum Modulus Theorem . . . . . . . . . . . . . . . 412.4.2 Harmonic Functions . . . . . . . . . . . . . . . . . . . . 432.4.3 Dirichlet Boundary Value Problem . . . . . . . . . . . . . 44

3 Series Representation of Analytic Functions 473.1 Series of Analytic Functions . . . . . . . . . . . . . . . . . . . . 473.2 Power Series and Taylor Theorem . . . . . . . . . . . . . . . . . 523.3 Laurent Series and Classification of Singularities . . . . . . . . . 56

4 Calculus of Residues 614.1 Calculus of Residues . . . . . . . . . . . . . . . . . . . . . . . . 61

4.1.1 Removable Singularities . . . . . . . . . . . . . . . . . . 614.1.2 Simple Poles. . . . . . . . . . . . . . . . . . . . . . . . . 614.1.3 Double Poles. . . . . . . . . . . . . . . . . . . . . . . . . 624.1.4 Higher-Order Poles. . . . . . . . . . . . . . . . . . . . . 634.1.5 Esssential Singularities. . . . . . . . . . . . . . . . . . . 63

4.2 Residue Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 644.3 Evaluation of Definite Integrals . . . . . . . . . . . . . . . . . . . 66

4.3.1 Rational Functions of Sine and Cosine . . . . . . . . . . . 664.3.2 Improper Integrals . . . . . . . . . . . . . . . . . . . . . 664.3.3 Fourier Transform. . . . . . . . . . . . . . . . . . . . . . 684.3.4 Cauchy Principal Value . . . . . . . . . . . . . . . . . . . 704.3.5 Integrals with Branch Cuts . . . . . . . . . . . . . . . . . 714.3.6 Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . 72

4.4 Evaluation of Infinite Series . . . . . . . . . . . . . . . . . . . . 73

5 Analytic Continuation 755.1 Analytic Continuation and Riemann Surfaces . . . . . . . . . . . 755.2 Roche Theorem and Princile of the Argument . . . . . . . . . . . 79

complanal.tex; December 8, 2009; 12:32; p. 1

CONTENTS III

5.2.1 Root and Pole Counting Formulas . . . . . . . . . . . . . 795.2.2 Principle of the Argument . . . . . . . . . . . . . . . . . 805.2.3 Injective Functions . . . . . . . . . . . . . . . . . . . . . 81

5.3 Mapping Properties of Analytic Functions . . . . . . . . . . . . . 83

Bibliography 83

Answers To Exercises 87

Notation 89


IV CONTENTS


Chapter 1

Analytic Functions

1.1 Complex Numbers

1.1.1 Complex Number System

•

Definition 1.1.1 The system of complex numbers, C, is the set R ofordered pairs of real numbers with two binary operations, addition andmultiplication, defined by

(x1, y1) + (x2, y2) = (x1 + x2, y1 + y2)

(x1, y1)(x2, y2) = (x1x2 − y1y2, x1y2 + y1x2),

and a scalar multiplication by a real number

a(x, y) = (ax, ay) .

• The numbers of the form (x, 0) are called the real numbers and the numbersof the form (0, y) are called pure imaginary numbers.

• The real part and the imaginary part of a complex number are real num-bers defined by

Re (x, y) = x , Im (x, y) = y .

• The number 1C = (1, 0) is the complex identity and the number i = (0, 1)is called the imaginary unit.

1

2 CHAPTER 1. ANALYTIC FUNCTIONS

• An arbitrary complex number can be represented as

(x, y) = x + iy .

• The property of the imaginary unit

i2 = −1 .

1.1.2 Algebraic Properties• Addition and multiplication of complex numbers are: associative, commu-

tative and distributive.

• Every non-zero complex number z = x + iy has the multiplicative inverse

z−1 =x

x2 + y2 − iy

x2 + y2

such thatzz−1 = 1 .

• The quotient of z by w is defined by

zw= zw−1 .

• All the usual algebraic rules of real numbers hold for complex numbers.

• The set of real numbers is a (complete ordered) field.

• The set of complex numbers forms a (complete but not ordered) field.

•

Proposition 1.1.1 For any complex number z = x + iy there is acomplex number w = a + ib with

a = ±

√x +

√x2 + y2

2, b = ±sign (y)

√−x +

√x2 + y2

2

such that w2 = z.

Proof: Check directly.�


1.1. COMPLEX NUMBERS 3

• The quadratic equationaz2 + bz + c = 0

with complex numbers a, b, c has solutions

z =−b ±

√b2 − 4ac

2a.

• Examples.

1.1.3 Properties of Complex Numbers• Complex numbers are represented as points on the real plane R2.

• Addition of complex numbers can be represented by the vector addition.

• Polar Representation of complex numbers

z = x + iy = r(cos θ + i sin θ) ,

where the real number r = |z| (called the absolute value (or the norm, orthe modulus))) is defined by

|z| =√

x2 + y2 ,

and the angle θ (called the argument) is defined by

cos θ =xr, sin θ =

yr.

• The argument arg z can be thought as an infinite set of values

arg z = {θ + 2πn|n ∈ Z}

• Specific values of the argument are called branches of the argument.

• Multiplication of Complex Numbers in Polar Form.

Proposition 1.1.2 For any complex numbers z1 and z2 there holds

r1(cos θ1 + i sin θ1)r2(cos θ2 + i sin θ2) = r1r2[cos(θ1 + θ2)+ i sin(θ1 + θ2)]

In other words,|z1z2| = |z1| |z2| ,

arg(z1z2) = arg z1 + arg z2 (mod 2π) .



Proof: Follows from the trigonometric identities.�

• De Moivre Formula.

Proposition 1.1.3 For any complex numbers z = r(cos θ+ i sin θ) anda positive integer n there holds

zn = rn[cos(nθ) + i sin(nθ)]

In other words,|zn| = |z|n ,

arg(zn) = n arg z (mod 2π) .

Proof: By induction.�

• Roots of Complex Numbers.

Proposition 1.1.4 Let w = r(cos θ + i sin θ) be a nonzero complexnumber and n be a positive integer. Then there are n nth roots of w,z0, z1, . . . , zn−1, given by

zk = r1/n[cos

(θ

n+

2πkn

)+ i sin

(θ

n+

2πkn

)]with k = 0, 1, . . . , n − 1.

Proof: Direct calculation.�

• Complex Conjugation. The complex conjugate of a complex number z =x + iy is a complex number z defined by

z = x − iy .


1.1. COMPLEX NUMBERS 5

•

Proposition 1.1.5

1. z + w = z + w,

2. zw = zw,

3.zw=

zw

for w , 0,

4. zz = |z|2,

5. z−1 =z|z|2

for z , 0,

6. Re z = 12 (z + z) ,

7. Im z = 12i (z − z),

8. ¯z = z ,

9. z = z if and only if z is real,

10. z = −z if and only if z is purely imaginary .


• Properties of the Absolute Value.



Proposition 1.1.6

1. |zw| = |z| |w| ,

2.∣∣∣∣∣ zw

∣∣∣∣∣ = |z||w| for w , 0 ,

3. |Re z| ≤ |z| , |Im z| ≤ |z| ,

4. |z| = |z| ,

5. Triangle inequality

|z + w| ≤ |z| + |w| ,

6.|z − w| ≥

∣∣∣∣|z| − |w|∣∣∣∣ ,7. Cauchy-Schwarz inequality∣∣∣∣∣∣∣

n∑k=1

zkwk

∣∣∣∣∣∣∣ ≤√√

n∑k=1

|zk|2

√√ n∑j=1

|w j|2 .

Proof: Do Cauchy-Schwarz inequality.

1. Let

a =n∑

k=1

zkwk , b =n∑

k=1

|zk|2 c =

n∑j=1

|w j|2 .

2. We haven∑

k=1

|czk − awk|2 = c2b + |a|2c − 2c|a|2 = bc2 − c|a|2 ≥ 0

3. Therefore|a|2 ≤ bc .

�

• Examples.

• Exercises: 1.1[10,12,13], 1.2[9,11,25]


1.2. ELEMENTARY FUNCTIONS 7

1.2 Elementary Functions

1.2.1 Exponential Function• Define

eiy =

∞∑k=0

(iy)k

k!

• Theneiy = cos y + i sin y

• The exponential function exp : C→ C is defined now by

exp(x + iy) = ex(cos y + i sin y) .

• A function f : C→ C is periodic if there is some w ∈ C such that f (z+w) =f (z).

•

Proposition 1.2.1 Let z,w ∈ C, x, y ∈ R, n ∈ Z. Then

1. ez+w = ezew.

2. ez , 0.

3. ex > 1 if x > 0 and 0 < ex < 1 if x < 0.

4. |ex+iy| = ex.

5. eπi/2 = i, eπi = −1, e3πi/2 = −i, e2πi = 1.

6. ez+2πni = ez.

7. ez = 1 if and only if z = 2πki for some k ∈ Z.

Proof: Easy.�

1.2.2 Trigonometric Functions

The sine and cosine functions are defined by

sin z =eiz − e−iz

2i, cos z =

eiz + e−iz

2



•

Proposition 1.2.2 Let z,w ∈ C. Then

1. sin2 z + cos2 z = 1.

2. sin(z + w) = sin z cos w + cos z sin w ,

3. cos(z + w) = cos z cos w − sin z sin w ,


1.2.3 Logarithm Function

•

Proposition 1.2.3 Let y0 ∈ R and Ay0 ⊂ C be a set of complex num-bers defined by

Ay0 = {z = x + iy|y0 ≤ y < y + 2π} .

Let f : Ay0 → C−{0} be a function defined by f (z) = ez for any z ∈ Ay0 ..Then the function f is an bijection (one-to-one onto map).

Proof: Show directly that it is one-to-one and onto.�

• Let y0 ∈ R. The branch of the logarithm function log : C − {0} → Ay0

lying in Ay0 is defined by

log z = log |z| + i arg z ,

where y0 ≤ arg z < y0 + 2π.

•

Proposition 1.2.4 Any branch of the logarithm satisfies: for any z ∈C, z , 0,

exp log z = z .

Also, for a branch of the logarithm lying in [y0, y0 + 2π) there holds: forany z = x + iy, y ∈ [y0, y0 + 2π),

log ez = z .

Proof: Obvious.�


1.2. ELEMENTARY FUNCTIONS 9

•

Proposition 1.2.5 Let z,w ∈ CC − {0}. Then

log(zw) = log z + log w mod (2πi)

Proof: Obvious.�

1.2.4 Complex Powers

• Let a, b ∈ C, a , 0. Let y0 ∈ R and log be the branch of the logarithm lyingin [y0, y0 + 2π). Then the complex power ab is defined by

ab = exp(b log a) .

•

Proposition 1.2.6 Let a, b ∈ C, a , 0.

1. Then ab is does not depend on the branch of the logarithm if andonly if b is an integer.

2. Let b be a rational number such that b = p/q with p, q ∈ Zrelatively prime. Then ab has exactly q distinct values, namely,the roots (ap)1/q.

3. If b is a real irrational number or Im b , 0, then ab has infinitelymany distinct values.

4. The distinc values of ab differ by factors of the form e2πnbi.

Proof: Obvious.�

1.2.5 Roots

• Let n be a positive integer. The a branch of the nth root function is definedby

z1/n = exp(log z

n

)where a branch of the logarithm is chosen.



•

Proposition 1.2.7 Let z ∈ C and n be a positive integer.

1. The nth root of z satisfies

(z1/n)n = z

2. If z = reiθ, thenz1/n = r1/neiθ/n ,

where θ lies within an interval of length 2π corresponding to thebranch choice.

Proof: Obvious.�

• Exercises: 1.3[11,16,17,21,23]


1.3. CONTINUOUS FUNCTIONS 11

1.3 Continuous Functions

1.3.1 Open Sets

•

Definition 1.3.1

1. A set A ⊂ C is called open if for any point z0 ∈ A there is ε > 0such that for any z ∈ A, if |z − z0| < ε, then z ∈ A.

2. For a ε > 0 and z0 ∈ C, the ε-neighborhood, (or disk, or ball)around z0 is the set

D(z0, ε) = {z ∈ C| |z − z0| < ε}

3. A deleted ε-neighborhood is a ε-neighborhood whose centerpoint has been removed, that is,

Ddel(z0, ε) = {z ∈ C| |z − z0| < ε, z , z0} .

4. A neighborhood of a point z0 ∈ C is a set containing some ε-neighborhood around z0.

• A set A is open if and only if each point z0 ∈ A has a neighborhood whollycontained in A.

Definition 1.3.2

1. The set C is open.

2. The empty set ∅ is open.

3. The union of any collection of open sets is open.

4. The intersection of any finite collection of open sets is open.

Proof: Exercise.�



1.3.2 Mappings, Limits and Continuity

• Let A ⊂ C. A mapping f : A→ C is a rule that assigns to every point z ∈ Aa point in C (called the value of f at z and denoted by f (z)). The set A iscalled the domain of f . Such a mapping defines a complex function of acomplex variable.

•

Definition 1.3.3 Let f : A → C. Let z0 ∈ C such that A contains adeleted r-neighborhood D0(z0, r) of z0. The f is said to have the limit aas z→ z0 if for any ε > 0 there is a δ > 0 such that for all z ∈ D0(z0, r),

if |z − z0| < δ, then | f (z) − a| < ε .

Then we writelimz→z0

f (z) = a .

• Proposition 1.3.1 If a function has a limit, then it is unique.

Proof: Easy.�

•

Proposition 1.3.2 Let A ⊂ C and f , g : A → C. Let z0 ∈ C havea deleted neighborhood in A. Suppose that f and g have limits at z0,limz→z0 f (z) = a and limz→z0 g(z) = b. Then:

1. limz→z0[ f (z) + g(z)] = a + b

2. limz→z0[ f (z)g(z)] = ab

3. limz→z0[ f (z)/g(z)] = a/b, if b , 0.

Proof: Easy.�

•

Definition 1.3.4 Let A ⊂ C be open, z0 ∈ A, and f : A → C be afunction.

1. f is continuous at z0 if

limz→z0

f (z) = f (z0) .

2. f is continuous on A if f is continuous at each point in A.



•

Proposition 1.3.3

1. The sum, the product and the quotient of continuous functions iscontinuous.

2. The composition of continuous functions is continuous.

Proof: Easy.�

1.3.3 Sequences

•

Definition 1.3.5 Let (zn)∞n=1 be a sequence in C and z0 ∈ C. Thesequence (zn)∞n=1 converges to z0 if for every ε > 0 there is a N ∈ Z+such that for any n ∈ Z+

if n ≥ N, then |zn − z0| < ε .

Then we writelimn→∞

zn = z0 or zn → z0 .

•

Proposition 1.3.4 Let (zn) and wn be sequences in C and z0,w0 ∈ C.Suppose that zn → z0 and wn → w0. Then:

1. zn + wn → z0 + w0

2. znwn → z0w0

3. zn/wn → z0/w0, if w0 , 0.

Proof: Same as for functions.�

•

Definition 1.3.6 Let (zn)∞n=1 be a sequence in C. The sequence (zn) iscalled a Cauchy sequence if for every ε > 0 there is a N ∈ Z+ such thatfor any n,m ∈ Z+

if n,m ≥ N, then |zn − zm| < ε .



•

Proposition 1.3.5 Completeness of R. Every Cauchy sequence of realnumbers is convergent. That is, for every Cauchy sequence (xn) in Rthere is x0 ∈ R such that xn → x0.

Proof: Real analysis.�

•

Proposition 1.3.6 Completeness of C. Every Cauchy sequence in C isconvergent. That is, for every Cauchy sequence (zn) in C there is z0 ∈ Csuch that zn → z0.

Proof: Follows from completeness of R.�

•

Proposition 1.3.7 Let A ⊂ C, z0 ∈ C, and f :→ C. Then f is contin-uous at z0 iff for every sequence (zn) in A such that zn → z0 there holdsf (zn)→ f (z0).

Proof: Exercise.�

1.3.4 Closed Sets

• The complement of a set A ∈ C is the set C \ A.

•Definition 1.3.7 A set A ∈ C is called closed if its complement isopen.

•

Proposition 1.3.8

1. The empty set is closed.

2. C is closed.

3. The intersection of any collection of closed sets is closed.

4. The union of a finite collection of closed sets is closed.

Proof: Follows from properties of open sets.�



•Proposition 1.3.9 A set is closed iff it contains the limit of every con-vergent sequence in it.

Proof: Do.�

•

Proposition 1.3.10 Let f : C → C. The following are equivalentstatements:

1. f is continuous.

2. The inverse image of every closed set is closed.

3. The inverse image of every open set is open.

Proof: Do the cycle of implications.�

•

Definition 1.3.8 Let A ∈ C and B ⊂ A.

1. B is open relative to A if B = A ∩ U for some open set U.

2. B is closed relative to A if B = A ∩ U for some closed set U.

•

Proposition 1.3.11 Let f : A → C. The following are equivalentstatements:

1. f is continuous.

2. The inverse image of every closed set is closed relative to A.

3. The inverse image of every open set is open relative to A.

Proof: Exercise.�



1.3.5 Connected Sets

•

Definition 1.3.9 Let A ∈ C.

1. The set A is path-connected iffor every two points z1, z2 ∈ Athere is a continuous map γ : [0, 1] → A such that γ(0) = z1 andγ(1) = z2. Then γ is called a path joining the points z1 and z2.

2. The set A is disconnected if there are open sets U and V suchthat

(a) A ⊂ U ∩ V,

(b) A ∩ U , ∅ and A ∩ V , ∅ ,

(c) (a ∩ U) ∩ (A ∩ V) = ∅ .

3. A set is connected if it is not disconnected.

•Proposition 1.3.12 Let A ⊂ C. The set A is connected iff the onlysubsets of A that are both open and closed relative to A are ∅ and A.

Proof: Follows from the definition.�

• Proposition 1.3.13 A path-connected set is connected.

Proof: Do.�

• Example. Let

A1 =

{z = x + iy | x , 0 y = sin

(1x

)}A2 = {z = x + iy | x = 0 , y ∈ [−1, 1]}

andA = A1 ∪ A2 .

Then A is connected but not path-connected.

•

Proposition 1.3.14 Every open connected set is path connected witha differentiable path. That is, if A ⊂ C is open connected and z1, z2 ∈ A.Then there is a differentiable map γ : [0, 1] → A such that γ(0) = z1

and γ(1) = z2.



Proof: See the book.�

•Definition 1.3.10 An open connected set in C is called a region (or adomain).

•Proposition 1.3.15 A continuous image of a connected set is con-nected.

Proof: Easy.�

1.3.6 Compact Sets

•

Definition 1.3.11

1. Let A be a set and O = {Uα}α∈A be a collection of open sets. LetK ⊂ C. Then the collection O is called an open cover of K if

K ⊂⋃α∈A

Uα .

2. A subcollection of an open cover is called a subcover if it is stillan open cover.

3. A set is compact if every open cover of it has a finite subcover.

•

Proposition 1.3.16 Let K ⊂ C. The follwoing are equivalent:

1. K is closed and bounded.

2. Every sequence in K has a convergent subsequence whose limitsis in K.

3. K is compact.

Proof: In real analysis courses.�

• Proposition 1.3.17 A continuous image of a compact set is compact.

Proof: Easy.



�

•

Theorem 1.3.1 Extreme value Theorem. Let K ⊂ C be compact andf : K → R be continuous. Then f attains maximum and minimumvalues.

Proof: Easy.�

•

Lemma 1.3.1 Distance Lemma. Let A,K ⊂ C be two disjoint sets.Suppose that K is compact and A is closed. Then the distance d(K, A) >0. That is, there is ε > 0 such that for any z ∈ K and w ∈ A, |z − w| > ε.

Proof: Do.�

1.3.7 Uniform Continuity

•

Definition 1.3.12 Let A ⊂ C and f : A → C. Then f is uniformlycontinuous on A if for any ε > 0 there is δ > 0 such that for anyz,w ∈ A,

if |z − w| < δ, then | f (z) − f (w)| < ε .

•Proposition 1.3.18 A continuous function on a compact set is uni-formly continuous.

Proof:�



1.3.8 Path Covering Lemma

•

Lemma 1.3.2 Path-Covering Lemma. Let a, b ∈ R and G ⊂ C be anopen set. Let γ : [a, b] → G be a continuous path. Then there is ρ > 0and a partition of [a, b], {t0 < t1 < · · · < tn} with t0 = a and tn = b, suchthat

1. D(γ(tk); ρ) ⊂ G ,

2. γ(t) ∈ D(γ(tk); ρ) for t ∈ [tk1 , tk+1] and k = 1, . . . , n − 1,

3. γ(t) ∈ D(γ(t0); ρ) for t ∈ [t0, t1],

4. γ(t) ∈ D(γ(tn); ρ) for t ∈ [tn−1, tn].

Proof: Do.�

1.3.9 Riemann Sphere and Point at Infinity

• The extended complex plane, C, is obtained by adding the point at infinity,∞, to C, C = C ∩ {∞}.

• Limits at infinity.

• A point is close to infinity if it lies outside a sufficiently large circle.

• The limit limz→∞ f (z) = a means that for any ε > 0 there is a R > 0 suchthat if |z| > R then | f (z) − a| < ε.

• The limit limz→z0 f (z) = ∞ means that for any R > 0 there is a δ > 0 suchthat if |z − z0| < δ then | f (z) > R.

• The limit limn→∞ zn = ∞ means that for any R > 0 there is a N ∈ Z+ suchthat if n ≥ N then |zn| > R.

• Riemann sphere and stereographic projection.

• S and C are compact.

• Exercises: 1.4[3,5,11,12,14,16,17,20,23,24]



1.4 Analytic Functions

•

Definition 1.4.1 Let A ⊂ C be an open set and f : A→ C.

1. Let z0 ∈ A. Then f is differentiable at z0 if the limit (called thederivative)

f ′(z0) = limz→z0

f (z) − f (z0)z − z0

exists.

2. The function f is analytic (or holomorphic) on A it is is differ-entiable at every point in A.

3. The function f is analytic at z0 if it is analytic in a neighborhoodof z0.

4. A function analytic on the whole complex plane is called entire.

• Remarks. A differentiable function must be continuous. However, a con-tinuous function does not have to be differentiable.

•Proposition 1.4.1 Let A ⊂ C be an open set, f : A → C, and z0 ∈ A.Suppose that f is differentiable at z0. Then f is continuous at z0.

Proof: Easy.�

• A polynomial of degree n is a function P : C→ C of the form

P(z) =n∑

k=0

akzk ,

where ak ∈ C.

• Let P and Q be two polynomials. Let A = {z ∈ C | Q(z) , 0} be an open setthat is the complement of the set of roots of the polynomial Q. A rationalfunction is a function R : A→ C of the form

R(z) =P(z)Q(z)

.


1.4. ANALYTIC FUNCTIONS 21

•

Proposition 1.4.2 Let A ⊂ C be an open set, f , g, h : A → C beanalytic on A, and a, b ∈ R. Suppose that h(z) , 0 for any z ∈ A. Thenthe functions (a f + bg), f g and f /g are analytic on A and ∀z ∈ C

(a f + bg)′(z) = a f ′(z) + bg′(z) ,

( f g)′(z) = f ′(z)g(z) + f (z)g′(z) ,

( f /h)′(z) =f ′(z)h(z) + f (z)h′(z)

[h(z)]2 .

Proof: Easy.�

•

Proposition 1.4.3

1. Every polynomial is an entire function.

2. Every rational function is analytic on its domain, that is, the com-plement of the set of roots of the denominator.

Proof: Easy.�

• The derivatives of polynomials:

ddz

c = 0 ,

ddz

zn = nzn−1 , for n ≥ 1 .

•

Proposition 1.4.4 Chain Rule. Let A, B ∈ C be open sets, f : A→ C,g : B → C. Suppose that f and g are analytic and f (A) ⊂ B. Leth = g ◦ f : A→ C be the composition of f and g. Then h is analytic onA and ∀z ∈ A

h′(z) = g′( f (z)) f ′(z) .

Proof:�



•

Proposition 1.4.5 Let A, B ⊂ C be open sets, and f : A → C. Leta, b ∈ R and γ : (a, b)→ B, be a curve in B. Let B ⊂ A and σ : (a, b)→C be the curve defined ∀t ∈ (a, b) by σ(t) = f (γ(t)). Suppose that f isanalytic and γ is differentiable. Then σ is also differentiable and

ddtσ(t) = f ′(γ(t))

ddtγ(t) .

Proof: Similar.�

•

Proposition 1.4.6 Let A ⊂ C be a region (an open connected set), andf : A→ C. Suppose that f is analytic on A and f ′(z) = 0 for any z ∈ A.Then f is constant on A.

Proof: Let z1, z2 ∈ A. Connect them by a differentiable path and use thechain rule.

�

1.4.1 Conformal Maps.

•

Definition 1.4.2 Let A ⊂ C be an open set and f : A→ C.

1. Let z0 ∈ A. Then f is a conformal at z0 if there exists θ ∈ [0, 2π)and r > 0 such that for any curve γ : (−ε, ε) → A that is: i)differentiable at t = 0, ii) γ(0) = z0, and iii) γ′(0) , 0, the curveσ : (−ε, ε)→ C defined by σ(t) = f (γ(t)) is differentiable at t = 0and

σ′(0)| = reiθγ′(0) ,

that is,

|σ′(0)| = r|γ′(0)| , argσ′(0) = arg γ′(0) + θ (mod 2π) .

2. The function f : A is called a conformal map if it is conformalat every point of A.

• A conformal map rotates and stretches tangent vectors to curves, that is, itpreserves angles between intersecting curves.



• The numbers r and θ are fixed at the point z0. They should apply for anycurve.

•

Theorem 1.4.1 Conformal Mapping Theorem. Let A ⊂ C be a openset, z0 ∈ A, and f : A → C. Suppose that f is analytic at z0 andf ′(z0) , 0. Then f is conformal at z0.

Proof: Follows from the chain rule.�

1.4.2 Cauchy-Riemann Equations

• Multiple real variables. A point in R2 is represented as a pair (x, y). It canbe also viewd as a column vector

(xy

), which can be added, multiplied by

scalar etc. A linear transformation J : R2 → R2 is represented by a matrix

J =(

J11 J12

J21 J22

)acting on column vectors

(xy

)by left multiplication, that is,

〈J, (x, y)〉 =(

J11 J12

J21 J22

) (xy

).

• Let A ⊂ R2 and f : A → R2 with f (x, y) = (u(x, y), v(x, y)). The Jacobianmatrix of f is defined by

D f =(∂xu ∂yu∂xv ∂yv

)

• The distance between the points (x, y) and (x0, y0) in R2 is given by ||(x, y)−(x0, y0)|| =

√(x − x0)2 + (y − y0)2.

• Let (x0, y0) ∈ A. The map f is differentiable at (x0, y0) ∈ A if there exists amatrix (called the derivative matrix) D f (x0, y0) such that for any ε > 0 thereis a δ > 0 such that if ||(x, y) − (x0, y0)|| < δ then

|| f (x, y) − f (x0, y0) − 〈D f (x0, y0), (x, y) − (x0, y0)〉 || < ε||x, y) − (x0, y0)|| .



• If f is differentiable, the the partial derivatives of u and v exist and thederivative matrix is given by the Jacobian matrix.

• If the partial derivatives of u and v exist and are continuous then f is differ-entiable.

•

Theorem 1.4.2 Cauchy-Riemann Theorem. Let A ∈ C be an openset, f : A→ C, and z0 ∈ A. The f = u+iv is differentiable at z0 = x0+iy0

if and only if f is differentiable at (x0, y0) in the sense of real variablesand the functions u and v satisfy the Cauchy-Riemann equations

∂xu = ∂yv , ∂yu = −∂xv .

Proof:�

• Cauchy-Riemann equations in polar form.

∂ru =1r∂θv , ∂rv = −

1r∂θu ,

1.4.3 Inverse Functions

• Real analysis. A continuously differentiable function on an open set isbijective and has a differentiable inverse in a neighborhood of a point if theJacobian matrix is not degenerate at that point.

•

Theorem 1.4.3 Real Variable Inverse Function Theorem. Let A ⊂R2 be an open set, f : A → R2, and (x0, y0) ∈ A. Suppose that f iscontinuously differentiable and the Jacobian matrix D f (x0, y0) is non-degenerate. Then there is a neighborhood U of (x0, y0) and a neighbor-hood V of f (x0, y0) such that f : U → V is a bijection, f −1 : V → U isdifferentiable, and

D f −1 f (x, y) = [D f (x, y)]−1 .

Proof: Advanced calculus.�



•

Theorem 1.4.4 Inverse Function Theorem. Let A ⊂ C be an openset, z0 ∈ A, and f : A→ C. Suppose that f is analytic, f ′ is continuousand f ′(z0) , 0. Then there is a neighborhood V of z0 such that thefunction f : U → V is a bijection with analytic inverse f −1 : V → Uwith the derivative

ddz

f −1(z) =1

f ′( f −1(z)).

Proof: Compute [D f ]−1 by using CR equations.�

• The Laplacian is a partial differential operator acting on functions on R2

defined by∆ = ∂2

x + ∂2y .

• Let A ∈ R2 be an open set and u : A→ R be a twice-differentiable function.Then f is harmonic if

∆u = 0 .

• An analytic function is infinitely differentiable (show later).

•

Proposition 1.4.7 Let A ⊂ C be an open set and f : A→ C. Supposethat f is analytic on A. Then the real and the imaginary parts of f areharmonic on A.

Proof: Use CR equations.�

• The real and the imaginary parts, u and v, of an analytic function f = u+ iv,are called harmonic conjugates of each other.

• The level curve u(x, y) = c is smooth if grad u , 0.

• Let f = u + iv be analytic. The level curves u(x, y) = c1 and v(x, y) = c2 aresmooth if f ′(z) , 0.

• The vector grad u is orthogonal to the level curve u(x, y) = c.

•

Proposition 1.4.8 Let A ⊂ C be a region, f = u + iv : A → C beanalytic on A. Suppose that the level curves u(x, y) = a and v(x, y) = bdefines smooth curves. Then they intersect orthogonally.




•

Proposition 1.4.9 Let A ⊂ C be an open set, u : A → R be a real-valued function and z0 ∈ A. Suppose that u is twice differentiable andharmonic on A. Then there is a neighborhood U of z0 and an analyticfunction f : U → C such that u = Re f .

Proof: Exercise.�

• Exercises: 1.5[7,8,10,11,13,14,15,16,17,25,30,31]


1.5. DIFFERENTIATION 27

1.5 Differentiation

1.5.1 Exponential Function and Logarithm

• Proposition 1.5.1 The exponential function f : C → C defined byf (z) = ez is entire and

dez

dz= ez

Proof: Verify CR conditions.�

•

Proposition 1.5.2 Let B ⊂ C be a closed set defined by

B = {z = x + iy | x ≤ 0, y = 0}

and A = C\B be its complement (which is open). The principal branchof the logarithm is a branch on A defined by

log z = log |z| + i arg z ,

where −π < arg z < π. Then log z is analytic on A and

d log zdz

=1z.

Proof: Verify CR equations.�

1.5.2 Trigonometric Functions

•

Proposition 1.5.3 The sine and the cosine functions are entire and

d sin zdz

= cos z ,d cos z

dz= − sin z .

Proof: Trivial.�



1.5.3 Power Function

•

Proposition 1.5.4 Let a ∈ C. Then the power function z 7→ za = ea log z

(defined with the principal branch of the logarithm) is analytic on thedomain of the logarithm and

dza

dz= aza−1 .

Proof: Use chain rule.�

1.5.4 The n-th Root Function

•

Proposition 1.5.5 Let n ∈ Z, n ≥ 1. Then the n-th root functionz 7→ z1/n = e(log z)/n (defined with the principal branch of the logarithm)is analytic on the domain of the logarithm and

dz1/n

dz=

1n

z(1/n)−1 .

Proof: Follows from above.�

• Exercises: 1.6[6,8,13,14]


Chapter 2

Cauchy Theorem

2.1 Contour Integrals

• Let h : [a, b]→ C. Let h(t) = u(t) + iv(t). The integral of h is defined by∫ b

ah(t)dt =

∫ b

au(t)dt + i

∫ b

av(t)dt .

• A continuous curve in C is a continuous map γ : [a, b] → C. A contouris a continuous curve. A curve is continuously differentiable if the mapγ is continuously differentiable. A curve is piecewise continuously differ-entiable if it is continuous and consists of a finite number of continuouslydifferentiable curves.

• AUsually in analysis a function is called smooth if it is differentiable in-finitely many times. However, in this course smooth will mean coninu-ously differentiable. This applies to functions, curves, maps, etc.

• Let A ⊂ C be an open set, γ : [a, b] → A be a smooth curve in A andf : A→ C be continuous. The integral of f along γ is defined by∫

γ

f (z)dz =∫ b

af (γ(t))

dγ(t)dt

dt .

• If γ is a closed curve the integral is denoted by∮γ

f (z)dz .

29

30 CHAPTER 2. CAUCHY THEOREM

• Let P,Q : A → R be real valued functions of two variables. Then the lineintegral along γ is defined by

∫γ

P(x, y)dx + Q(x, y)dy =∫ b

a

[P(x(t), y(t))

dxdt+ Q(x(t), y(t))

dydt

]dt .

•

Proposition 2.1.1 Let A ⊂ C be an open set, γ : [a, b] → A be asmooth curve in A and f : A → C be continuous. Let f (z) = u(x, y) +iv(x, y). Then∫

γ

f (z)dz =∫γ

[u(x, y)dx − v(x, y)dy] + i∫γ

[u(x, y)dy + v(x, y)dx]

Proof: Easy.�

• Let γ : [a, b] → C be a curve. The opposite curve −γ : [a, b] → C isdefined by

(−γ)(t) = γ(a + b − t) .

It is the same curve traversed in the opposite direction.

• The zero curve at the point z0 is the curve γ0 : [a, b] → C defined byγ0(t) = z0 for any t. It is just the point z0 itself.

• Let γ1 : [a, b] → C and γ2 : [b, c] → C be two curves. The sum of twocurves is the curve (γ1 + γ2) : [a, c]→ C defined by

(γ1 + γ2)(t) =

γ1(t) if a ≤ t ≤ b

γ2(t) if b ≤ t ≤ c


2.1. CONTOUR INTEGRALS 31

•

Proposition 2.1.2 Let f , g : A → C be continuous, a, b ∈ C, and γ1,γ2 be two piecewise smooth curves. Then

1. ∫γ

(a f + bg) = a∫γ

f + b∫γ

g ,

2. ∫−γ

= −

∫γ

f ,

3. ∫γ1+γ2

f =∫γ1

f +∫γ2

f

Proof: Easy.�

• Let γ : [a, b] → C be a piecewise smooth curve. A curve γ : [c, d] → C isa reparametrization of γ if there is a smooth function α : [a, b] → [c, d]such that α′(t) > 0, α(a) = c, α(b) = d, and γ(t) = γ(α(t)).

•

Proposition 2.1.3 Let A ⊂ C be an open set, γ : [a, b] → A be asmooth curve in A and f : A→ C be continuous. Then∫

γ

f =∫γ

f

Proof: Easy.�

• The arc length of the curve γ : [a, b]→ C is defined by

l(γ) =∫ b

a

∣∣∣∣∣dγdt

∣∣∣∣∣ dt .

• More generally, we define the integral∫γ

| f (z)| |dz| =∫ b

a| f (γ(t))|

∣∣∣∣∣dγdt

∣∣∣∣∣ dt .

Thenl(γ) =

∫γ

|dz|



•

Proposition 2.1.4 Let A ⊂ C be an open set, γ : [a, b] → A be asmooth curve in A and f : A→ C be continuous. Then∣∣∣∣∣∣

∫γ

f

∣∣∣∣∣∣ ≤∫γ

| f | |dz|

In particular, suppose that there is M ≥ 0 such that for any z on thecurve γ there holds | f (z)| ≤ M. Then∣∣∣∣∣∣

∫γ

f

∣∣∣∣∣∣ ≤ Ml(γ) .

Proof:�

•

Theorem 2.1.1 Fundamental Theorem of Calculus for Contour In-tegrals. Let A ⊂ C be an open set, γ : [a, b] → A be a smooth curve inA and F : A→ C be analytic in A. Then∫

γ

dFdz

dz = F(γ(b)) − F(γ(0))

In particular, if γ is a closed curve then∮γ

dFdz

dz = 0

Proof:�

•

Proposition 2.1.5 Let A ⊂ C be an open connected set, and F : A →C be analytic in A such that F′(z) = 0 for any z ∈ A. Then F is constanton A.

Proof:�


2.1. CONTOUR INTEGRALS 33

•

Theorem 2.1.2 Path Independece Theorem. Let A ⊂ C be an openconnected set, and f : A → C be continuous on A. Then the followingare equivalent:

1. Integrals of f are path independent.

2. Integrals of f around closed curves are equal to zero.

3. There is an antiderivative F for f on A such that F′ = f .

Proof:�

• Exercises: 2.1[3,5,7,8,12,13,14]



2.2 Cauchy Theorem

• A continuous curve is simple if it does not have self-intersection (exceptpossibly the endpoints). A simple closed curve is called a loop.

• Let A be an open set, z1 and z2 be two points in A and γ1, γ2 : [0, 1] → Abe two continuous curves in A connecting the points z1 and z2 such thatγ1(0) = γ2(0) = z1 and γ1(1) = γ2(1) = z2. Then γ1 is homotopic with fixedendpoints to γ2 in A if there is a continuous functions H : [0, 1]×[0, 1]→ A,such that: H(0, t) = γ1(t), H(1, t) = γ2(t), H(s, 0) = z1 and H(s, 1) = z2.For a fixed s the γs(t) = H(s, t) is a continuous one-parameter family ofcontinuous curves.

• A homotopy H : [0, 1] × [0, 1] → A is smooth if H(s, t) is smooth functionof both s and t.

• Two homotopic closed curves γ1 and γ2 are just called homotopic as closedcurves in A. That is, the loop γ1 can be continuously deformed to the loopγ2.

• A loop in A is contractible if it is homotopic to a point (zero loop) in A,that is, it can be deformed to a point.

• Let A ⊂ C be an open set. Then A is simply connected if A is connectedand every loop in A is contractible. A simply connected region does nothave any holes.

•

Theorem 2.2.1 Green’s Theorem. Let A be a simply connected regionwith a boundary γ = ∂A that is a loop oriented counterclockwise. LetP,Q be smooth functions defined on an open set that contains A. Then∮

γ

[P(x, y)dx + Q(x, y)dy] =∫

A

[∂xQ(x, y) − ∂yP(x, y)

]dx dy .

Proof: Vector analysis.�


2.2. CAUCHY THEOREM 35

•

Theorem 2.2.2 Cauchy’s Theorem. The integral of an analytic func-tion in a simply connected domain along any loop is equal to zero. Moreprecisely, let A ⊂ C be a simply connected region, γ be a loop in A andf : A→ C be an analytic function. Then∮

γ

f = 0 .


•

Theorem 2.2.3 Deformation Theorem. The integrals of an analyticfunction along homotopic loops are equal to each other. More precisely,let A ⊂ C be a region, f : A→ C be an analytic function and γ1 and γ2

be two homotopic loops in A. Then∮γ1

f =∮γ2

f .

Proof: Use Cauchy theorem.�

•

Theorem 2.2.4 Path Independence Theorem. The integral of an an-alytic function in a simply connected domain along any contours con-necting two points are equal to each other. More precisely, let A ⊂ C bea simply connected region, f : A → C be an analytic function, γ1 andγ2 be two contours in A connecting the points z1 and z2 in A. Then∫

γ1

f =∫γ2

f .

Proof: Use Cauchy theorem.�

•

Theorem 2.2.5 Antiderivative Theorem. An analytic function on asimply connected domain has an antiderivative. More precisely, let A ⊂C be a simply connected region, f : A → C be an analytic function.Then there is a function F : A→ C such that F′(z) = f (z) for any z ∈ A.

Proof: Use path independence theorem.�



•

Proposition 2.2.1 Let A ⊂ C be a simply connected region not con-taining 0. Then there is an analytic function F : A → C (called abranch of the logarithm) such that exp[F(z)] = z. This function isunique modulo 2πi.

Proof: In the book.�

• Exercises: 2.2[2,4,5,9,11]


2.3. CAUCHY INTEGRAL FORMULA 37

2.3 Cauchy Integral Formula• Example. Let z0 ∈ C and γ be a loop (without selfcrossings) not passing

through z0. Then

12πi

∫γ

dzz − z0

=

1 if z0 is inside γ and γ is traversed counterclockwise around z0

−1 if z0 is inside γ and γ is traversed clockwise around z0

0 if z0 is outside γ

• Let z0 ∈ C and γ be a closed curve not passing through γ. Then the index(or winding number) of γ with respect to z0 is defined by

I(γ; z0) =1

2πi

∫γ

dzz − z0

.

•

Proposition 2.3.1

1. The index of a closed curve counts the number of times the curvewinds around the point. More precisely, let z0 ∈ C, γ : [0, 2πn]→C be the circle of radius r around z0, γ(t) = z0 + reit and gamma :[0, 2πn]→ C, (−γ)(t) = z0 + re−it, be the opposite curve. Then

I(γ; z0) = −I(−γ; z0) = n .

2. Indices of homotopic curves are equal. More precisely, let z0 ∈ Cand γ1 and γ2 be two homotopic curves in C \ {z0} not passingthrough z0. Then

I(γ1; z0) = I(γ2; z0) .

Proof: Easy.�

•

Proposition 2.3.2 The index of a curve is an integer. More precisely,let z0 ∈ C and γ : [a, b] → C be a smooth closed curve not passingthrough z0. Then I(γ; z0) is an integer.

Proof: Use change of variables.�



•

Proposition 2.3.3 Cauchy’s Integral Formula. Let A be a region,z0 ∈ A, γ be a closed contractible curve in A not passing through z0,and f : A→ C be analytic on A. Then

12πi

∫γ

dzf (z)

z − z0= I(γ; z0) f (z0) .

In particular, if γ is a loop and z0 is inside γ then

12πi

∫γ

dzf (z)

z − z0= f (z0) .

Proof:

1. Letg(z) =

f (z) − f (z0)z − z0

if z , z0 and g(z0) = f ′(z0).

2. Then g is analytic except at z0 and is continuous at z0.

3. Therefore∫γ

g = 0.

4. The statement follows.

�

•

Theorem 2.3.1 Differentiability of Cauchy Integrals. Let A be aregion, γ be a curve in A, z be a point not on γ, and f : A → C becontinuous on A. Let F : C \ γ([a, b]) be a function defined by

F(z) =1

2πi

∫γ

dζf (ζ)ζ − z

.

Then F is analytic on C \ γ([a, b]).Moreover, F is infinitely differentiable with the kth derivative given by

F(k)(z) =k!

2πi

∫γ

dζf (ζ)

(ζ − z)k+1 ,

for k = 1, 2, . . . .

Proof:�


2.3. CAUCHY INTEGRAL FORMULA 39

•

Theorem 2.3.2 Cauchy’s Integral Formula for Derivatives. Let Abe a region, z0 ∈ A, γ be a closed contractible curve in A not passingthrough z0, and f : A → C be analytic on A. Then f is infinitelydifferentiable and

k!2πi

∫γ

dzf (z)

(z − z0)k+1 = I(γ; z0) f (k)(z0) .

In particular, if γ is a loop and z0 is inside γ then

k!2πi

∫γ

dzf (z)

(z − z0)k+1 = f (k)(z0) .


•

Theorem 2.3.3 Cauchy’s Inequalities. Let A be a region, z0 ∈ A, γ bea circle of radius R centered at z0 such that the whole disk |z − z0| < Rlies in A, and f : A → C be analytic on A such that for all z ∈ γ,| f (z)| ≤ M for some M > 0. Then for any k = 0, 1, 2, . . .∣∣∣ f (k)(z0)

∣∣∣ ≤ k!Rk M .

Proof: Follows from Cauchy integral formula.�

•

Theorem 2.3.4 Liouville Theorem. An entire bounded function isconstant. More precisely, let f : C → C be an entire function such thatfor any z ∈ C, | f (z)| ≤ M for some M.


•

Theorem 2.3.5 Fundamental Theorem of Algebra. Every non-constant polynomial has at leat one root. More precisely, let n ≥ 1,a0, . . . , an ∈ C, an , 0 and P(z) =

∑nk=0 akzk. Then there exists z0 ∈ C

such that P(z0) = 0.

Proof: Use Liouville theorem.�



•

Theorem 2.3.6 Morera’s Theorem. Let f : A → C be a continuousfunction such that

∫γ

f = 0 for every closed curve in A. Then f isanalytic on A and has an analytic antiderivative on A, that is, f = F′

for some analytic function F.

Proof:�

•Corollary 2.3.1 LetA be a region, z0 ∈ A and f : A → C be continu-ous on A and analytic on A \ {z0}. Then f is analytic on A.

Proof:�

• Exercises: 2.4[5,6,8,13,16,17,21]


2.4. MAXIMUM MODULUS THEOREM AND HARMONMIC FUNCTIONS41

2.4 Maximum Modulus Theorem and HarmonmicFunctions

2.4.1 Maximum Modulus Theorem

•

Proposition 2.4.1 Mean value Property. Let A be a simply connectedregion, z0 ∈ A, and γ be a circle (in A) of radius r centered at z0. Letf : A→ CC be analytic. Then

f (z0) =1

2π

∫ 2π

0dθ f (z0 + reiθ) .

Proof: Use Cauchy formula.�

•

Theorem 2.4.1 Local Maximum Modulus Theorem. Let A be a re-gion, z0 ∈ A, and f : A → C be analytic on A. Suppose that themodulus | f | has a relative maximum at z0. Then f is constant in someneighborhood of z0.

Proof:

1. Assume | f (z)| ≤ | f (z0)| for any z in a disk around z0.

2. Then there is no point z1 such that | f (z1)| < | f (z0)|.

3. So, | f (z)| = | f (z0)| for any z.

4. Thus | f | is constant.

5. Then by using CR equations show that f is constant,

�

• Let A be a set. A point z ∈ A is an interior point of A it A contains aneighborhood of z. A point z is an exterior point of A if the complementC \ A contains a neighborhood of A (that is, if z is an interior point of thecomplemenet). A point z is a boundary point of A if it is neigher interiornor exterior. The set of all interior points of A is the interior of A, denotedby int(A). The set of all exterior points of A is the exterior of A, denoted byext(A). The set of all boundary points of A is the boundary of A, denotedby ∂A. The closure of the set A is the set cl(A) = A ∪ ∂A.



•

Definition 2.4.1

1. Let A be a set. The closure of the set A is the set A (or cl(A))consisting of A together with all limit points of A (that is, thelimits of all converging sequences in A).

2. The boundary of A is the set ∂A defined by

∂A = cl(A) ∩ cl(C \ A) ,

so that cl(A) = A ∪ ∂A.

•

Proposition 2.4.2 Let A, B ⊂ C.

1. Ever set is a subset of its closure, that is, A ⊂ cl(A).

2. The closure of any set is closed, that is, cl(A) is closed.

3. The interior of any set is open, that is, int(A) is open.

4. A set is closed if and only if it coincides with its closure. That is,the set A is closed if and only if A = cl(A).

5. A set is open if and only if it coincides with its interior. That is,the set A is open if and only if A = int(A).

6. A closed set contains the closure of all its subsets. That is, ifA ⊂ B and B is closed, then cl(A) ⊂ B.

Proof:�

•

Theorem 2.4.2 Maximum Modulus Principle. Let B and f : B→ Cbe analytic. Let A ⊂ B be a closed bounded connected region in B.Then the modulus | f | has a maximum value on A, which is attained onthe boundary of A. If it is also attained in the interior of A, then f isconstant on A.

Proof:�



•

Lemma 2.4.1 Schwarz Lemma. Let A = D(0; 1) be the open unit diskcentered at the origin, f : A → C be analytic. Suppose that f (0) = 0and | f (z)| ≤ 1 for any z ∈ A. Then:

1. | f ′(0)| ≤ 1 and | f (z)| ≤ |z| for any z ∈ A.

2. If | f ′(0)| = 1 or if there is a point z0 , 0 such that | f (z0)| = |z0|,then f (z) = cz for all z ∈ A with some constant c such that |c| = 1.

•

Lemma 2.4.2 Lindelof Principle. Let A = D(0; 1) be the open unitdisk centered at the origin, f : A → f (A), g : A → g(A) be analytic.Suppose that g is a bijection, f (A) ⊂ g(A), f (0) = g(0). Then:

1. | f ′(0)| ≤ |g′(0)| and | f (z)| ≤ |z| for any z ∈ A.

2. The image of an open disk D(0, r) of radius r < 1 centered at 0under f is contained in its image under g, that is, f (D(0, r)) ⊂g(D(0, r)).

Proof:�

2.4.2 Harmonic Functions

•

Proposition 2.4.3 Let A be a region, u ∈ C2(A) be a harmonic functionin A. Then:

1. u ∈ C∞(A).

2. For any z0 ∈ A there is a neighborhood V of z0 and an analyticfunction f : V → C such that u = Re f .

3. If A is simply connected then there is an analytic function g : A→C such that u = Re g.

Proof: By construction.�

• If f = u + iv is analytic, then u and v are harmonic conjugates.



•

Theorem 2.4.3 Mean Value Property for Harmonic Functions. LetA be a simply connected region, z0 = x0 + iy0 ∈ A such that A containsthe circle of radius r centered at z0, and u be aharmonix function on A.Then

u(x0, y0) =1

2π

∫ 2π

0dθ u(x0 + r cos θ, y0 + r sin θ) .

Proof:�

•

Theorem 2.4.4 Local Maximum Principle for Harmonic Func-tions. Let A be a region, u : A → R be harmonic on A. Suppose that uhas a relative maximum at z0 ∈ A. Then u is constant in a neighborhoodof z0.

Proof:�

•

Theorem 2.4.5 Global Maximum Principle for Harmonic Func-tions. Let B and u : B → R be continuous harmonic function. LetA ⊂ B be a closed bounded connected region in B and M be the max-imum of u on the boundary of A and m be the minimum of u on theboundary of A.. Then:

1. m ≤ u(x, y) ≤ M for all (x, y) ∈ A.

2. If u(x, y) = M for some interior point (x, y) ∈ A or if u(x, y) = mfor some interior point (x, y) ∈ A, then u is constant on A.

Proof:�

2.4.3 Dirichlet Boundary Value Problem

•

Theorem 2.4.6 Uniqueness of Solution of Dirichlet Problem. Let Abe a closed bounded region and u0 : ∂A → R be a continuous functionon the boundary of A. If there is a continuous function u : A→ R that isharmonic in the interior of A and equals u0 on the boundary of A, thenit is unique.

Proof:



�

•

Theorem 2.4.7 Poisson’s Formula. Let D(0; r) be an open disk ofradius r centered at the origin and A = cl(D(0; r)) be its closure. Letu : A → R be a continuous function that is harmonic in the interior ofA. Then for ρ < r and any ϕ

u(ρeiϕ) =r2 − ρ2

2π

∫ 2π

0dθ

u(reiθ)r2 − 2rρ cos(ϕ − θ) + ρ2 .

Alternatively, let γ = ∂A be the circle of radius r centered at the origin.Then for any z ∈ A

u(z) =1

2π

∮γ

dζ u(ζ)r2 − |z|2

|ζ − z|2

=1

2π

∫ 2π

0dθ u(reiθ)

r2 − |z|2

|reiθ − z|2

Proof:�

• Exercises: 2.5[1,6,7,10,12,18]


Chapter 3

Series Representation of AnalyticFunctions

3.1 Series of Analytic Functions

•

•

Definition 3.1.1

1. Let (zn) be a sequence inC and z0 ∈ C. The sequence zn convergesto z0 (denoted by zn → z0, or lim zn = z0) if for any ε > 0 there isN ∈ Z+ such that for any n ∈ Z+

if n ≥ N, then |zn − z0| < ε .

2. Let∑∞

k=1 ak be a series and (sn) be the corresponding sequenceof partial sums defined by sn =

∑nk=1 ak, and S ∈ C. The series∑∞

k=1 ak converges to S (denoted by∑∞

k=1 ak = S ) if sn → S .

• The limit of a convergent sequence is unique.

• A sequence (zn) is Cauchy if for any ε > 0 there is N ∈ Z+ such that for anyn,m ∈ Z+

if n,m ≥ N, then |zn − zm| < ε .

• A sequence converges if and only if it is Cauchy.

47

48 CHAPTER 3. SERIES REPRESENTATION OF ANALYTIC FUNCTIONS

• Cauchy Criterion for Series. A series∑∞

k=1 ak converges if and only if forany ε > 0 there is N ∈ Z+ such that: for any n,m ∈ Z+

if n,m ≥ N, then |sn − sm| < ε ,

or

if n ≥ N, then

∣∣∣∣∣∣∣n+m∑

k=n+1

ak

∣∣∣∣∣∣∣ < ε ,• In particular, if the series

∑∞k=1 ak converges, then the sequence (ak) con-

verges to zero, ak → 0.

• A series∑∞

k=1 ak converges absolutely if the series∑∞

k=1 |ak| converges.

• Proposition 3.1.1 If a series converges absolutely, then it converges.

Proof:�

•

Proposition 3.1.2 Tests for Convergence.

1. Geometric series. If |r| < 1, then

∞∑k=0

rk =1

1 − r.

If |r| ≥ 1, then∑∞

k=0 rk diverges.

2. Comparison test. If the series∑∞

k=1 bk converges and 0 ≤ ak ≤

bk, then the series∑∞

k=1 ak converges. If the series∑∞

k=1 ck divergesand 0 ≤ ck ≤ dk, then the series

∑∞k=1 dk diverges.

3. p-series. If p > 1, then the series∑∞

k=1 k−p converges. If p ≤ 1,then

∑∞k=1 k−p diverges to∞.

4. Ratio test. Suppose that limn→∞

∣∣∣∣an+1an

∣∣∣∣ = L exist. If L < 1, then∑∞k=1 ak converges absolutely. If L > 1, then

∑∞k=1 ck diverges. if

L = 1, then the test is inconclusive.

5. Root test. Suppose that limn→∞ |an|1/n = L exist. If L < 1, then∑∞

k=1 ak converges absolutely. If L > 1, then∑∞

k=1 ck diverges. ifL = 1, then the test is inconclusive.


3.1. SERIES OF ANALYTIC FUNCTIONS 49

Proof: In calculus.�

• Let A be a set and fn : A → C be a sequence of functions on A. Thesequence ( fn) converges pointwise if it converges for any point z ∈ A.

•

Definition 3.1.2 Uniform Convergence. Let A be a set, f : A → C,and fn : A → C be a sequence of functions on A. The sequence ( fn)converges uniformly to f (denoted by “ fn → f uniformly on A”) if forany ε > 0 there is an N ∈ Z+ such that for any n ∈ Z+ and any z ∈ A

if n ≥ N, then | fn(z) − f (z)| < ε .

• Similarly a series∑∞

k=1 gn converges uniformly on A if the sequence of par-tial sums converges uniformly.

• Uniform convergence implies pointwise convergence.

•

Theorem 3.1.1 Cauchy Criterion. Let A ⊂ C, and gn, fn : A → C besequences of functions on A.

1. Then fn converges uniformly if and only if for any ε > 0 there isN ∈ Z+ such that for any n,m ∈ Z+, and any z ∈ A

if n ≥ N, then | fn(z) − fn+m(z)| < ε .

2. The series∑∞

k=1 gn converges uniformly on A if and only if for anyε > 0 there is N ∈ Z+ such that for any n,m ∈ Z+, and any z ∈ A

if n ≥ N, then

∣∣∣∣∣∣∣n+m∑

k=n+1

gk(z)

∣∣∣∣∣∣∣ < ε .Proof:

�



•

Proposition 3.1.3

1. The limit of a uniformly convergent sequence of continuous func-tions is continuous. Let A ⊂ C, f : A → C and fn : A → C bea sequence of functions on A. Suppose that all functions fn arecontinuous on A and fn → f uniformly on A. Then the function fis continuous on A.

2. The sum of a uniformly convergent series of continuous functionsis continuous. That is, if g =

∑∞k=1 gk uniformly on A and gk are

continuous on A then g is continuous on A.

Proof:�

•

Theorem 3.1.2 Weierstrass M Test. Let A ⊂ C and gn : A → C. LetMn be a sequence in R such that:

1. |gn(z)| ≤ M for any z ∈ A ,

2.∑∞

k=1 Mk converges.

Then the series∑∞

k=1 gk converges absolutely and uniformly on A.

Proof: By Cauchy criterion.�

•

Theorem 3.1.3 Analytic Convergence Theorem.

1. Let A be an open set, f : A → C, fn : A → C be a sequenceof functions on A. Suppose that fn are analytic on A and fn →

f uniformly on every closed disk D contained in A. Then f isanalytic on A. Moreover, f ′n → f ′ uniformly on every closed diskin A.

2. Let A be an open set, g : A → C, gn : A → C be a sequenceof functions on A. Suppose that gn are analytic on A and g =∑∞

k=1 gk uniformly on every closed disk D contained in A. Theng is analytic on A. Moreover, g′ =

∑∞k=1 g′k uniformly on every

closed disk in A.

Proof:�


3.1. SERIES OF ANALYTIC FUNCTIONS 51

•

Proposition 3.1.4 Let A be a region, γ : [a, b] → A be a curve in A,f : A → C, and fn : A → C. Suppose that fn are continuous on thecurve γ([a, b]) and fn → f uniformly on γ([a, b]). Then∫

γ

fn →

∫γ

f .

Similarly, if the series∑∞

k=1 gk converges uniformly on γ, then∫γ

∞∑k=1

gk =

∞∑k=1

∫gk .

Proof:�

• Exercises: 3.1[2,4,5,12,19]



3.2 Power Series and Taylor Theorem• A power series is a series of the form

∞∑k=0

ak(z − z0)k .

•

Lemma 3.2.1 Abel-Weierstrass Lemma. Let (an)∞n=1 ∈ C and r0,M ∈R such that |an|rn

0 ≤ M for any n ∈ Z+. Then for any r such that 0 ≤ r <r0 the series

∑∞k=0 ak(z − z0)k converges uniformly and absolutely on the

closed disk cl(D(z0, r)) = {z ∈ C | |z − z0| ≤ r}.

Proof:�

•

Theorem 3.2.1 Power Series Convergence Theorem.Let

∑∞k=0 ak(z − z0)k be a power series.

1. There is a unique real number R ≥ 0, (which can also be ∞),called the radius of convergence, such that the series convergeson the open disk D(z0,R) = {z ∈ C | |z − z0| < R} of ra-dius R centered at z0 and diverges in the exterior of this diskext(D(z0,R)) = {z ∈ C | |z − z0| > R}.

2. The convergence is uniform and absolute on any closed diskcl(D(z0, r)) = {z ∈ C | |z − z0| ≤ r} of radius r smaller than R,r < R.

3. The series may diverge or converge on the boundary ∂D(z0,R) ofthe disk, i.e. the circle |z − z0| = R, called the circle of conver-gence. There is a least one point on this circle such that the seriesdiverges.

Proof:�

•Theorem 3.2.2 Analyticity of Power Series. Every power series isanalytic inside its circle of convergence.

Proof:�


3.2. POWER SERIES AND TAYLOR THEOREM 53

•

Theorem 3.2.3 Differentiation of Power Series. Let∑∞

k=0 ak(z − z0)k

be a power series and A be the inside of its circle of convergence. Letf : A→ C be the analytic function defined by

f (z) =∞∑

k=0

ak(z − z0)k .

1. Then for any z ∈ A

f ′(z) =∞∑

k=1

kak(z − z0)k−1 =

∞∑k=0

(k + 1)ak+1(z − z0)k .

2. Moreover, for any k = 0, 1, . . . ,

ak =f (k)(z0)

k!,

and, therefore,

f (z) =∞∑

k=0

f (k)(z0)k!

(z − z0)k .

This series is called the Taylor series of f around z0.

Proof:�

•Theorem 3.2.4 Uniqueness of Power Series. Power series expan-sions around the same point are unique.

Proof:�

•

Proposition 3.2.1 Let∑∞

k=0 ak(z − z0)k be a power series.

1. Ratio test.R = lim

n→∞

∣∣∣∣∣ an

an+1

∣∣∣∣∣ .2. Root test.

R = limn→∞|an|−1/n .



Proof:�

•

Theorem 3.2.5 Taylor’s Theorem. Let A be an open set and f : A→C be analytic on A. Let z0 ∈ A and Aρ = D(z0, ρ) be the open disk ofradius ρ around z0. Let r be the largest value of the radius ρ such thatAr ⊂ A. Then for every z ∈ Ar the series

f (z) =∞∑

k=0

f (k)(z0)k!

(z − z0)k

converges.

1.Proof:

�

•

Corollary 3.2.1 Let A be an open set and f : A → C be analyticon A. Then f is analytic on A if and only if for each point z0 ∈ A theTaylor series of f around z0 has a non-zero radius of convergence andf is equal to the sum of that series.

Proof:�

• Let A be an open set, z0 ∈ A and f : A→ C be analytic on A. We say that fhas a zero of order n at z0 if

f (k)(z0) = 0 for k = 0, 1, . . . , (n − 1), and f (n)(z0) , 0 .

• If f has a zero of order n at z0 then

f (z) = (z − z0)ng(z) ,

for some analytic function g such that g does not have zero at z0, g(z0) , 0.

• If a function f has a zero at z0 and does not have zeroes in a neighborhoodof z0 then the zero of f is called isolated.

•

Proposition 3.2.2 Let A be an open set and f : A → C be analyticon A. Suppose that f has a zero at a point z0 ∈ A. Then either f isidentically zero or f has an isolated zero of some order n at z0.


3.2. POWER SERIES AND TAYLOR THEOREM 55

Proof:�

•

Proposition 3.2.3 Local Isolation of Zeroes. Let A be an open set,z0 ∈ A, and f : A → C be analytic on A. Let zk be a sequence of zeroesof f in A converging to z0, zk → z0. Then f is identically zero in A.

Proof:�

• Exercises: 3.2[1c,2b,2d,12,17,22]



3.3 Laurent Series and Classification of Singulari-ties

•

Theorem 3.3.1 Laurent Expansion. Let 0 ≤ r1 < r2, z0 ∈ C, andA = {z ∈ C | r1 < |z − z0| < r2}. (It is possible that r1 = 0 and r2 = ∞).Let f : A→ C be an analytic function. Then:

1. There is an expansion (called Laurent series around z0 in A)

f (z) =∞∑

n=−∞

an(z − z0)n .

2. The series converges absolutely on A and uniformly on every setB = {z ∈ C | ρ1 ≤ |z − z0| ≤ ρ2} with r1 < ρ1 < ρ2 < r2.

3. The coefficients are given by

an =1

2πi

∮γ

dζf (z)

(z − z0)n+1 , n ∈ Z ,

where γ is a circle centered at z0 of radius r with r1 < r < r2.

4. The Laurent series is unique.

Proof:�

• The Laurent series can also be written in the form

f (z) =∞∑

n=0

an(z − z0)n +

∞∑k=1

a−k(z − z0)−k .

• Example.


3.3. LAURENT SERIES AND CLASSIFICATION OF SINGULARITIES 57

•

Definition 3.3.1 Let A be a region and z0 ∈ A. Let f be a function andan be the coefficients of the Laurent series around z0.

1. If f if f is analytic on a deleted neighborhood of z0, then z0 iscalled an isolated singularity.

2. If there are only a finite number of coefficients an with negative n,then z0 is a pole of f . The order of the pole is the largest integerk such that bk , 0. A pole of order one is called a simple pole.

3. If there are infinitely many coefficients an with negative n, then z0

is called an essential singularity.

4. The coefficient a−1 is called the residue of f at z0.

5. If an = 0 for any negative n, then z0 is called removable singu-larity.

6. If the only singularities of a function f are poles, then f is calledmeromorphic.

7. If z0 is a pole of order k, then the principal part of f at z0 is

k∑n=0

a−n(z − z0)−n =ak

(z − z0)k + · · · +a−1

(z − z0).

• If z0 is a removable singularity, then by defining f (z0) = a0, f becomesanalytic at z0.

•

Proposition 3.3.1 Let f be an analytic function with an isolated sin-gularity at z0. Let γ be a circle centered at z0 of a sufficiently smallradius so that f is analytic on an annulus containing γ. Then

a−1 =1

2πi

∮γ

dz f (z) .

Proof:�



•

Proposition 3.3.2 Let z0 ∈ C Let f and g be analytic in a neighbor-hood of z0 with zeroes of order n and k respectively. Let h = f /g. Then

1. If k > n, then h has a pole of order (k − n) at z0.

2. If k = n, then h has a removable singularity at z0.

3. If k < n, then h has a removable singularity at z0, and by definingh(z0) = 0 we get an analytic function h with zero of order (n − k)at z0.

Proof:�

•

Proposition 3.3.3 Let z0 ∈ C. Let f be an analytic function with anisolated singularity at z0.

1. z0 is a removable singularity if and only if one of the followingholds:

(a) f is bounded in a deleted neighborhood of z0.

(b) f has a limit at z0.

(c) limz→z0(z − z0 f (z) = 0.

2. z0 is a simple pole if and only if limz→z0(z− z0 f (z) exists and is notequal to 0.

3. z0 is a pole of order m (or a removable singularity) if and only ifone of the following holds

(a) There are M > 0 and k ∈ Z+ such that for any z in a deletedneighborhood of z0 | f (z)| ≤ M

|z−z0 |k.

(b) for any k ≥ m, limz→z0(z − z0)k+1 f (z) = 0.

(c) for any k ≥ m, limz→z0(z − z0)k f (z) exists.

4. z0 is a pole of order m ≥ 1 if and only if there is an analyticfunction g on a neighborhood of z0 such that g(z0) , 0 and forany z , z0

f (z) =g(z)

(z − z0)m .


3.3. LAURENT SERIES AND CLASSIFICATION OF SINGULARITIES 59

Proof:�

• For a simple pole the residue is equal to

a−1 = limz→z0

(z − z0 f (z) .

•

Theorem 3.3.2 Picard Theorem. Let z0 ∈ C and U be a deletedneighborhood of z0. Let f be an analytic function with an essentialsingularity at z0. Then for all w ∈ C (except possibly one value), theequation f (z) = w has infinitely many solutions in U.

Proof:�

•

Theorem 3.3.3 Casorati-Weierstrass Theorem Let z0,w ∈ C. Let fbe an analytic function with an essential singularity at z0. Then there isa sequence (zn) in C such that zn → z0 and f (z0)→ w.

Proof:�

• Exercises: 3.3[4,7,8,10,18,19]


Chapter 4

Calculus of Residues

4.1 Calculus of Residues

4.1.1 Removable Singularities

• Proposition 4.1.1 Let f and g be analytic in a deleted neighborhoodof z0 ∈ C. Suppose that f and g have zeros of the same order at z0. Thenthe function h = f /g has a removable singularity at z0.

Proof:�

• Examples.

4.1.2 Simple Poles.

•

Proposition 4.1.2 Let f and g be analytic in a deleted neighborhoodof z0 ∈ C. Suppose that f (z0) , 0 and g′(z0) , 0. Then the functionh = f /g has a simple pole at z0 and

Res(h, z0) =f (z0)g′(z0)

.

Proof:�

61

62 CHAPTER 4. CALCULUS OF RESIDUES

•

Proposition 4.1.3 Let f and g be analytic in a deleted neighborhoodof z0 ∈ C. Suppose that f has a zero of order k at z0 and g has a zero oforder (k + 1) at z0. Then the function h = f /g has a simple pole at z0

and

Res(h, z0) = (k + 1)f (k)(z0)

g(k+1)(z0).

Proof:�

• Examples.

4.1.3 Double Poles.

•

Proposition 4.1.4 Let f and g be analytic in a deleted neighborhoodof z0 ∈ C. Suppose that f (z0) , 0 and g(z0) = g′(z0) = 0 and g′′(z0) , 0.Then the function h = f /g has a double pole at z0 and

Res(h, z0) = 2f ′(z0)g′′(z0)

−23

f (z0)g′′′(z0)[g′′(z0)]2 .

Proof:�

•

Proposition 4.1.5 Let f and g be analytic in a deleted neighborhoodof z0 ∈ C. Suppose that f (z0) , 0 and g(z0) = g′(z0) = g′′(z0) = 0 andg′′′(z0) , 0. Then the function h = f /g has a double pole at z0 and

Res(h, z0) = 3f ′′(z0)g′′′(z0)

−32

f ′(z0)g(4)(z0)[g′′′(z0)]2 .

Proof:�

• Examples.


4.1. CALCULUS OF RESIDUES 63

4.1.4 Higher-Order Poles.

•

Proposition 4.1.6 Let f be analytic in a deleted neighborhood ofz0 ∈ C. Let k be the smallest non-negative integer such that the limitlimz→z0(z − z0)k f (z) exists. Let ϕ be a function in the deleted neighbor-hood of z0 defined by ϕ(z) = (z − z0)k f (z). Then f has a pole of order kat z0 and ϕ has a removable singularity at z0 and

Res( f , z0) =ϕ(k−1)(z0)(k − 1)!

.

Proof:�

•

Proposition 4.1.7 Let f and g be analytic in a deleted neighborhoodof z0 ∈ C. Suppose that f (z0) , 0 and g(z0) = g′(z0) = · · · = g(k−1)(z0) =0 and g(k)(z0) , 0. Then the function h = f /g has a pole of order k at z0

and

Res(h, z0) =(

k!g(k)(z0)

)k

det

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

g(k)

k! 0 · · · 0 f (z0)0!

g(k+1)

(k+1)!g(k)

k! · · · 0 f ′(z0)1!

......

. . ....

...

g(2k−1)

(2k−1)!g(2k−2)

(2k−2)! · · ·g(k+1)

(k+1)!g(k−1)(z0)

(k−1)!

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣Proof:

�

• Summary.

• Examples.

4.1.5 Esssential Singularities.

• Exercises: 4.1[2,5,8,11,14]



4.2 Residue Theorem•

•

Theorem 4.2.1 Residue Theorem. Let A be a region and z1, . . . , zn ∈

A be n distinct points in A. Let B = A − {z1, . . . , zn} and f : B → Cbe analytic with isolated singularities at z1, . . . , zn. Let γ be a closedcontractible curve in A that does not pass through any points z1, . . . , zn.Then ∮

γ

dz f (z) = 2πin∑

k=1

I(γ, zk)Res( f , zk) ,

where I(γ, zk) is the winding number of γ with respect to zk.In particular, if γ is a loop traversed counterclockwise and z1, . . . , zp arethe points inside γ, then∮

γ

dz f (z) = 2πip∑

k=1

Res( f , zk) ,

Proof:�

•

Definition 4.2.1 Let F(z) = f (1/z). Then:

1. f has a pole of order k at∞ if F has a pole of order k at 0;

2. f has a zero of order k at∞ if F has a zero of order k at 0;

3. The residue of f at∞ is defined by

Res( f ,∞) = −Res[F(z)/z2; 0] .

• The R-neighborhood of∞ is the set {z | |z| > R} for some R > 0.

•

Proposition 4.2.1 Let f be analytic in a R0-neighborhood of infinity.Let γ be a sufficiently large circle centered at 0 with radius R, R > R0,traversed once counterclockwise. Then∮

γ

dz f (z) = −2πiRes( f ,∞) .

Proof:


4.2. RESIDUE THEOREM 65

�

•

Proposition 4.2.2 Let γ be a simple closed curve in C traversed oncecounterclockwise. Let f be analytic along γ and have only finitely manyisolated singularities, z1, . . . , zn, outside γ. Then∮

γ

dz f (z) = −2πi

n∑k=1

Res( f , zk) + Res( f ,∞)

.Proof:

�

• Exercises: 4.2[4,5,7,12,15]



4.3 Evaluation of Definite Integrals

4.3.1 Rational Functions of Sine and Cosine

• Proposition 4.3.1 Let R(cos θ, sin θ) be a rational function of cos θand sin θ for any θ ∈ [0, 2π]. Then∫ 2π

0dθ R(cos θ, sin θ) = 2π

∑|zk |<1

Res( f , zk) ,

where

f (z) =1z

R[12

(z +

1z

),

12

(z −

1z

)].

Proof:�

• Example.

4.3.2 Improper Integrals

•

Proposition 4.3.2 Let a ∈ R and f , g : [a,∞) → C are two complex-valued functions of a real variable. Suppose that | f (x)| ≤ |g(x)| for anyx ∈ [a,∞) and the integral

∫ ∞a

dx |g(x)| converges. Then the integral∫ ∞a

dx f (x) also converges and∣∣∣∣∣∫ ∞

adx f (x)

∣∣∣∣∣ ≤ ∫ ∞

adx |g(x)| .

This also holds if, formally, a = −∞.

Proof:�

• Remark. The convergence of the integral∫ ∞−∞

dx f (x) means that bothintegrals

∫ ∞a

dx f (x) and∫ a

−∞dx f (x) exist for any a, that is,∫ ∞

−∞

dx f (x) = limR1,R2→∞

∫ R2

−R1

dx f (x) .


4.3. EVALUATION OF DEFINITE INTEGRALS 67

• If the integral∫ ∞−∞

dx f (x) exists then it is equal to the symmetric limit

∫ ∞

−∞

dx f (x) = limR→∞

∫ R

−Rdx f (x) .

• However, the existence of this symmetric limit does not imply the conver-gence of the integral. If this limit exists, it is called the (Cauchy) principalvalue integral and denoted by P

∫ ∞−∞

, p.v.∫

, or −∫

, that is, by definition,

P∫ ∞

−∞

dx f (x) = limR→∞

∫ R

−Rdx f (x) .

•

Proposition 4.3.3 Let f : R → C. Suppose that the limitlimR1,R2→∞

∫ R2

−R1dx f (x) exists. Then the integral

∫ ∞−∞

dx f (x) exists and∫ ∞

−∞

dx f (x) = limR1,R2→∞

∫ R2

−R1

dx f (x) .

Proof:�

•

Proposition 4.3.4 Let f be an function that is analytic on an openset containing the closed upper half plane except for a finite number ofisolated singularities which do not lie on the real axis. Suppose thatthere are M,R0 > 0 and p > 1 such that for any z in the upper halfplane, if |z| > R0, then

| f (z)| ≤M|z|p.

Then ∫ ∞

−∞

dx f (x) = 2πi∑

Im zk>0

Res( f , zk) .

Proof:�



•

Proposition 4.3.5 Let f be an function that is analytic on an openset containing the closed lower half plane except for a finite number ofisolated singularities which do not lie on the real axis. Suppose thatthere are M,R0 > 0 and p > 1 such that for any z in the lower halfplane, if |z| > R0, then

| f (z)| ≤M|z|p.

Then ∫ ∞

−∞

dx f (x) = −2πi∑

Im zk<0

Res( f , zk) .

Proof:�

• Let Pn and Qm be polynomials such that m ≥ n + 2. Suppose that Qm doesnot have any zeroes on the real line. Let f = P/Q. Then both of the aboveformulas hold.

• Example.

4.3.3 Fourier Transform.

• Let f : R → C. The Fourier transform of the function f is a functionf : R→ C defined by

f (ω) =1√

2π

∫ ∞

−∞

dx e−iωx f (x) .

The Fourier cosine transform and Fourier sine transform are defined by

fc(ω) =1√

2π

∫ ∞

−∞

dx cos(ωx) f (x) .

fs(ω) = −1√

2π

∫ ∞

−∞

dx sin(ωx) f (x) .

Obviously,f = fc + i fs



• The inverse Fourier transform is defined by

f (x) =1√

2π

∫ ∞

−∞

dω eiωx f (ω) .

•

Proposition 4.3.6 Let f be analytic on an open set containing theclosed upper half plane except for a finite number of isolated singular-ities not lying on the real axis. Suppose that f (z) → 0 as z → ∞ in theupper half plane, that is, for any ε > 0 there is R > 0 such that if |z| > Rin the upper half plane, then | f (z)| < ε. Then for ω < 0∫ ∞

−∞

dx e−iωx f (x) = 2πi∑

Im zk>0

Res(e−iωz f (z), zk) .

Proof:�

•

Proposition 4.3.7 Let f be analytic on an open set containing theclosed lower half plane except for a finite number of isolated singular-ities not lying on the real axis. Suppose that f (z) → 0 as z → ∞ in thelower half plane, that is, for any ε > 0 there is R > 0 such that if |z| > Rin the lower half plane, then | f (z)| < ε. Then for ω > 0∫ ∞

−∞

dx e−iωx f (x) = −2πi∑

Im zk<0

Res(e−iωz f (z), zk) .

Proof:�

• Remark. Let Pn and Qm be polynomials and m > n. Suppose that Q has nozeros on the real axis. Let f = P/Q. Then both of the above formulas arevalid.

•

Lemma 4.3.1 Jordan’s Lemma. Let R > 0 and f be analytic outsidea semicircle |z| > R in the upper half plane. Suppose that f → 0 asz → ∞ uniformly in the upper half plane. Let ω < 0 and γr be asemicircle of radius r > R centered at the origin in the upper half plane.Then as r → ∞ ∫

γr

dz e−iωz f (z)→ 0 .



Proof: Advanced books.�

• Example.

4.3.4 Cauchy Principal Value

• Let c ∈ (a, b). Let f : [a, b] → C be continuous on [a, c) and (c, b] but notnecessarily at c. The integral integral

∫ b

adx f (x) converges if both integrals∫ c

adx f (x) and

∫ b

cdx f (x) exist, that is,∫ b

adx f (x) = lim

ε1,ε2→0

(∫ c−ε1

adx f (x) +

∫ b

c+ε2dx f (x)

).

• If the integral∫ b

adx f (x) exists then it is equal to the symmetric limit∫ b

adx f (x) = lim

ε→0

(∫ c−ε

adx f (x) +

∫ b

c+εdx f (x)

).

• However, the existence of this symmetric limit does not imply the conver-gence of the integral. If this limit exists, it is called the (Cauchy) principalvalue integral and denoted by P

∫ ∞−∞

, p.v.∫

, or −∫

, that is, by definition,

P∫ b

adx f (x) = lim

ε→0

(∫ c−ε

adx f (x) +

∫ b

c+εdx f (x)

). .

•

Proposition 4.3.8 Let f be analytic in a deleted neighborhood of z0.Suppose that f has a simple pole at z0. Let γr be an arc of a circle ofradius r and angle ϕ centered at z0. Then as r → 0

limr→0

∫γr

dz f (z) = ϕi Res ( f , z0) .

Proof:�



•

Proposition 4.3.9 Let f be analytic on an open set containing theclosed upper (lower) half-plane except for a finite number of isolatedsingularities, with the isolated singularities on the real axis being sim-ple poles. Suppose that:

1. there are M,R0 > 0 and p > 1 such that for any z in the upper(lower) half-plane if |z| > R0 then

| f (z)| ≤M|z|p,

or,

2. there is ω < 0 (ω > 0) and a function g such that f (z) = e−iωzg(z),where g is analytic on an open set containing the closed upper(lower) half-plane except for a finite number of isolated singular-ities and g(z)→ 0 as z→ ∞ in the upper (lower) half-plane.

Then the principal value integral exists and

P∫ ∞

−∞

dz f (z) = ±2πi∑

upper (lower) half−plane

Res ( f , zk)±πi∑

real axis

Res ( f , zk) .

Proof:�

• Example.

4.3.5 Integrals with Branch Cuts

• Mellin transform of a function f : [0,∞) → R is a function f of a complexvariable s defined by

f (s) =∫ ∞

0dt ts−1 f (t) .



•

Proposition 4.3.10 Let f be analytic on C except for a finite numberof isolated singularities none of which lie on the positive real axis. Lets ∈ R such that s > 0 and s is not a positive integrer. Suppose that:

1. there are M > 0,R > 0 and p > 1 such that for any z ∈ C, if|z| > R, then

|zs−1 f (z)| ≤M|z|p.

2. there are m > 0, r > 0 and q < 1 such that for any z ∈ C, if |z| < r,then

|zs−1 f (z)| ≤m|z|q.

Let zs−1 = exp[(s − 1) log z] be the branch with 0 < arg < 2π. Then thefollowing integral converges absolutely and is equal to∫ ∞

0dt ts−1 f (t) = −π [cot(πs) − i]

∑all singularities except 0

Res (zs−1 f (z), zk) .

Proof:�

• Example. Rational functions.

4.3.6 Logarithms

• Example.

• Exercises: 4.3[3,5,6,8,9,12,13,16,19,20b]


4.4. EVALUATION OF INFINITE SERIES 73

4.4 Evaluation of Infinite Series•

•

Proposition 4.4.1 Summation Theorem. Let f be a function that isanalytic in a region containing the real axis. Let ak = f (k) , k ∈ Z . LetG be a function that is analytic in a region containing the real axis withsimple poles at the real integers. Let gk = Res (G, k) , k ∈ Z . Supposethat the series

∑∞k=−∞ gkak converges absolutely. Let γ = γ1 + γ2, where

γ1 is a line above the real axis and γ2 is a line below the real axis, suchthat f is analytic in the strip between γ1 and γ2. Suppose that γ1 goesfrom∞+ iε to −∞+ iε and γ2 goes from −∞− iε to∞− iε, where ε > 0.Then there holds

∞∑k=−∞

gkak =1

2πi

∫γ

dz G(z) f (z) .

Proof: Let γ be a closed contour oriented counterclockwise enclosing thepoints −N1, . . . ,−1, 0, 1, . . . ,N2. Then

12πi

∮γ

dz G(z) f (z) =N2∑

k=−N1

gkak .

As N1,N2 → ∞ we have γ → γ and we obtain the result.�

• Similarly we can obtain a summation formula for the series∑∞

k=0 ak.

•

•

Proposition 4.4.2 Let f be a function that is analytic in a region con-taining the positive real axis. Let ak = f (k) , k = 0, 1, 2, . . . . Let Gbe a function that is analytic in a region containing the positive realaxis with simple poles at the nonnegative integers. Let gk = Res (G, k) ,k = 0, 1, 2, . . . . Suppose that the series

∑∞k=0 gkak converges absolutely.

Let γ be counter encircling the positive real axis in the counterclock-wise direction, that is, γ goes from ∞ + iε around 0 to ∞ − iε, whereε > 0. Then there holds

∞∑k=0

gkak =1

2πi

∫γ

dz G(z) f (z) .



Proof: Let γ be a closed contour oriented counterclockwise enclosing thepoints 0, 1, . . . ,N. Then

12πi

∮γ

dz G(z) f (z) =N∑

k=0

gkak .

As N → ∞ we have γ → γ and we obtain the result.�

• If the functions f and g are meromorphic and decrease sufficiently fast atinfinity, then these integrals can be computed by residue calculus.

• Examples.

• Exercises: 4.4[1,3,4,5]


Chapter 5

Analytic Continuation

5.1 Analytic Continuation and Riemann Surfaces

• A function analytic in a domain is uniquely determined by the values insome set with at least one limit point, such as a curve, a subdomain etc. Itdoes not matter how small that set is.

•

Lemma 5.1.1 Let A be a region in C and h : A → C. Let B ⊂ A be asubset of A with a limit point. Suppose that h is analytic on A and h = 0on B. Then h = 0 everywhere on A.

Proof: By isolation of zeros theorem or by using Taylor series.�

• In particular, B can be a region, a curve, a neighborhood of a point etc.

•

Corollary 5.1.1 Principle of Analytic Continuation. Let f , g : A →C be two analytic functions and B ⊂ A be a subset of A with a limitpoint. Suppose that f = g on B. Then f = g everywhere in A.

Proof: By using Taylor series.�

75

76 CHAPTER 5. ANALYTIC CONTINUATION

•

Proposition 5.1.1 Let A1 and A2 be two disjoint simply connecteddomains whose boundaries share a common contour γ. Let A = A1 ∩

γ∩A2. Let f be analytic in A1 and continuous in A1∩γ and g be analyticin A2 and continuous in A2∩γ. Suppose that f and g coincide on γ. Leth be a function on A defined by

h(z) =

f (z), z ∈ A1

f (z) = g(z), z ∈ γg(z), z ∈ A2

.

Then h is analytic on A.

• The function g is called the analytic continuation of the function f fromA1 to A2 through γ.

Proof: Show that for any contour C in A,∫

Ch = 0.

�

•

Proposition 5.1.2 Let A1 and A2 be two overlapping regions so thatA1 ∩ A2 , ∅. Let f : A1 → C and g : A2 → C. Suppose f = g onA1 ∪ A2. Then there is a unique analytic function h : A1 ∪ A2 → C suchthat h = f on A1 and h = g on A2 defined by

h(z) ={

f (z), z ∈ A1

g(z), z ∈ A2.

• The function h is called the analytic continuation of the function f fromA1 and of the function g from A2.

Proof: Show that for any contour C in A,∫

Ch = 0.

�

• Examples.

• Analytic continuation can be done along curves by using Taylor series.


5.1. ANALYTIC CONTINUATION AND RIEMANN SURFACES 77

•

Proposition 5.1.3 Monodromy Theorem. Let A be a simply con-nected domain and D be a disk in A. Let f be analytic in D. Supposethat the function f can be analytically continued from D along two dis-tinct contours γ1 and γ2 to functions f1 and f2 a point z in A. Supposethat in the region enclosed by the contours γ1 and γ2 the function fcontains at most isolated singularities. Then the result of each analyticcontinuation is the same, that is, f1(z) = f2(z).

Proof:�

• Example.

• There are singularities of analytic functions that prevent analytic continua-tion to a larger domain. Such singularities are called natural boundary, ornatural barrier.

• Example. Let f be defined by

f (z) =∞∑

k=0

z2n.

This series converges in the unit disk |z| < 1 and defines an analytic functionthere. However, it cannot be analytically continued to a larger domain.

• One can show that f satisfies the equation

f (z2) = f (z) − z .

Thereforef (z4) = f (z) − z − z2 ,

and, by induction,

f (z2m) = f (z) −

m−1∑k=0

z2k.

• Let zm be defined by z2m

m = 1. These are roots of unity that are uniformlyspread on the unit circle. As m→ ∞ they form a dense set on the unit circle.

• On another hand, for any zm, f (zm) = ∞. That is, all these points are singularpoints of f . Since they dense on the unit circle, f cannot be analyticallycontinued to any z such that |z| ≥ 1.



• Riemann Surfaces. Square root and logarithm.

• Exercises: 6.1[1,2,3,4,8,11]


5.2. ROCHE THEOREM AND PRINCILE OF THE ARGUMENT 79

5.2 Roche Theorem and Princile of the Argument

5.2.1 Root and Pole Counting Formulas

•

Proposition 5.2.1 Root-Pole Counting Theorem. Let f be an ana-lytic function on a region A except for a finite number of poles at b1,. . . , bm of orders p1, . . . , pm. Suppose that f has zeros at a1, . . . , an oforders q1, . . . , qn. Let γ be a closed contractible curve in A not passingthrough the points Ak and b j. Then

12πi

∮γ

f ′

f=

n∑k=1

qkI(γ, ak) −m∑

j=1

p jI(γ, bk) .

Proof:�

•

Corollary 5.2.1 Let γ be a simple closed contractable curve. Let f bean analytic function on a region A containing γ and its interior exceptfor a finite number of poles. Suppose that the poles and zeros of f donot lie on γ. Then

12πi

∮γ

f ′

f= Z − P ,

where Z and P are the number of zeros and poles counted with multi-plicities inside γ.

Proof:�

•

Corollary 5.2.2 Root Counting Formula. Let γ be a simple closedcontractable curve. Let f be an analytic function on a region A con-taining γ and its interior. Let w ∈ C. Suppose that f , w on γ. Then

N =1

2πi

∮γ

f ′

f − w

is the number of roots of the equation f (z) = w inside γ counted withtheir multiplicities.

Proof:�



5.2.2 Principle of the Argument

• Let z0 ∈ C and γ a closed contour. Recall the definition of the windingnumber (or index)

I(γ, z0) =1

2πi

∮γ

dzz − z0

.

• The change of the argument of z as z traverses γ is equal to

∆γ arg z = 2πI(γ, 0) .

• Now let f be an analytic function and γ : [a, b]→ C be a closed curve. Letus choose a branch of arg f (γ(t)) that varies continuously with t. Then thechange in the argument of w = f (z) as z = γ(t) traverses γ is equal to

∆γ arg f = arg[ f (γ(b))] − arg[ f (γ(a))] .

Note that f (γ(b)) = f (γ(a)).

• Equivalently, ∆γ f is equal to the change in the argument of z along theimage curve f ◦ γ, that is,

∆γ arg f = 2πI( f ◦ γ, 0) .

•

Proposition 5.2.2 Principle of the Argument. Let f be an analyticfunction on a region A except for a finite number of poles at b1, . . . , bm

of orders p1, . . . , pm. Suppose that f has zeros at a1, . . . , an of ordersq1, . . . , qn. Let γ be a closed contractible curve in A not passing throughthe points Ak and b j. Then

∆γ arg f = 2π

n∑k=1

qkI(γ, ak) −m∑

j=1

p jI(γ, bk)

.In particular, if γ is a simple closed curve, then

∆γ arg f = 2π(Z − P) ,

where Z and P are the number of zeros and poles counted with multi-plicities inside γ.

Proof:�


5.2. ROCHE THEOREM AND PRINCILE OF THE ARGUMENT 81

•

Proposition 5.2.3 Rouche’s Theorem. Let f and g be analytic on aregion A except for a finite number of poles. Suppose that f and g havea finite number of zeros in A. Let γ be a closed contractible curve in Anot passing through zeros and poles of f and g. Let ak be the zeros of fof order qk and

Z f =

n∑k=1

qkI(γ, ak) ;

and let Zg, P f , Pg be defined correspondingly for zeros and poles offunctions f and g. Suppose that on γ

| f (z) − g(z)| < | f (z)| .

Then:

1.∆γ arg f = ∆γ arg g ,

2.Z f − P f = Zg − Pg .

Proof:�

•

•

Corollary 5.2.3 Let γ be a simple closed contractible curve. Let fand g be analytic on a region A containing γ. Suppose that on γ

|g(z)| < | f (z)| .

Then f and f + g have the same number of zeros inside γ.

Proof:�

5.2.3 Injective Functions

•

Proposition 5.2.4 Let f : A → C be an analytic function. If f islocally one-to-one, then f ′ , 0 in A. If f is globally one-to-one, thenf −1 : f (A)→ A is analytic function.



Proof:�

•

Proposition 5.2.5 Injection Theorem. Let f be analytic on a regionA. Let γ be a closed contractible contour in A. Suppose that for anyz ∈ A, I(γ, z) = 0 or 1. Let B ⊂ A be a subset of A (the inside of γ)defined by

B = {z ∈ A | I(γ, z) , 0} .

Suppose that for any point w ∈ f (B)

I( f ◦ γ,w) = 1 .

Then f is one-to-one on B.

Proof: Let z0 ∈ B and w0 = f (z0) ∈ f (B). Then

N =1

2πi

∮γ

f ′(z)f (z) − w0

dz

is the number of solutions of the equation f (z) = w0 on B. Further,

N =1

2πi

∮f◦γ

dww − w0

= 1 .

Thus f is injective in B.�

• To show that an analytic function is injective on a region it is sufficient toshow that it is injective on the boundary.

• Exercises: 6.2[2,5,13,17]


5.3. MAPPING PROPERTIES OF ANALYTIC FUNCTIONS 83

5.3 Mapping Properties of Analytic Functions


84 Bibliography


Bibliography

[1] J. E. Marsden and M. J. Hoffman, Basic Complex Analysis, Freeman, NewYork, 1999

85

86 BIBLIOGRAPHY


Answers to Exercises

87

88 Answers To Exercises


Notation

89

90 Notation


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