anal is is comple jo

7/29/2019 Anal is is Comple Jo

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Complex Analysis

Istvn Mez, PhD


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Contents

I Complex numbers, sequences, series, topology 1

1 Complex numbers 31.1 The definition of the complex numbers and the basic operations 3Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Polar form of complex numbers . . . . . . . . . . . . . . . . . 6

1.2.1 The argument function . . . . . . . . . . . . . . . . . . 8Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2 Complex sequences and series 102.1 Complex sequences . . . . . . . . . . . . . . . . . . . . . . . . 10Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.2 Complex series . . . . . . . . . . . . . . . . . . . . . . . . . . 13Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.2.1 Complex power series . . . . . . . . . . . . . . . . . . . 152.2.2 The Polylogarithm functions . . . . . . . . . . . . . . . 18

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3 Topological basics 203.1 Metric spaces and their properties . . . . . . . . . . . . . . . . 203.2 Continuous functions on metric spaces . . . . . . . . . . . . . 23Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

II Complex Differentiability and the Integral For-mula of Cauchy 25

4 Differentiability 274.1 The Cauchy-Riemann equations . . . . . . . . . . . . . . . . . 28

5 The exponential and logarithm function and their deriva-tives 315.1 The sine and cosine functions . . . . . . . . . . . . . . . . . . 35Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385.2 The logarithm function . . . . . . . . . . . . . . . . . . . . . . 38

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5.3 General powers . . . . . . . . . . . . . . . . . . . . . . . . . . 41

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

6 Path integrals 43

6.1 Curves in the complex plane . . . . . . . . . . . . . . . . . . . 43

6.2 The path integral . . . . . . . . . . . . . . . . . . . . . . . . . 45

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

7 The theorem of Cauchy 49

7.1 The theorem of Cauchy for triangle paths . . . . . . . . . . . . 49

7.2 Cauchys theorem for starlike domains . . . . . . . . . . . . . 51

7.3 Cauchys theorem on simply connected domains . . . . . . . . 53

8 Cauchy Integral formula 56

8.1 A special version of the Cauchy Integral Formula . . . . . . . 56

8.2 Example circle integrals . . . . . . . . . . . . . . . . . . . . . 58

8.3 The general version of Cauchys integral formula . . . . . . . . 61

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

III Local and global analysis 65

9 The Laurent series 67

9.1 The Laurent series. Main part of a function . . . . . . . . . . 67

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

10 Singularities and zeroes of analytic functions 73

10.1 Classification of singularities . . . . . . . . . . . . . . . . . . . 73

10.1.1 The definition of zeros . . . . . . . . . . . . . . . . . . 77

10.2 The connection between zeros and poles . . . . . . . . . . . . 77

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

10.3 The Theorem of Casorati and Weierstrass . . . . . . . . . . . 8010.4 Further properties of zeros and poles . . . . . . . . . . . . . . 81

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

11 The Residue Theorem of Cauchy and its applications 84

11.1 The residue . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

11.2 The winding number . . . . . . . . . . . . . . . . . . . . . . . 87

11.3 The Residue Theorem of Cauchy . . . . . . . . . . . . . . . . 88

11.3.1 Applications of the Residue Theorem . . . . . . . . . . 89

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

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12 Some additional theorems on zeros and poles. . . 94

12.1 The number of zeros and poles . . . . . . . . . . . . . . . . . . 9412.2 Entire functions - the theorem of Liouville . . . . . . . . . . . 96Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

13 The theorems of Weierstrass and Mittag-Leffler 9813.1 Infinite products . . . . . . . . . . . . . . . . . . . . . . . . . 9813.2 The Weierstrass Product Theorem . . . . . . . . . . . . . . . . 101Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10413.3 The Mittag-Leffler expansion . . . . . . . . . . . . . . . . . . . 104Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

IV Special functions 108

14 Analytic extension 11014.1 The notion of analytic extension . . . . . . . . . . . . . . . . . 11014.2 The Riemann zeta function . . . . . . . . . . . . . . . . . . . 110

15 The Riemann zeta function 11315.1 The values of at positive even integers . . . . . . . . . . . . 11315.2 The connection between primes and the Riemann zeta function 115

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

16 The Euler Gamma function 11716.1 The connection between the and functions . . . . . . . . . 120Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

17 Conformal maps 12217.1 Some examples of transformations on the complex plane . . . 122Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12317.2 Conformal mappings . . . . . . . . . . . . . . . . . . . . . . . 12417.3 The Mbius transformations . . . . . . . . . . . . . . . . . . . 124

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

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Part I

Complex numbers, sequences,

series, topology

1


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2

Complex function theory is a widely applicable branch of mathematics.

Its powerful results can be applied to solve problems in other parts of math-ematics and even physics.

Complex analyis is based on complex numbers, which are extensions ofthe real ones. Many notions of real analysis can be carried to this more gen-eral situation. Therefore in the foregoing chapter we introduce the complexnumbers and (re)introduce many notions from the real analysis, such thatsequences, series, convergence, open and closed sets, compactness, etc.


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1

Complex numbers

1.1 The definition of the complex numbers and

the basic operations

If we try to solve the quadratic equation

x2 + x + 1 = 0,

we find that

x1,2 = 1

1

4

2 .

In the square root there is 3, which is not a square of any real number.Instead of saying that this equation cannot be solvable, we can try to calculatewith these type of numbers.

x1,2 =1 1 4

2= 1

2

3

2

1.

As it is easy to see, any solution of a quadratic equation can be rewritten as

a + bi,

where a, b R and i is the symbol 1. Can we calculate with thesenumbers as we did with the reals? Can we add, multiply divide them? Letus take an example:

(5 + 3i) + (4 + 2i) = 5 + 31 + 4 + 21 = 9 + 51 = 9 + 5i.

(5 + 3i) (4 + 2i) = 5 4 + 5 2i + 3i 4 + 3i 2i = 20 + 10i + 12i + 6i2.Since i =

1, i2 must be 1. Therefore

(5 + 3i) (4 + 2i) = 20 + 22i + 6(1) = 14 + 22i.

3


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4 1. chapter. Complex numbers

Now let us try the division. This is a bit more tricky:

5 + 3i

4 + 2i=

5 + 3i

4 + 2i 4 2i

4 2i =(5 + 3i)(4 2i)(4 + 2i)(4 2i) =

20 10i + 12i 6i216 8i + 8i 2i2 =

26 + 2i

18=

26

18+

2

18i =

13

9+

1

9i.

We can write the result of the division in the form a + bi.Therefore we can see, that we can calculate with these quantities as with

the usual real numbers, so we have the right to call these quantities as "num-bers".

1.1.1. Definition. The algebraic expressions of the forma + bi,

where a, b R and i = 1 are called complex number. The symbol i iscalled the imaginary unit.

The set of complex numbers is denoted by C. That is,

C = {a + bi | a, b R}.The next task after the definition is to give the rules, how we calculate

with these new numbers. As we have already seen, the four basic opeara-

tions (addition, substraction, multiplication, division) are easy to describe.Formally, we can have the following proposition.

1.1.2. Proposition. Leta+bi C andc+di C are two complex numbers.Then

(a + bi) + (c + di) = (a + c) + (b + d)i,

(a + bi) (c + di) = (a c) + (b d)i,(a + bi) (c + di) = (ac bd) + (ad + bc)i,

a + bi

c + di=

ac + bd

c2 + d2 + bc adc2 + d2 i.

In the division at least one of c and d is not zero.

For one who is familiar with abstract algebra, we remark that the setC is a field with respect to the above defined addition and multiplication,therefore C is often called as the field of complex numbers.

It will be useful to define three functions on the set of complex numbers:

: C R; (a + bi) = a,

: C

R;

(a + bi) = b,

: C C; a + bi = a bi.


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1.1. The definition of the complex numbers and the basic operations 5

The first function is called real part, the second one is the imaginary part.

The third function is the conjugate function.It is worth to note the following simple identities.

1.1.3. Proposition. For any complex number z, w Cz = z,

zz = |z|2,or, in algebraic form,

(a + bi)a + bi = (a + bi)(a bi) = a2 + b2,|z| = |z|,

z+ w = z+ w,

zw = zw,

(z) = 12

(z+ z),

(z) = 12i

(z z),(z+ w) = (z) + (w),(z+ w) = (z) + (w),

1

z =1

|z|2 z.Some useful inequalities can also be easily proven.

1.1.4. Proposition. For any complex number z, w C|(z)| |z|,|(z)| |z|,

||z| |w| | |z+ w| |z| + |w|,|z| | (z)| + |(z)|.

Exercises

1.1.5. Exercise. Calculate (3 + 2i) + (5 7i), (2 + 3i)(4 6i), and 132i78i .

1.1.6. Exercise. Calculate z w2, where z = 12 4i and w = 5 + 5i.

1.1.7. Exercise. Calculate (2 + 3i)4.

1.1.8. Exercise. What is the square root of i?

1.1.9. Exercise. Determine i2013.

1.1.10. Exercise. Calculate the sum i + i2 + i3 + + i100.


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1.2 Polar form of complex numbers

Since the complex number z = a + bi have two coordinates, we can drawthem in coordinate system.

The angle is called the angle or argument. The quantity r is the absolutevalue of z. The absolute value is also often called as the modulus of z, andit is denoted by |z|.

Since

cos =a

r

and

sin =b

r,

we can write that

z = a + bi = r cos + ir sin = r(cos + i sin ).

Thus we have got the following proposition.

1.2.1. Theorem. Every complex number z = a + bi can be written in theform

z = r(cos + i sin ),

where r 0 is a real number (the length of z), and [0, 2[ is the angleof z.

1.2.2. Definition. The above representation of z is called polar form. Therepresentation z = a + bi is the algebraic representation.

The polar form is extremely useful if we want to calculate the powers androots of a complex number. Let us fix two complex numbers in polar form:

z1 = r1(cos 1 + i sin 1),

z2 = r2(cos 2 + i sin 2).


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1.2. Polar form of complex numbers 7

Then their product is

z1z2 = r1r2(cos 1 + i sin 1)(cos 2 + i sin 2) =

r1r2[(cos 1 cos 2 sin 1 sin 2) + i(sin 1 cos 2 + cos 1 sin 2)].Then we can use the well known addition theorems for the trigonometricfunctions:

cos(1 + 2) = cos 1 cos 2 sin 1 sin 2,sin(1 + 2) = sin 1 cos 2 + cos 1 sin 2.

Then we get that

z1z2 = r1r2[cos(1 + 2) + i sin(1 + 2)].By an obvious reason, the division is as follows:

z1z2

=r1r2

[cos(1 2) + i sin(1 2)].

We arrived at the next proposition.

1.2.3. Proposition. In polar form we multiply and divide the complex num-bers

z1 = r1(cos 1 + i sin 1),

z2 = r2(cos 2 + i sin 2).as

z1z2 = r1r2[cos(1 + 2) + i sin(1 + 2)], (1.1)

and

z1z2

=r1r2

[cos(1 2) + i sin(1 2)] (r2 = 0). (1.2)

In special, thenth (n 1 is an integer) powers and roots of a complex numberz = r(cos + i sin ) become

zn = rn[cos(n) + i sin(n)], (1.3)

and

n

z = n

r

cos

+ 2k

n

+ i sin

+ 2k

n

(k = 0, 1, . . . , n 1).

(1.4)

The penultimate formula in the theorem

zn = rn[cos(n) + i sin(n)]

is the so-called de Moivres formula1.

1Abraham de Moivre (1667-1754) french mathematician.


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1.2.1 The argument function

The angle of a complex number is important enough to introduce a newfunction. We have seen, that for every z C there is an angle , which isalso called as argument. This is what the argument function arg maps to acomplex number. To be more formal, if we have

z = r(cos + i sin ),

then

arg : C R; arg(z) = .Note that this angle is not unique, since + 2k represents the same angle,

if k Z. Therefore the arg function maps infinite values to a same complexnumber z. For example, if z = 1 + i, then its argument is = 45 =

4.

Hence

arg(1 + i) =

4,

4 2,

4 4 , . . .

.

If we restrict us to the interval [0, 2[, the argument becomes unique. Tobe more general, we can say that if we restrict us to any interval of the form[, + 2[, the argument becomes unique. This phenomenon is connectedto an important notion, which is called branch. Any interval on which the

argument function is unique, is a branch of this function. Hence the intervals

[0, 2[, [2, 4[, [4, 6[, . . . , [2, 0[, 4, 2[, . . .

are the branches of arg. From these branches it is very natural to choose[0, 2[ which we shall call as the principal branch. On the principal branch

arg(1 + i) =

4.

Remark that we can index the branches with the integers. That is, by an

obvious reason on the 1th brancharg(1 + i) =

4 2 = 7

4,

while on the 1th branch

arg(1 + i) =

4+ 2 =

9

4.

The zeroth branch is the principal branch.

Now we list the two most basic functional identities of the argumentfunction which help to calculate its value.


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1.2. Polar form of complex numbers 9

1.2.4. Proposition. For any two complex numbers z1, z2 we have

arg(z1z2) = arg(z1) + arg(z2),

arg

z1z2

= arg(z1) arg(z2).

The proof of these identities comes from Proposition 1.2.3.Finally we remark that we have already met with branches in the defini-

tion of the root function, see (1.4). The nth root function have n branches,since a complex number have n pieces of nth roots.

Exercises

1.2.5. Exercise. Determine the absolute value, the argument and finallythe polar form of the complex numbers 1 + i, i, 1 + 2i.1.2.6. Exercise. Determine |3 2i|.1.2.7. Exercise. Find the algebraic and polar form of 1/i.

1.2.8. Exercise. Calculate

1 + i and (1 + i)100.

1.2.9. Exercise. Draw in the coordinate system the following sets:

A = {z C | (z) < 3},B = {z C | (z) 2},C = {z C | |z| < 1},D = {z C | |z| 1},E = {z C | |z 1| < 3},F = {z C | |z+ 3 2i| 4}.

1.2.10. Exercise. What is the value of arg(i) on the principal branch, and

on the second branch?1.2.11. Exercise. Determine the function value

arg

1 ii

on the principal branch.

1.2.12. Exercise. Draw in the coordinate system the following sets:

A =

z C|z| > 1, 0 < arg(z) <

4

,

B =

z C |z| arg(z) =

2

.


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2

Complex sequences and series

2.1 Complex sequences

Since we have the absolute value of complex numbers, which serve as a basefor measure distance, it is easy to define the convergence of sequences in theset C.

2.1.1. Definition. The function z : N C is called complex sequence. Byconvenience, we write zn in place of z(n).

This sequence converges if there is a z C such that for every > 0there is a natural number n

0such that for any n > n

0 |zn

z|

< . In thiscase we say that zn converges to z, and we denote this fact by

limn

zn = z.

If the sequence converges, this existing z is called the limit of the sequencezn.

Fortunately, we can trace back finding the complex limit of the sequenceto finding the limit of its real and imaginary part. This is the statement ofthe following theorem.

2.1.2. Theorem. Let zn be a complex sequence with the limit z. Then

limn

zn = z

holds if and only if

limn

(zn) = (z) and limn

(zn) = (z).

Proof. First let us suppose that

limn

zn = z.

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2.1. Complex sequences 11

Then we will prove that the real and imaginary part ofzn converges to

(z)

and (z), respectively. The limit means that for a fixed > 0 there is anindex n0 for which

|zn z| < for all n > n0.Then

|(zn) (z)| = |(zn z)| |zn z| < .The first equality comes from Proposition 1.1.3., the inequality comes fromPorposition 1.1.4. We have proven that for any fixed > 0 |(zn)(z)| < for any n > n0. Thus

limn(zn) = (z).The same can be proven for the imaginary parts.

To complete the proof, we have to show that the second statement impliesthe first one, that is, the convergence of the real and imaginary parts to (z)and (z), respectively, imply the convergence ofzn to z. Let > 0 such that

|(zn) (z)| < 2

,

and

|(zn) (z)| < 2

.

and let n0 and m0 be the indices for this . Then

|zn z| = |(zn) + i(zn) ((z) + i(z))|= |(zn) (z) + i((zn) (z))| |(zn) (z)| + |i((zn) (z))|=

|(z

n)

(z)|

+|i||

(zn

)

(z)|= |(zn) (z)| + |(zn) (z)|

max(n0, m0). We have proven the proposition.

As a demonstration, we show a plot of a convergent and a divergentsequence. The first plot belongs to the sequence

z = 1 + i

n

n2 :


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12 2. chapter. Complex sequences and series

The second plot below belongs to

z = (1 + i)n :


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2.2. Complex series 13

Exercises

2.1.3. Exercise. Find the limit of the complex sequence

zn =

n + i(n + 1)

n.

2.1.4. Exercise. Find the limit if exists of the following complex se-quences

zn =n + in

n,

zn =(n + i)(1

ni)

n2 ,

zn = i1

n .

(In the last example take the principal branch of the root function.)

2.1.5. Exercise. Show that the complex sequence zn = (1 + i)n diverges.

What is the limit of zn =(1+i)n

2n?

2.1.6. Exercise. Let zn be a sequence such that limn zn = z. What isthe limit of zn?

2.2 Complex series

The definition of the convergence for complex series is totally the same asfor the real ones.

2.2.1. Definition. Let zn be a complex sequence, and let us define the fol-lowing (complex) sequence:

sn := z1 + z2 + + zn.If s

nconverges as a sequence, we say that the infinite sum

z1 + z2 + exists and the limit ofsn is the value of this sum. For short, this sum is alsodenoted by

n=1

zn.

Ifsn does not converge, we say that the sum z1 + z2 + diverges. This factis often symbolized by

n=1 z

n = .


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Theorem 2.1.2. makes it clear that the sum or divergence of a complex

sequence can be determined by the behaviour of the real and imaginary partof sn. That is, let

n=1

zn

be a convergent series with sum z. Then

n=1

(zn) = (z) andn=1

(zn) = (z).

Closing this section we plot the partial sums of the infinite series

n=0

1

k + 1 + i 1

k + i

.

It can be seen, that the values of the partial sums has real part close to0, and imaginary part close to 1. (We have plotten up to n = 50.) So theseries seemingly converges ti i. One of the exercises is to prove this simplefact.

Exercises

2.2.2. Exercise. Find the sum of the following complex series:

n=1

1n +

i

n2

,

n=1

15n +

4i

3n

,

n=1

1 + in

6n .


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2.2.3. Exercise. Show that

n=0

1

k + 1 + i 1

k + i

= i.

2.2.4. Exercise. Let n=1

zn = z.

What is the sum of

n=1

zn ?

2.2.5. Exercise. For which z C will be the seriesn=1

1

n2 + z2and

n=1

arg(z+ n)

1 + in

convergent?

2.2.1 Complex power series

An infinite power series connected to a real sequence n (n = 0, 1, 2, . . . ) isdefined for a domain of the variable x for which the series around

f(x) =n=0

n(x )n

is convergent.From real analysis we know the next theorem.

2.2.6. Theorem. Let f be defined by a power series as above. Then thedomain of f is always is one of the following types.

1. {},2. {x R | |x | < } for some > 0 union a (possibly empty) subset

of the boundary.

3. The whole setR.

2.2.7. Definition. The number in the theorem is called radius of conver-gence of the power series.

Now we introduce a notation which we will frequently use as an abbrevi-ation.


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2.2.8. Definition. The open circle on the complex plane with radius and

with center will be denoted by B(, ), that is,

B(, ) = {z C | |z | < }.

A real power series obviously generalizes to complex ones.

2.2.9. Definition. Let n be a complex series, and C. Then the seriesn=0

n(z )n

is a function f of z. The series is called complex power series and f(z) iscalled the sum of the series in a given point z for which the above seriesconvergent.

The theorem above obviously generalizes to the complex case, mutatismutandis: we have to change the absolute value to the complex absolutevalue, and the set R to C.

The radius of convergence can be calculated as

=1

lim supn |

n|1

n

.

If the first case is valid in the theorem, then = 0, while if the third, = .2.2.10. Example. The following power series have radius of convergence = 0, = 1 and = , respectively:

n=0

n!zn,n=1

zn

n,

n=0

zn

n!.

In the first case

=1

lim sup n!1

n

= 0,

in the second case

=1

lim sup(1/n)1

n

=1

limsup1/ n

n= 1,

while in the third case

=1

lim sup(1/n!)1

n

=1

limsup1/ n

n!= .

Now we present a very useful theorem which helps us to derive new func-tions from old ones.


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2.2.11. Theorem. Let > 0 be the radius of convergence of the power series

f(z) =n=0

n(z )n.

Then the functionf is differentiable infinitely many times on the setB(, ),and

f(z) =n=0

nn(z )n1,

f(z) =

n=0

n(n

1)n(z

)n2,

...

f(k)(z) =n=0

n(n 1) (n k + 1)n(z )nk.

Here z B(, ) and k = 1, 2, 3 . . . .2.2.12. Example. Let

f(z) =

n=0

zn.

It is easy to find a closed form for this function:

f(z) zf(z) = 1 + z+ z2 + z3 + z z2 z3 ,so

f(z) =1

1 z (z B(0, 1)).Therefore, for example,

f12 = 1 + 1

2+

1

4+

=

1

1 12= 2.

Or, considering a complex number,

f

1 + i

2

= 1 +

1 + i

2+

1 + i

2

2+ = 1

1 1+i2

= 1 + i.

(Note that 1+i2

B(0, 1).)2.2.13. Example. To see how we use the above theorem, we deduce theformula

n=0nzn =

z

(1 z)2 .


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First, we start from the above proved result

f(z) =n=0

zn =1

1 z (z B(0, 1)).

Then we derive f:

f(z) =n=0

nzn1 =1

(1 z)2 (z B(0, 1)).

Finally we multiply by z:

zf(z) =

n=0

nzn = z(1 z)2 (z B(0, 1)).

2.2.2 The Polylogarithm functions

The Polylogarithm functions are important, often appearing special func-tions. Hence we give a short introduction of them.

2.2.14. Definition. Let k be a fixed integer. The kth Polylogarithm func-tion is defined by

Li(z) =

n=1

zn

nk .

By the above considerations we have the following special polylogarithms:

Li1(z) =n=1

nzn =z

(1 z)2 ,

Li0(z) =n=1

zn =z

1 z,

Li1(z) =

n=1

zn

n

=

log(1

z).

The Polylogarithm functions of higher positive order does not have a simpleclosed form. The most what we can do is to find some special values. Forexample, we shall see that

Li2(1) =2

6.

Note that for any parameter k

Lik(z) = zLik1(z)

for all z within the domain of the convergence.


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Exercises

2.2.15. Exercise. Calculate the value of the sums

Li1

i

2

=

n=1

n

i

2

n, Li2

1

2

=

n=1

n2

1

2

n.

2.2.16. Exercise. Theorem 2.2.11. works reversely. Find the general ex-pression and radius of convergence for the power series

Li1(z) =

n=1

zn

n

.

Then find the sum

Li1(1) =n=1

(1)nn

.

2.2.17. Exercise. Determine the radius of convergence for the power seriesof the polylogarithm function for all integer k.


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3

Topological basics

In this short chapter we reconsider the basic topological notions which areneeded in what follows. We suppose that the reader is already more or lessfamiliar with the foregoing topological facts, therefore we shall not prove thepresented statements. We will not go into the details and we do not try tobe general; the topological notions will be defined via the metric.

3.1 Metric spaces and their properties

3.1.1. Definition. A metric space is a pair (X, d), where X

=

and d :

X X [0, +[ is a function with the following properties1. d(x, y) = d(y, x)

2. d(x, y) = 0 if and only if x = y,

3. d(x, z) d(x, y) + d(y, z)for any x,y,z X. The function d is called metric.

Often if we are not specifically interested in the form of the metric d, inplace of (X, d) we simply write X.

We know that if X = R and d(x, y) =

|x

y

|then (X, d) is a metric

space. It is also true that X = C and d(x, y) = |x y| then (X, d) is ametric space. This metric is the standard one for the purpose of complexanalyisis, therefore in the following we shall deal with this metric space indetail. However, before we introduce some important notions on generalmetric spaces.

3.1.2. Definition. Let r > 0 is a fixed real number. We define the sets

B(x, r) = {y X | d(x, y) < r},B(x, r) = {y X | d(x, y) r}.

These sets are called the open and closed balls with center x and radius r,respectively.

20


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3.1. Metric spaces and their properties 21

3.1.3. Definition. A point x

A

X in a subset of a metric space is an

inner point of A if it is contained in an open ball which is entirely in A. Inother words, the point of A is an inner point of A if there is an r > 0 suchthat

B(x, r) A.

3.1.4. Definition. For a metric space (X, d) a subset A X is open if allof its points are inner points. If the complement of A is open, we say that Ais closed.

3.1.5. Proposition. Let(X, d) be a metric space. Then the following propo-

sitions are true:

1. The sets X and are open and close.2. Any finite intersection of open sets are open.

3. Any union (finite or infinite) of open sets are open.

4. Any finite union of closed sets are closed.

5. Any intersection (finite or infinite) of closed sets are closed.

3.1.6. Exercise. Which of the following sets are open in C?

The diskA = {z C | |z| < 1}

with center 0 and radius 1,

the real line, B = {z C | z4 = 1}. C = {z C | zn = 1 for some n N}.

3.1.7. Definition. The smallest closed set which contains A is called theclosure of A, and it is denoted by A. Moreover, the set of inner points of Ais called the interior of A and denoted by Int A. The boundary of a set A ina metric space is denoted by A and is defined by

A = A \ Int A.

3.1.8. Proposition. LetA be a subset of a metric space X. Then:

1. A is open if and only if A = Int A.

2. A is closed if and only if A = A.


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22 3. chapter. Topological basics

3. Int A = X

\X

\A.

4. A = X\ Int(X\ A).5. A B = A B.

3.1.9. Exercise. Find the closure, the interior and the boundary of the setsin the above exercise.

3.1.10. Definition. A subset A X of a metric space X is dense if itsclosure equals to X, that is,

A = X.

It is obvious that X = X.

3.1.11. Exercise. Give examples for dense sets in C.

3.1.12. Definition. An open ball B(a, r) is a neighbourhood of a point x ina metric space if x B(a, r).

3.1.13. Definition. A point x of A X is a limit point if for any r > 0there is a neighbourhood of x which contains an element of A.

For example, all the points ofC are limit points ofC.

3.1.14. Proposition. A point x is a limit point of A X if and only ifthere is a non-constant sequence xn in A which tends to x.

Note that the limit point does not necessarrily belong to the set A.

3.1.15. Exercise. Let (C, | |) be the already considered metric space. Findthe limit point(s) of the sets

Q,

Q

Q,

QR, [0, 1[ {i}, 1

n+ i 1

n2

Cn = 1, 2, 3, . . ..Now we deal with connectedness of sets in metric spaces.

3.1.16. Definition. Let A X be a subset of the metric space X. Then Ais connected if it is not possible decompose A in the form A = B C whereB and C are disjoint, nonempty and open. IfA is not connected, we say thatdisconnected.


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3.2. Continuous functions on metric spaces 23

3.1.17. Example. The whole set C is connected in the metric space (C,

||).

The open and closed balls with any center and any radius in C are connected.In special, any singleton set is connected.

3.1.18. Definition. If a set A is disconnected, it is possible represent it asa union of maximal connected subsets. In this case these sets are calledcomponents of A.

3.1.19. Example. The set A = [5, 2] [3, 4] {i, i + 2} is disconnected.Its components are

[

5, 2], [3, 4],

{i

},

{i + 2

}.

Finally, we introduce a notion which turned to be very important inanalysis.

3.1.20. Definition. Let A be a set in a metric space X, and let be anindex set. A collection of sets B is a covering system for A if

A

B.

If all of the B sets are open, we say that the covering system is an open one.

A is compact if it has a finite covering system (i.e., is finite).

3.1.21. Definition. A subset A of the metric space X is bounded if there isa radius r > 0 such that

A B(x, r)for some x X.

Here is a very useful characterization of compactness in the complex plane.

3.1.22. Theorem. A subset A ofC is compact if and only if it is boundedand closed.

3.2 Continuous functions on metric spaces

3.2.1. Definition. Let X and Y be two metric spaces. The function f :X Y is continuous in a point x X if for any sequence xn such that inthe case limn xn x in X we have that limn f(xn) f(x). If f iscontinuous in any x X we say that f is continuous.

Now we list some basic and important equivalents of continuity.

3.2.2. Theorem. Let X and Y be two metric spaces, and let f : X Y bea function. Then the followings are equivalent.


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24 3. chapter. Topological basics

1. f is continuous,

2. if A Y is open then f1(A) is open,3. if A Y is closed then f1(A) is closed.It is also easy to see iff and g are two continuous functions from X into

C then the functionsf + g (, C)

are also continuous. Moreover, f /g is continuous provided g(x) = 0 for everyx X.

The next theorem connects compactness and connectedness to continuity.

3.2.3. Theorem. Let f : X Y be a continuous function between themetric spaces X and Y. Then

1. If X is compact then f(X) is a compact subset of Y.

2. If X is connected then f(X) is a connected subset of Y.

Exercises

3.2.4. Exercise. Let us consider the metric space (Z, | |). Characterize thenull sequences in this space. (A null sequence is a sequence with limit 0.)3.2.5. Exercise. Let us consider the metric space (Z, | |). Decide whetherthe subset

{0, 2, 4, . . . }is closed or open.

3.2.6. Exercise. Let us consider the metric space (Z, | |). Look for thedense sets in this space.

3.2.7. Exercise. We define the set K R as

K =

2n2m + 1

n, m Z .What is the closure of this set (with respect to the usual topology of R)?

3.2.8. Exercise. We define the set K C as

K =

2n

2m + 1+ i

2k + 1

2l

n,m,k,l Z

.

What is the closure of this set (with respect to the usual topology of C)?

3.2.9. Exercise. Give example for a subset in a metric space X which is

closed and bounded, but not compact.


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Part II

Complex Differentiability and the

Integral Formula of Cauchy

25


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26

From now on we deal with the real topics of complex analysis. First we

turn to the question of differentiability of complex functions. We shall seethat the complex differential behaves a bit differently than the real differen-tiability, because the complex differentiability gives a rather strict restrictionfor the function. Then we turn to the complex integral. We shall presentvery powerful results which help to calculate integrals.

The well known real functions, like the exponential function, the sin, cosand their derivatives will be extended to complex variable in this part.


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4

Differentiability

The definition of differentiability does not differ in the complex case.

4.0.10. Definition. Let E be a nonempty set with a nonepty interior. Letf : E C and z Int A. We say that f is differentiable in z if the limit

limh0

f(z+ h) f(z)h

exists. If this limit exists, we denote it by f(z) and we say that this is thederivative of f in the point z.

We shall see the surprising fact that differentiability in a class of functions(the analytic functions) implies the infinite differentiability.

It is convenient to suppose that the set E is itself open (instead of suppos-ing that has a nonempty interior) and it is connected (in contrary, we shouldalways handle the different components of E). Therefore we introduce thenext definition.

4.0.11. Definition. If a set E C is nonempty, open and connected, wecall it as domain.

4.0.12. Definition. Let D be a domain in C. The function f : D C isanalytic on D if it is differentiable in all the points of D. The set of analyticfunctions on this domain D will be denoted by A(D).

The definition of differentiability immediately yields the following propo-sition.

4.0.13. Proposition. Letf, g A(D) be two complex functions and C.Then f, f + g , f g

A(D). Moreover, (f) = f, (f + g) = f + g and

(f g) = fg + gf.

27


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28 4. chapter. Differentiability

4.1 The Cauchy-Riemann equations

It is very interesting to check that what causes on the real and imaginarypart of a function if we suppose the differentiability. Let f A(D) and letus write f as

f(z) = u(z) + iv(z), where u = (f), v = (f).We have the following very important theorem. This theorem in many caseshelps us to check the differentiability.

4.1.1. Theorem. Let f A(D) with the above decomposition, and let z

D. The function f is differentiable in z if and only if u and v are differen-tiables and they satisfy the partial differential equations

u

x(z) =

v

y(z),

u

y(z) = v

x(z).

These equations are called Cauchy-Riemann equations.In other words, f is analytic on D if u and v are differentiables and on

this domain they satisfy the Cauchy-Riemann equations.

Proof. We prove the only if part. Let z = x + iy D andu(x, y) =

f(x + iy), and v(x, y) =

f(x + iy).

Then

f(z) = limh0

f(z+ h) f(z)h

.

The fraction becomes

f(z+ h) f(z)h

=f(x + iy + h) f(x + iy)

h

=u(x + h, y) u(x, y)

h+ i

v(x + h, y) v(x, y)h

.

If h 0, the fractions tend to the respective partial derivatives:

f(z) = ux

(x, y) + i vx

(x, y). (4.1)

Calculating f(z) in an other way, we take the fraction

f(z+ ih) f(z)ih

= iu(x, y + h) u(x, y)h

+v(x, y + h) v(x, y)

h.

If h tends to 0, the left hand side tends to f(z), on the right hand side thefollowing partial derivatives appear:

f(z) = iuy

(x, y) +v

y(x, y).

Comparing this with (4.1), we are ready.


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4.1. The Cauchy-Riemann equations 29

4.1.2. Example. The absolute value function is nowhere differentiable. To

show thsi, let f(z) = |z| = x2 + y2. Thenu(x, y) =

x2 + y2, and v(x, y) = 0

Then it is straightforward to see that the partial derivatives ofu are not zero(where it is exists) but the partial derivatives of v are identically zero, so theCauchy-Riemann equations does not satisfy.

4.1.3. Example. The arg function is nowhere differentiable. The reason isthe same as above.

4.1.4. Example. The function f(z) = z2 is differentiable on the whole setC and

f(z) = 2z.

If f(z) = z2 then

u(x, y) = f(x + iy) = (x + iy)2 = x2 y2,

and

v(x, y) =

f(x + iy) =

(x + iy)2 = 2xy.

Then (4.1) gives that

f(z) =u

x(x, y) + i

v

x(x, y)

= 2x + i2y = 2(x + iy) = 2z.

4.1.5. Remark. It can be derived in the same manner as the above examplethat if f(z) = zn then

f(z) = nzn1,

or, more generally,

(anzn + an1zn1 + + a1z+ a0) = nanzn1 + (n 1)an1zn2 + + a1.

To close this section, we also show that the function f(z) = z is nowheredifferentiable. Let

u(x, y) = x + iy = (x iy) = x,

and

v(x, y) = x + iy = (x iy) = y.


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30 4. chapter. Differentiability

Then the Cauchy-Riemann equations says that

u

x(z) =

v

y(z),

u

y(z) = v

x(z).

Substituting the actual u and v, we get that

1 = 1, and 0 = 0.

The first equation is a contradiction, so the function f(z) = z does notdifferentiable, as we stated.


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5

The exponential and logarithm

function and their derivatives

We shall extend the well known exponential function to complex variable.The real exponential function is defined via a power series:

ex = exp(x) =n=0

xn

n!.

This series has convergence radius = . Without any problem, we canextend the definition to complex variable, because in place of a real x, wecan put a complex number z: the nth power is unique.

5.0.6. Definition. The complex exponential function exp : C C is definedby the power series

ez = exp(z) =

n=0zn

n!.

This definition is valid for all z C. What gives the special importanceof this function? That its derivative is itself:

exp(z) =

n=0

zn

n!

=

n=1

nzn1

n!=

n=1

zn1

(n 1)! =n=0

zn

n!.

Now we deal with the basic properties of this function. The real expo-nential function has the familiar graph

31


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32 5. chapter. The exponential and logarithm function and their derivatives

To get a picture on the complex exponential, we plot its real and imagi-nary part when z = x + iy and x, y [10, 10]:


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33

On the same intervals we plot the absolute value | exp(z)| to see how thisfunction is growing:


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The constant

exp(1) = e =n=0

1

n!= 2.718281828 . . .

has fundamental importance in mathematics. This number is called Euler1

constant hence the letter e - or Napier2 constant.That the well known identity exp(x + y) = exp(x) exp(y) holds for com-

plex numbers as well, we can check easily with the Cauchy product and thebinomial theorem.

5.0.7. Theorem. The complex exponential function exp : C C satisfy theidentity exp(z+ w) = exp(z)exp(w).

Proof.

exp(z)exp(w) =

n=0

z

n!

n=0

w

n!

=

n=0

nk=0

zk

k!

wnk

(n k)! =

n=0

1

n!

nk=0

n!zk

k!

wnk

(n k)! =n=0

1

n!

nk=0

n

k

zkwnk =

n=0

(z+ w)n

n! = exp(z+ w).

5.0.8. Corollary. The exp function does not have zero, that is, there is noz C for which exp(z) = 0.Proof. Let us suppose that exp(z) = 0 for a z. Then

exp(z) exp(z) = exp(0) = 1.This is impossible, if exp(z) = 0.

Now let us check the real and imaginary parts of the exponential function.5.0.9. Theorem. Let z = x + iy be a complex number. The real and imag-inary parts of exp are

(exp(z)) = exp(x)n=0

(1)n(2n)!

y2n,

(exp(z)) = exp(x)n=0

(1)n(2n + 1)!

y2n+1.

1Leonhard Euler (1707-1783) swiss mathematician.2

John Napier () scottish mathematician, physicist, astronomer. He introduced thelogarithm function to simplify calculations.


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5.1. The sine and cosine functions 35

Proof. Using the above proved identity for the exp function,

exp(z) = exp(x)exp(iy) = exp(x)n=0

(iy)n

n!.

The number exp(x) is real, moreover, (iy)n = inyn is real if n is even, andpurely imaginary if n is odd. Hence

exp(z) = exp(x)

n=0

(1)n(2n)!

y2n

+ i

n=0

(1)n(2n + 1)!

y2n+1

.

Now the result follows.

5.1 The sine and cosine functions

Now we define the complex sine and cosine functions. They naturally comefrom the real and imaginary parts of the complex exponential function.

5.1.1. Definition. The functions

(exp(z))/ exp(x) and (exp(z))/ exp(x)

are called cosine and sine function, respectively. They sign are cos and sin.

By the above theorem we have that cos, sin : C C, and

cos(z) =

n=0

(1)n

(2n)!

y2n,

sin(z) =n=0

(1)n(2n + 1)!

y2n+1.

These functions are the complex extensions of the real cosine and sinefunctions. From the definition and by Theorem 2.2.11. it is obvious that

sin, cos A(C).

Here comes the graph of the real and imaginary part of the complex sinefunction on the square [2, 2] [2i, 2i]:


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With these functions we have the following fundamental identity.


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5.1. The sine and cosine functions 37

5.1.2. Corollary. Letz = x + iy. Then

exp(z) = ex(cos(y) + i sin(y)).

In special, if z = i, we have that

exp(i) = 1.

The other two familiar functions, the tangent and cotangent now will bedefined.

5.1.3. Definition. The function sin(z)/ cos(z) is called complex tangent func-tion and it is denoted by tan(z). Its reciprocal is the complex cotangentfunction, ctg(z).

However the real exponential function is bijective, the complex one is veryfar from satisfying this property. To get a deeper picture on exp, we dealwith its mapping properties.

Let us fix a set of complex numbers:

A := {x + iy C | y R}.

(The real part x is fixed.) The image of A by exp is

exp(A) = {ex(cos(y) + i sin(y)) | y R}.

This set is nothing else but a circle with center 0 and radius r = ex. Theoriginal set A is a vertical line.

If x is running and y is fixed, that is, we have a horizontal line, a similarargument gives that the image of this line is a half line which starts from theorigin (this point is not the part of the half line). The angle of this line is y.

Therefore we have got the next proposition.

5.1.4. Proposition. The complex exponential function maps vertical linesof the complex plane with abscissa x to circles with center 0 and radius ex.The images of horizontal lines with ordinate y are half line starting from theorigo with angle y. The origo is not a part of the image.

We have the next corollary.

5.1.5. Corollary. Th ecomplex functionexp is defined for all complex z andits codomain is the whole setC except 0:

exp : C C \ {0}.


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Exercises

5.1.6. Exercise. Show that for the complex sine and cosine functions

sin(z) =1

2i(exp(iz) exp(iz)),

cos(z) =1

2(exp(iz) + exp(iz)).

For all z C.Let us define the hyperbolic sine and hyperbolic cosine functions as

sinh(z) = 12(exp(z) exp(z)),cosh(z) =

1

2(exp(z) + exp(z)).

5.1.7. Exercise. Deduce the closed form value

cos(i) =1

2(e + e) = cosh() = 11.59195328 . . . .

5.1.8. Exercise. Show that sin = cos, cos = sin, sinh = cosh andcosh = sinh.

5.1.9. Exercise. Derive the addition theorems for the complex sine andcosine:

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

cos(z+ w) = cos(z) cos(w) sin(z)sin(w).for all z, w C.

5.2 The logarithm function

In real analysis the logarithm function is the inverse of the exponential. Thisinverse exists, since the real exponential function is strictly increasing. But,the complex exponential is not bijective as we had seen, hence the complexlogarithm as a function cannot be defined. What we can do? Let us try tosolve the equation

exp(z) = w. (5.1)

Let us suppose that z = x + iy, Then via Corollary 5.1.2. |w| = exp(x)and y = arg(w) + 2k for som k Z. Hence we have that the solution of(5.1) is not a definite number but a set:

{log

|w

|+ i(arg(w) + 2k)

|k

Z

}.

Now we can define the logarithm functions:


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5.2. The logarithm function 39

5.2.1. Definition. Let

logk(w) = log |w| + i(arg(w) + 2k), (5.2)

where k Z.

Note that since exp(z) is never zero, we cannot define log function asan inverse of exp at zero.

In addition, we have infinitely many logarithm functions or if we wantto use the notion of branch the log function have infinitely many branches.From now on we consider the branch belonging to k = 0 as the principal

branch and we denote this function with log

5.2.2. Example. Let us calculate log(1 + i) and log(i) and log(1).Since |1 + i| = 2, and arg(1 + i) = /4, we have that

log(1 + i) = log(

2) + i

4=

1

2log(2) + i

4.

Next, log(i) = i2

, because |i| = 1 and arg(i) = /2.Finally, log(1) = i, because | 1| = 1 and arg(1) = .All of these values are on the principal branch.

We can see that the derivative of log has the same form as in the realanalysis:

5.2.3. Proposition. We have that

log(z) =1

z

for any branch.

Proof. Since exp(log(z)) = z, the well known rule for the derivative of theinverse function

log(z) =1

exp(log(z))=

1

exp(log(z))=

1

z.

This has an important corollary:

5.2.4. Theorem. Thelog function (with the principal branch) has the powerseries expansion

log(z) =

n=1

(

1)n+1

n (z 1)n

.


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Proof. Let us define the function g(z) as

log(z) n=1

(1)n+1n

(z 1)n.

The series on the right has convergence radius 1, so this function is welldefined on the open ball B(1, 1). Its derivative is

g(z) =1

z

n=1

(1)n+1(z 1)n1.

Here we used the above theorem and Theorem 2.2.11. The series on the rightis a geometric series with sum 1

1+(z1) , whence g(z) = 0. This gives thestatement.

5.2.5. Corollary. We have the famous expansion for the constant log(2):

log(2) =n=1

(1)n+1n

= 0.6931471806 . . . .

This series is called alternating harmonic series.

In general, the series

log(1 + z) =n=1

(1)n+1n

zn

is the Mercator3 series.

We close this section with pointing out that the well known identity

log(ab) = log(a) + log(b)

is not valid in general for the complex logarithm (on the principal branch).

5.2.6. Theorem. If z and w are two complex numbers such that (z) > 0,(w) > 0 and(zw) > 0, then

log(zw) = log(z)log(w).

This relation is not valid if the above assumptions do not hold.

3Nicholas Mercator (1620-1687) netherlandic mathematician and inventor. He pub-

lished firstly this series for the logarithm in 1668, but Isaac Newton and Gregory Saint-Vincent are also revealed this formula independently.


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5.3. General powers 41

Proof. We begin with the definition of the log function:

log(z) = log |z| + i arg(z).

Then

log(zw) = log |zw| + i arg(zw) = log(|z||w|) + i(arg(z) + arg(w)).

This latter equality comes from Proposition 1.2.4. We continue:

log(|z||w|) + i(arg(z) + arg(w)) = log |z| + log |w| + i(arg(z) + arg(w)) =

log(z) + log(w).

This relation holds, but if the assumptions are not valid the argument stepsover 2, and we arrive at an other branch.

5.3 General powers

It is known, how we define the power zn, when z is an arbitrary complexnumber and n is a nonnegative integer. This definition can be extended tonegative integers: zn = 1/zn. Fractions also can be involved via formula(1.4): z1/n = nz. But how can we define the power for arbitrary exponent?What is the value, for example, ii? In this section we deal with this question.

In real analysis we defined powers as eb log(a), since

eb log(a) =

elog(a)b

= ab.

We can use this definition in the complex case as well.

5.3.1. Definition. Let z and w be two complex number. The power of zwith exponent w is defined as

(zw)k = exp(zlogk(w)).

Here logk(z) is the logarithm function defined in (5.2). If k = 0, i.e. we usethe principal branch, then we omit the zero and write

zw = exp(zlog(w)).

5.3.2. Example. We calculate ii. Already we know that log(i) = i2

, so wehave that

ii = exp

i2 2

= exp2 = 0.2078795764 . . . .


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Exercises

5.3.3. Exercise. Determine the value log(2) and log(2 + i).

5.3.4. Exercise. Calculate ii on the branch k = 1.

5.3.5. Exercise. What is the value of the power (1)i?

5.3.6. Exercise. Describe the set {z C | ez = i}.

5.3.7. Exercise. We know that the real sin function have codomain [1, 1].However, the complex sin takes the value, for example, 2. Therefore solvethe equation

sin(z) = 2.


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6

Path integrals

If we would like to calculate the area of a shape under a function with morethan one variable, we use path integrals. Introducing the complex pathintegral which is a central notion in complex function theory we shallextensively use the notions introduced in real analysis. This also means thatwe will not prove results which are straight generalizations of the real analiticcase (like additivity of the path integral, parametrization invariance, etc.)

6.1 Curves in the complex plane

6.1.1. Definition. A function of the form : [a, b] C is called curve onthe complex plane or simply complex curve or complex path. (The domaininterval can be opened, we can exclude the point a or b as well.)

6.1.2. Example. The curve 1 : [0, 1] C,1(t) = (2 + i)t + 3(1 t)

is a straight line connecting the points 3 and 2 + i on the plane C.The curve 2 : [0, 2[ C,

2(t) = 1 + i + 3 exp(it)

is a circle with radius 1 + i and radius 3.

6.1.3. Definition. The complex curve : [a, b] C is called smooth if asa function can be differentiated continuously infinitely many times.

If is not smooth, but it is smooth except a finitely many point, we saythat it is piecewise smooth. The restrictions of the curve to the intervals onwhich it is smooth called the smooth components of

6.1.4. Example. The curves in the above example are smooth. But thecurve : [0, 2] C,

(t) =

t(1 + i) t [0, 1],(2 t)(1 + i) + (t 1)2 t ]1, 2]

43


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44 6. chapter. Path integrals

is not a smooth one. It is piecewise smooth, and it has two smooth compo-

nents.

Now we define a very important class of curves. The real variant shouldbe familiar to the reader, therefore we just declare our statements, we willnot prove them.

Let : [a, b] C is a curve on the complex plane. Let P = {t0, t1, . . . , tn}be a partition of the interval [a, b], which means that

a = t0 < t1 < < tn1 < tn = b.

Let P[a, b] be the set of all the partitions of [a, b]. We introduce the notationT(, P) =

nj=1

|(tj) (tj1)|.

6.1.5. Definition. The total variation of is the value

V() = sup{T(, P) | P P[a, b]}

(if it exists).If the total variation exists for , it is said to be of bounded variation or

rectifiable.The set of the curves with bounded variation is denoted by BV[a, b]:

BV[a, b] = { : [a, b] C | V(f) < }.

The total variation is nothing else but the arc length of the curve. Fromthe definition directly is very hard to calculate the total variation. Fortu-nately there is a very useful theorem, which helps.

6.1.6. Theorem. Let : [a, b] C be a complex curve such that

BV[a, b]. ThenV() =

ba

|(t)|dt.

6.1.7. Example. Let : [0, 2[ C be the curve which represents the unitcircle, that is, (t) = exp(it). Then

V() = 2,

as we expect. To prove this, we just calculate the integral

20

| exp(it)|dt = 20

|i exp(it)|dt = 10

1dt = 2i.


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6.2. The path integral 45

Now we calculate the arc length of the curve in Example 6.1.4. : [0, 2]

C,(t) =

t(1 + i) t [0, 1],(2 t)(1 + i) + (t 1)2 t ]1, 2]

We have that

V() =

20

|(t)|dt =10

|1 + i|dt +21

|(1)(1 + i) + 2|dt =10

2dt +

21

|1 i|dt =

2 +

21

2 = 2

2.

This result is also obvious if we sketch the graph of our function.

The orientation of a curve is a very picturesque notation.

6.1.8. Definition. If : [a, b] C is a curve, then its reverse orientationis the curve

(t) = (a + b t).It is obvious that and have the same arc length.Two more notations will come.

6.1.9. Definition. A curve : [a, b]

C is closed if

(a) = (b).

It is simple if it is injective.

6.1.10. Definition. The curve

(t) = exp(it) (t [0, 4])is closed but not simple.

6.2 The path integral

In this section we shall extensively use the notion of Riemann-Stieltjes inte-gral, which is considered to be familiar to the reader.

6.2.1. Definition. Let f : A C C be a continuous function and :[a, b] C be a complex rectifiable path such that ([a, b]) A. Then thepath integral of f with respect to the curve is denoted by

f(z)dz anddefined by the Riemann-Stieltjes integral

f(z)dz =

ba (f )(t)d(t).


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By a well known theorem of the theory of Riemann-Stieltjes integrals we

have that the path integral equals to

f(z)dz =

ba

(f )(t)(t)dt.

By utilizing the basic properties of Riemann-Stieltjes integrals, we canstate the next theorem, which collects the basic properties of path integrals.(We often omit the part (z)dz if we do not use other variables.)

6.2.2. Theorem. Let f : A C C and g : B C C be two contin-uous functions and : [a, b] C BV[a, b] a complex rectifiable path suchthat ([a, b]) A, B. Then

1.

f =

f ( C),

2.

(f + g) =

f +

g,

3.

f =

f.

Proof. The first two statements follow from the properties from the Riemann-Stieltjes integral, so we focus just on the third one.

f =

ba

(f ())(t)()(t)dt =

b

a

(f )(a + b t)(a + b t)dt =

ba

(f )(t)(t)dt =

f.

A useful approximation can be proven easily.

6.2.3. Theorem. We have the next estimation (estimacin) for path inte-grals

f(z)dz

V() supz([a,b]) |f(z)|.


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6.2. The path integral 47

Proof. The well known property of the Riemann-Stieltjes integrals give that

f(z)dz

=ba

(f )(t)(t)dt

ba

|(f )(t)||(t)|dt supz([a,b])

|f(z)|ba

|(t)|dt = supz([a,b])

|f(z)|V().

6.2.4. Definition. An analytic function F is a primitive function of f ifF = f.

Any function which has a primitive function has path integral zero, if thepath is closed. This is a special consecuence of the next theorem.

6.2.5. Theorem. If f : A C has a primitive function F, and : [a, b] C is a curve such that ([a, b]) A, then

f(z)dz = F((b)) F((a)).

Proof. We have

f(z)dz = b

a

(f )(t)(t)dt = b

a

(F )(t)(t)dt =ba

(F )(t)dt = F((b)) F((a)).

6.2.6. Example. Let f(z) = z, and (t) = exp(2it) is the closed unitcircle, where t [0, 1]. Then

zdz =1

2exp2(2i 1) 1

2exp2(2i 0) = 1

2 1

2= 0.

Note that this is true not just for f(z) = z but for any polynomial.

Exercises

6.2.7. Exercise. Calculate the next path integrals:

(z)dz, where (t) = t (t [0, 2]),

(z)dz, where (t) = it (t [0, 3]),

zdz, where (t) = exp(2it) (t [0, 1]).


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6.2.8. Exercise. Taking care on the branches of the log(z) function, let us

calculate the integral i0

log(z)dz.

(This is a path integral, where the path is the line segment [0, i].)


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7

The theorem of Cauchy

In this chapter we present the most important, central theorem of Cauchy.This theorem says, roughly speaking, that if a function f is analytic on adomain and the closed path is contained in this domain, then

f(z)dz = 0.

To prove the theorem, we first prove it for an easier closed path, a triangle.Then we shall extend this and specify the domain of f.

We follow the idea of the proof what can be found in the book of J.

Duncan (The Elements of Complex Analysis, John Wiley & Sons, 1968).

7.1 The theorem of Cauchy for triangle paths

On the complex plane we define a triangle T by three non collinear point1, 2, 3 C:

T = {11 + 22 + 33 | 1, 2, 3 0; 1 + 2 + 3 = 1}.The path running around this triangle (with the orientation 1, 2, 3) isdenoted by T.

7.1.1. Theorem. If T is a triangle in the domain D and f : D C isanalytic, then

T

f(z)dz = 0.

The assumption that T is in D implies that D is convex. Later we shallgeneralize this theorem to starlike sets and in place of triangles we can haveany simple closed paths.

Proof. Let

A =

T

f(z)dz .

49


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50 7. chapter. The theorem of Cauchy

We decompose the triangle to four smaller ones:

1 =1

2(2 + 3), 2 =

1

2(3 + 1), 3 =

1

2(1 + 2).

Let the triangles (1, 3, 2), (2, 1, 3), (3, 2, 1) and (1, 2, 3) be de-noted by T11 , T

21 , T

31 , T

41 , respectively.

Then we have thatT

f(z)dz =

[1,3]

f(z)dz+

[3,2]

f(z)dz+

[2,1]

f(z)dz+

[1,3] f(z)dz+

[3,2] f(z)dz+

[2,1] f(z)dz.

Since [j ,k]

f(z)dz = [k,j ]

f(z)dz

for all indices, we have thatT

f(z)dz =

T11

f(z)dz+

T21

f(z)dz+

T31

f(z)dz+

T41

f(z)dz.

Now we choose a triangle Tr1 from Ti1 (i = 1, 2, 3, 4) such that

A 4Tr

1

f(z)dz

.This can be done by the previous integral decomposition. This triangle willbe denoted by T1 (so we leave the upper index, because in what follows wedoes not deal with them).

This triangle T1 has sidelengths a/2, b/2, c/2. Now consider this triangleas a whole triangle. We repeat the process, and win a sequence of trianglesT1, T2, . . . such that

1. Tm+1 Tm,2. the sidelengths of Tm are 2

ma, 2mb, 2mc,

3. A 4mTm f(z)dz

.If we take the intersection of all of these triangles, we have that

n=1

Tm = {}.

That is, it contains only one point. (This is not a trivial statement. Thereis a theorem, which says that if we have a decreasing sequence of compact


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7.2. Cauchys theorem for starlike domains 51

sets such that the limit of the diameters tends to zero, then the intersection

contains one and only one point.)This point is in T. Since D is open, there is a > 0 such that B(, )

D. Since f is differentiable on D, it is differentiable in the point , so thereis a function : B(, ) C such that

f(z) f()z = f

() + (z).

This is continuous and () = 0.Hence for all > 0 there is a such that

|(z)| < for all z B(, ).It is possible choose an m such that Tm B(, ). Hence, utilizing thedefinition of ,

Tm

f(z)dz =

Tm

(f() + (z )f(z))dz+Tm

(z )(z)dz.

The first integrand has a primitive function:

f()z+

1

2f()(z )2

,

whence it follows that the first integral vanishes by Theorem 6.2.5. Using theinequality for A, and the estimation of Theorem 6.2.3. we have that

0 A 4mTm

f(z)dz

= 4mTm

(z )(z)dz

4mV(Tm) supzTm

{|z ||(z)|}

4m2m(a + b + c)2m(a + b + c) = (a + b + c)2.

The last line comes, since the two farest points in the triangle Tm is notgreater than 2m(a + b + c). Since was arbitrary, we have that A = 0, whatgives the statement.

7.2 Cauchys theorem for starlike domains

Now we go to present and prove a more general version of the theorem of

Cauchy. We have said that the theorem remains valid for any simple closedpath in a starlike domain. Now we specify what does it mean.


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7.2.1. Definition. A domain D

C is starlike if there is a point x0 in D

such that any point x in D can be connected to x0 by a line segment.

7.2.2. Example. The ball B(0, 1) is starlike in C but the annulus B(0, 2) \B(0, 1) is not.

Now we prove a theorem, from which the more general version of Cauchystheorem will immediately follow.

7.2.3. Theorem. If D is a starlike domain on the plane C, then any f A(D) has a primitive function.Proof. Let be a point from which every point can be seen (this exists, since

D is starlike). Then we define a function g as

g(z) =

[,z]

f (z D).

Let D \ {}. Since D is open, there is a h such that + h D. Wehave that

g(+ h) =

[,+h]

f (z D).

Let us choose r such that

[, + h]

B(, r)

D.

Since D is starlike, for any point in w [, +h] the line segment [, w] D.Hence the triangle given by , , +h is entirely in D. Hence Theorem 7.1.1.gives that

[,]

f +

[,+h]

f +

[+h,]

f = 0.

Hence

g(+ h) g() =[,+h]

f[,]

f(t)dt =

[,+h]

f.

After an algebraic rearrangement we take absolute value and win thatg(+ h) g()h f() =

1h[,+h]

f f()h

[,+h]

1

==

1h[,+h]

(f(z) f())dz 1h

[,+h]

|(f(z) f())| dz

1h

h supz[,+h]

|f(z) f()|.

If h tends to 0, f(z) tends to f(), because f is analytic, so differentiable,so continuous. Hence we get that g is differentiable and its derivative in

is f(). This means that g is a primitive function of f, so the theorem isproven.


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7.3. Cauchys theorem on simply connected domains 53

Now the generalized version of Cauchys theorem follows.

7.2.4. Theorem (Theorem of Cauchy). LetD be a starlike domain inCand letf A(D) be an analytic function on D. If the path : [a, b] is simpleand closed such that ([a, b]) D, then

f(z)dz = 0.

Proof. The proof now comes from Theorem 6.2.5., because f has a primitivefunction.

7.3 Cauchys theorem on simply connected do-

mains

We present a more general version of Cauchys theorem. This says that thepath integral of an analytic function vanishes if the curve is closed; not justwhen the domain of the function is starlike, but it can have a more generalshape: simply connected.

Two curves are called homotopic, if , informally, it is possible to deformone to the other with a continuous transformation. Here comes the mathe-

matically precise definition.

7.3.1. Definition. Let D C be a domain. Two paths 0, 1 : [a, b] Dare called homotopic if there is a continuous function H : [a, b] [0, 1] Dsuch that

H(t, 0) = 0(t), and H(t, 1) = 1(t).

In this case we write that 0 1.

A sample of homotopic curves can be seen on the following graphic:

(This picture is come from the WikiMedia Commons; homotopy curves.svg)Note that the homotopy is an equivalence relation.


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7.3.2. Definition. If the path 0 : [a, b]

D is homotopic to a point, it is

called null-homotopic.If in the domain D C every closed path is null-homotopic, this domain

is called simply connected.

Intuitively, a simple connectivity means that in the domain there are noholes.

7.3.3. Example. A circle, or more general, an ellipse is always null-homoto-pic on convex domains. But, for example, the circle with origin 0 and radius 1is not nullhomotopic on the domain C\{0}. For this reason the domain C\{0}is not simply connected. The same is true for the annulus B(0, 2) \ B(0, 1).

The above example leads to the next proposition.

7.3.4. Proposition. Any two closed paths 0, 1 : [a, b] D are homotopicif D is convex.

Proof. Let H : [a, b] [0, 1] D is defined by

H(t, s) = s0(t) + (1 s)1(t).

Since D is convex, all the points in the codomain ofH(which are the connect-ing line segments between two points of the curves) is in D. H is continuousand it is also holds that

H(t, 0) = 0(t), and H(t, 1) = 1(t).

So the two paths are homotopic.

7.3.5. Proposition. Every starshaped domain is simply connected. (Thusconvex domains are simply connected.)

The above two proposition shows that the homotopy is at least as stronglyconnected to the domain D as to the curves itself.

The above theorem is very important. It says that for any two homotopiccurves the path integrals of an analytic function are equal.

7.3.6. Theorem (Deformation Theorem). LetD C be a domain andf A(D). Moreover, let 0, 1 : [a, b] D are homotopic curves. Then

0

f(z)dz =

1

f(z)dz.

The corollary is the generalized version of Cauchys theorem:


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7.3. Cauchys theorem on simply connected domains 55

7.3.7. Theorem (Cauchys theorem on simply connected domains).

Let D C is a simply connected domain. Moreover, let f A(D) be ananalytic function on this domain, and let : [a, b] D be a closed path.Then

f(z)dz = 0.

Proposition 7.3.5. tells us why this is a generalization of Theorem 7.2.4.

Proof. D is a simply connecte domain, so our curve is homotopic to a pointp D. If we consider this point p as a curve p : [a, a] {p}, it is obviousthat the theorem holds by the Deformation Theorem.

If in the domain where our path runs and our function is defined, thereare holes, the theorem loses its validity. For example, the function f(z) = 1

z

is defined on C\{0}. If we take the circle (t) = exp(2it), the path integralis

1

zdz =

10

1

exp(2it)exp(2it)dt =

log(exp(2it))|t=1t=0 = 2it|t=1t=0 = 2i.And this is not zero. The reason as we said is that there is a hole in the

domain C \ {0}, that is, it is not simply connected.


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8

Cauchy Integral formula

In what follows we shall present some important consequences of Cauchystheorem. The first application, with which this chapter deals, is the CauchyIntegral Formula, which helps us to calculate complicated circle integrals(circle integral is a path integral where the path is a circle). This theoremsays that the values of a function in a ball is determined by the values of thefunction on the boundary of this ball.

8.1 A special version of the Cauchy Integral

Formula

To have shorter phrasing of our theorems, we introduce a technical definition.

8.1.1. Definition. The closed curve : [0, 1] C which is defined by

(t) = a + r exp(2it) (t [0, 1]),

describes a circle with centre a and radius r. This curve is shortly will bedenoted by

(a, r).

8.1.2. Theorem (Cauchys Integral Formula). Let f : D C is ananalytic function and let r > 0 such that B(r, a) D for some a D. Thenfor any z B(r, a)

1

2i

(a,r)

f(w)

w zdw = f(z).

Proof. Let us fix a point z B(a, r). The function

f(w)w z

56


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8.1. A special version of the Cauchy Integral Formula 57

is differentiable on D

\ {z

}, since f and 1

wz

are. Now we choose an 0 < 0 be a radius such that B(z, ) B(a, R2) \B(a, R1). Then by Cauchystheorem

f(z) =1

2i

(z,)

f(w)

w zdw =1

2i

(a,R2)

f(w)

w zdw 1

2i

(a,R1)

f(w)

w zdw.

With the same argument as in Theorem 8.3.2., we can see that

1

2i(a,R2)

f(w)

w zdw =n=0

(z

a)n

2i(a,R2)

f(w)

(w a)n+1 dw.

67


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68 9. chapter. The Laurent series

To calculate the other integral, we use the modification of (8.4):

1w z =

n=0

(w a)n(z a)n+1 (|w a| < |z a|).

Now we continue with the second integral:

12i

(a,R1)

f(w)

w zdw =

1

2i

(a,R1)

f(w)

n=0(w a)n

(z a)n+1 =

n=0

(z a)n1 12i

(a,R1)

f(w)

(w a)n =n=1

(z a)n 12i

(a,R1)

f(w)

(w a)n+1 .

Altogether,

f(z) =n=0

(z a)n2i

(a,R2)

f(w)

(w a)n+1 dw+n=1

(z a)n1

2i(a,R1)

f(w)

(w a)n+1 .It can be seen that the value of the path integrals is independent from thechoice of R1 and R2 iw we stay in the domain B(a, R). Therefore we canchoose a common r ]0, R[. For this reason let us define the coefficients anas

an =1

2i

(a,r)

f(w)

(w a)n+1 (n Z).

Then we can rewrite the expression for f(z) as

f(z) =

n=0

(z a)nan +n=1

(z a)nan =

n=an(z a)n.

The proof of the first point is done.To prove the second part, let us separate the sum

f(z) =

n=an(z a)n

to two parts:

n=

an(z a)n =n=0

an(z a)n +n=1

an(z a)n.


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9.1. The Laurent series. Main part of a function 69

The first part as a function is analytic on B(a, R), because this is the Taylor

part of the function. The second sum is convergent on C \ {a}. Thus, if wedefine the functions f1 and f2 as

f1(z) : B(a, R) C; f1(z) =n=0

an(z a)n,

f1(z) : C \ {a} C; f2(z) =n=1

an(z a)n,

then f(z) = f1(z) + f2(z) on B(a, R), as we stated.

9.1.2. Definition. The Laurent coefficients are the coefficients

an =1

2i

(a,r)

f(w)

(w a)n+1 dw (0 < r < R, n Z) (9.1)

in the Laurent expansion of the function f.

The function f2 in the theorem can be considered as the complement partof the Taylor series for a function, which is not analytic in a point just in a

punctured disk around this point. This function f2 will play an importantrole later, so we introduce the next definition.

9.1.3. Definition. The function f2 in the Laurent theorem is called themain part of the function f.

9.1.4. Example. We said at the begin of this chapter that the function1/z does not have a Taylor expansion, since it is not analytic on a discwhich contain the origin. But it is analytic on any punctured disk of theform B(0, R) \ {0}, 0 < R < . Hence it has a Laurent series expansion.Of course, this Laurent expansion coincide with 1/z, since it is already the

Laurent form of this function. The Laurent coefficients are all zero, excepta1. Indeed,

an =1

2i

(0,R)

1/w

(w 0)n+1 dw =

1

2i

(0,R)

1

wn+2dw =

1

2i

20

1

(eRit)n+2RieRitdt =

R

2

20

1

(eRit)n+1dt =

R

2

20

eRi(n+1)tdt =

R2

eRitRi(n + 1)

20

= 12i(n + 1)

eRit

20 = 12i(n + 1) (1 1) = 0.


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Of course, if n =

1 we cannot use this primitive function. Instead, we go

back to the integral:

a1 =1

2i

(0,R)

1

wdw.

This integral, by Cauchys theorem equals to 1, since f(z) 1 is analytic onour domain.

So1

z=

n=

anzn = 1 z1 = 1

z.

9.1.5. Example. To take a more advanced example, let us look for the

Laurent series for the function

f(z) =1

1 z2 .

This function is analytic on C \ {1}. Thus, it has a Taylor series in anydisk not containing 1. For example, around a = 0 it has the Taylor seriesexpansion

f(z) =n=0

z2n.

Now we look the Laurent expansion around a = 1, for example. We canseparate the expression for f(z) to two parts:

f(z) =1

1 z2 =1

2

1

1 + z+

1

2

1

1 z.

The first part has a Taylor expansion (since it is analytic around a = 1):

1

2

1

1 + z=

1

2

1

2 + (z 1) =1

4

1

1 z12

=

=

1

4

n=0

12 n

(z 1)n

.

Hence we have found an where n 0. Now we find out that the negativelyindexed ans are all zero, except a1. This is so, because the term

1

2

1

1 z = 1

2(z 1)1

is yet a Laurent series around a = 1.Altogether, we have that

f(z) =1

1 z2 =n=0

1

412

n

(z 1)n + 12 (z 1)1.


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9.1. The Laurent series. Main part of a function 71

Hence the Laurent coefficients are

an =1

4

12

n

if n = 0, 1, 2, . . . and

an = 12

if n = 1 andan = 0

for n =

1,

2, . . . .

To see how the Laurent theorem works, let us calculate the Laurent co-efficients by the formula (9.1) as well. This formula says that

an =1

2i

(1,R)

1

2

11w

(w 1)n+1 dw

for all n Z and for an radius R > 0. This radius cannot be greater than2, since otherwise we reach the point 1, where the function has an othernon-analyticity point. For the sake of simplicity, let R = 1.

1

2i(1,R)

1

2

1

1w(w 1)n+1 dw =

1

4i(1,1)

1

(w 1)n+2 dw =

14i

20

1

(1 + eit 1)n+2 ieitdt =

14

20

e(n+1)itdt =

14

e

(n+1)it

i(n + 1)

20

=1

4

1

n + 1(1 1) = 0

if n = 1. If n = 1, the integral is 2, so a1 = 12 , as we have seen before.Using the definition of the principal part, we say that it is

f2(z) =1

2

1

1 z (z C \ {1}).

Exercises

9.1.6. Exercise. Find the main part of the functions with respect to thegiven points.

f(z) = 11+z2

, a = i,

f(z) = exp z+ 1z

, a = 0.


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9.1.7. Exercise. Look for all the points in which the following functions are

not analytic:

|z|, (z) (z), 1

1+z4,

1sin(z)

.

9.1.8. Exercise. Find the Laurent expansion of the function

f(z) =1

z2 z 2in the point a = 2. What is the principal part of this function at this point?


f(z) = exp

1

z

in the point a = 0. What is the principal part of this function at this point?


f(z) = exp

z+

1

z

in the point a = 0. What is the principal part of this function at this point?


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10

Singularities and zeroes of a-

nalytic functions

10.1 Classification of singularities

The zeroes and singularities are important points of a function. In whatfollows, we deal with these specific points of functions. This theory hasimportant applications what we shall present as well.

On singular point we mean that the function is not analytic in this point.For example, the function f(z) = 1/z is singular in the point zero. If we

go close to zero from the left (z = , where > 0 a small quantity),the function tends to . If we go close to zero from the right (z = ),the function diverges to +. If we consider complex values, like z = i, thefunction tends to 1

i. We can see that the different directions give different

infinities.Now we define precisely what we mean on singularity.

10.1.1. Definition. If a function f : D C C is analytic on its domainbut not in the point D, we say that f is singular in this point. Thepoint is called singular point of f.

We note that it could be happen that a function is not analytic on adomain like B(a, R) \ B(a, r), where 0 < r < R. In this case we also coulddefine singularity. But we will not go into these details. However, consideringthese possibilities, we could distinguish our above defined singularity and wecould say that is an isolated singularity because is an isolated point inwhich our function is not analytic.

For example, the function 1/z is singular in the point = 0. To take another example,

f(z) =1

z+ 2+

1

z

3

is singular in the points = 2 and = 3.

73


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74 10. chapter. Singularities and zeroes of analytic functions

We know that if we have an analytic function f on a domain D, then

f has Taylor series expansion, or, in other words, the principal part of f isidentically zero (on the domain of analyticity D). This shows that singularityis connected to the principal part of the function. Using this observation, wedefine the order of singularity.

10.1.2. Definition. Let f : D \ {} C C be an analytic function withsingularity in . Then by the theorem of Laurent f has the Laurent seriesexpansion

f(z) =1

n=n(z )n +

n=0n(z )n =

+ 2 1(z )2 + 1

1

(z )1 + 0 + 1(z ) + 2(z )2 +

The singular point can be of three different types:

1. is a removable singularity if 12 = 3 = = 0.2. is pole of order n if n = 0 but m = 0 if m > n.3. is an essential singularity if infinitely many of n = 0.

Hence we have an important observation: a singularity can appear if theprincipal part is zero, but if and only if the singularity is removable.

10.1.3. Example. Let f(z) be defined as

f(z) =sin(z)

z.

The function is not defined in z, but the limit exists:

limz0

f(z) = 1,

as it can be easily seen by the Taylor expansion of sin (see Section 5.1):

sin(z) =n=0

(1)n(2n + 1)!

z2n+1 = z z3

6+ .

If we divide by z, we have that

f(z) =sin(z)

z= 1 z

2

6+ .

And the limit really converges to 1.


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10.1. Classification of singularities 75

The function g(z) = 1z+1

has a singularity of order 1 in the point =

1,

since this is yet a Laurent expansion

1

z+ 1= 1 1

z (1) ,

so 1 = 1 = 0 but n = 0 if n > 1.The function

h(z) =1

(z 3)3 +3

(z+ 1)2+

5

z

has three singular points: = 3 with order 3, = 1 with multiplicity 2and = 0, which is a first order singularity.

To see an example of essential singularity, we can consider the function

exp

1

z

=

n=0

1

znn!= 1 +

1n=

1

n!zn = 1 +

1

1 1

z+

1

2 1

z2+

1

6 1

z3+

The removable singularity can be caracterized as follows: the function fhas a removable singularity in the point if and only if f is bounded in thispoint. This is so, because if is removable, the principal part is zero of faround , and there is no fractions which make the series giving f divergent.

The essential singularity is the most curious one, as we shall see later. Thetheorem of Casorati and Weierstrass will discribe how a function behaves near

to its essential singularity. Now we just plot the absolute value of exp(1/z)and the real part of this function, respectively.


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10.2. The connection between zeros and poles 77

10.1.1 The definition of zeros

In this point we define the zero of a function, because it is strongly connectedto singularities. In the following sections after dealing with singularities we turn back to zeros and we shall give a deeper description of them.

10.1.4. Definition. Let f : D C C be a function. We say that thepoint is a zero or root of f if f() = 0.

We can define the order of a zero. is a zero of f of order n if

f() = 0, f() = 0, f() = 0, but f(n) = 0.

Iff is analytic on the domain D, then f has a Taylor series in this domainwith the center (since is contained in the domain D, we can have a Taylorseries expansion around this point):

f(z) =n=0

f(n)()

n!(z )n (z D).

If D is a zero of order n, the first n term of the Taylor series cancel, sowe have that

f(z) = f(n)

()n!

(z )n + f(n+1)

(n + 1)!(z )n+1 +

We can see that it is possible carry the common factor (z )n. Hence wehave the following proposition.

10.1.5. Proposition. The function f : D C C is analytic on D andhas a zero of order n if and only if then there is a functiong : D C Cwhich is analytic on D, g() = 0 and

f(z) = (z

)ng(z).

10.2 The connection between zeros and poles

The very strong connection between the zeros and poles now will be discuted.

10.2.1. Theorem. Let f : D C C be analytic on D \ {}. Then thefollowing two propositions are true

1. If f has a pole of order n, then 1f

has a zero of order n.

2. If f has a zero of order n, then 1f has a pole of order n.


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