lecture notes complex analysis math 435infohost.nmt.edu/~iavramid/notes/complanal.pdf · lecture...
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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
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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
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IV CONTENTS
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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.�
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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) .
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4 CHAPTER 1. ANALYTIC FUNCTIONS
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 .
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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 .
Proof: Direct calculation.�
• Properties of the Absolute Value.
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6 CHAPTER 1. ANALYTIC FUNCTIONS
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]
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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
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8 CHAPTER 1. ANALYTIC FUNCTIONS
•
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 ,
Proof: Direct calculation.�
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.�
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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.
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10 CHAPTER 1. ANALYTIC FUNCTIONS
•
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]
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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.�
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12 CHAPTER 1. ANALYTIC FUNCTIONS
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.
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1.3. CONTINUOUS FUNCTIONS 13
•
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 .
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14 CHAPTER 1. ANALYTIC FUNCTIONS
•
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.�
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1.3. CONTINUOUS FUNCTIONS 15
•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.�
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16 CHAPTER 1. ANALYTIC FUNCTIONS
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.
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1.3. CONTINUOUS FUNCTIONS 17
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.
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18 CHAPTER 1. ANALYTIC FUNCTIONS
�
•
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:�
complanal.tex; December 8, 2009; 12:32; p. 20
1.3. CONTINUOUS FUNCTIONS 19
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]
complanal.tex; December 8, 2009; 12:32; p. 21
20 CHAPTER 1. ANALYTIC FUNCTIONS
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)
.
complanal.tex; December 8, 2009; 12:32; p. 22
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:�
complanal.tex; December 8, 2009; 12:32; p. 23
22 CHAPTER 1. ANALYTIC FUNCTIONS
•
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.
complanal.tex; December 8, 2009; 12:32; p. 24
1.4. ANALYTIC FUNCTIONS 23
• 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)|| .
complanal.tex; December 8, 2009; 12:32; p. 25
24 CHAPTER 1. ANALYTIC FUNCTIONS
• 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.�
complanal.tex; December 8, 2009; 12:32; p. 26
1.4. ANALYTIC FUNCTIONS 25
•
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.
Proof: Use CR equations.�
complanal.tex; December 8, 2009; 12:32; p. 27
26 CHAPTER 1. ANALYTIC FUNCTIONS
•
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]
complanal.tex; December 8, 2009; 12:32; p. 28
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.�
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28 CHAPTER 1. ANALYTIC FUNCTIONS
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]
complanal.tex; December 8, 2009; 12:32; p. 30
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
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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|
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32 CHAPTER 2. CAUCHY THEOREM
•
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:�
complanal.tex; December 8, 2009; 12:32; p. 33
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]
complanal.tex; December 8, 2009; 12:32; p. 34
34 CHAPTER 2. CAUCHY THEOREM
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.�
complanal.tex; December 8, 2009; 12:32; p. 35
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 .
Proof: Use CR equations.�
•
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.�
complanal.tex; December 8, 2009; 12:32; p. 36
36 CHAPTER 2. CAUCHY 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]
complanal.tex; December 8, 2009; 12:32; p. 37
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.�
complanal.tex; December 8, 2009; 12:32; p. 38
38 CHAPTER 2. CAUCHY THEOREM
•
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:�
complanal.tex; December 8, 2009; 12:32; p. 39
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) .
Proof: Follows from above.�
•
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.
Proof: Follows from above.�
•
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.�
complanal.tex; December 8, 2009; 12:32; p. 40
40 CHAPTER 2. CAUCHY 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]
complanal.tex; December 8, 2009; 12:32; p. 41
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.
complanal.tex; December 8, 2009; 12:32; p. 42
42 CHAPTER 2. CAUCHY THEOREM
•
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:�
complanal.tex; December 8, 2009; 12:32; p. 43
2.4. MAXIMUM MODULUS THEOREM AND HARMONMIC FUNCTIONS43
•
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.
complanal.tex; December 8, 2009; 12:32; p. 44
44 CHAPTER 2. CAUCHY THEOREM
•
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:
complanal.tex; December 8, 2009; 12:32; p. 45
2.4. MAXIMUM MODULUS THEOREM AND HARMONMIC FUNCTIONS45
�
•
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]
complanal.tex; December 8, 2009; 12:32; p. 46
46 CHAPTER 2. CAUCHY THEOREM
complanal.tex; December 8, 2009; 12:32; p. 47
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.
complanal.tex; December 8, 2009; 12:32; p. 48
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:
�
complanal.tex; December 8, 2009; 12:32; p. 49
50 CHAPTER 3. SERIES REPRESENTATION OF ANALYTIC FUNCTIONS
•
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:�
complanal.tex; December 8, 2009; 12:32; p. 50
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]
complanal.tex; December 8, 2009; 12:32; p. 51
52 CHAPTER 3. SERIES REPRESENTATION OF ANALYTIC FUNCTIONS
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:�
complanal.tex; December 8, 2009; 12:32; p. 52
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 .
complanal.tex; December 8, 2009; 12:32; p. 53
54 CHAPTER 3. SERIES REPRESENTATION OF ANALYTIC FUNCTIONS
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.
complanal.tex; December 8, 2009; 12:32; p. 54
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]
complanal.tex; December 8, 2009; 12:32; p. 55
56 CHAPTER 3. SERIES REPRESENTATION OF ANALYTIC FUNCTIONS
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.
complanal.tex; December 8, 2009; 12:32; p. 56
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:�
complanal.tex; December 8, 2009; 12:32; p. 57
58 CHAPTER 3. SERIES REPRESENTATION OF ANALYTIC FUNCTIONS
•
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 .
complanal.tex; December 8, 2009; 12:32; p. 58
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]
complanal.tex; December 8, 2009; 12:32; p. 59
60 CHAPTER 3. SERIES REPRESENTATION OF ANALYTIC FUNCTIONS
complanal.tex; December 8, 2009; 12:32; p. 60
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.
complanal.tex; December 8, 2009; 12:32; p. 61
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]
complanal.tex; December 8, 2009; 12:32; p. 62
64 CHAPTER 4. CALCULUS OF RESIDUES
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:
complanal.tex; December 8, 2009; 12:32; p. 63
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]
complanal.tex; December 8, 2009; 12:32; p. 64
66 CHAPTER 4. CALCULUS OF RESIDUES
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) .
complanal.tex; December 8, 2009; 12:32; p. 65
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:�
complanal.tex; December 8, 2009; 12:32; p. 66
68 CHAPTER 4. CALCULUS OF RESIDUES
•
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
complanal.tex; December 8, 2009; 12:32; p. 67
4.3. EVALUATION OF DEFINITE INTEGRALS 69
• 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 .
complanal.tex; December 8, 2009; 12:32; p. 68
70 CHAPTER 4. CALCULUS OF RESIDUES
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:�
complanal.tex; December 8, 2009; 12:32; p. 69
4.3. EVALUATION OF DEFINITE INTEGRALS 71
•
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) .
complanal.tex; December 8, 2009; 12:32; p. 70
72 CHAPTER 4. CALCULUS OF RESIDUES
•
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]
complanal.tex; December 8, 2009; 12:32; p. 71
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) .
complanal.tex; December 8, 2009; 12:32; p. 72
74 CHAPTER 4. CALCULUS OF RESIDUES
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]
complanal.tex; December 8, 2009; 12:32; p. 73
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.
complanal.tex; December 8, 2009; 12:32; p. 74
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.
complanal.tex; December 8, 2009; 12:32; p. 75
78 CHAPTER 5. ANALYTIC CONTINUATION
• Riemann Surfaces. Square root and logarithm.
• Exercises: 6.1[1,2,3,4,8,11]
complanal.tex; December 8, 2009; 12:32; p. 76
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:�
complanal.tex; December 8, 2009; 12:32; p. 77
80 CHAPTER 5. ANALYTIC CONTINUATION
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:�
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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.
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82 CHAPTER 5. ANALYTIC CONTINUATION
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]
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5.3. MAPPING PROPERTIES OF ANALYTIC FUNCTIONS 83
5.3 Mapping Properties of Analytic Functions
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84 Bibliography
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Bibliography
[1] J. E. Marsden and M. J. Hoffman, Basic Complex Analysis, Freeman, NewYork, 1999
85
86 BIBLIOGRAPHY
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Answers to Exercises
87
88 Answers To Exercises
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Notation
89
90 Notation
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