1

Functions of A Complex Variables

Functions of a complex variable provide us some powerful and widely useful tools in in theoretical physics.

• Some important physical quantities are complex variables (the wave-function ) • Evaluating definite integrals.

• Obtaining asymptotic solutions of differentials equations.• Integral transforms

2

Complex Algebra

We here go through the complex algebra briefly.A complex number z = (x,y) = x + iy, Where. We will see that the ordering of two real numbers (x,y) is significant, i.e. in general x + iy y + ix

X: the real part, labeled by Re(z); y: the imaginary part, labeled by Im(z)

Three frequently used representations:

(1) Cartesian representation: x+iy

(2) polar representation, we may write z=r(cos + i sin) or

r – the modulus or magnitude of z - the argument or phase of z

1i

ierz

3

r – the modulus or magnitude of z - the argument or phase of z

The relation between Cartesian

and polar representation:

The choice of polar representation or Cartesian representation is a

matter of convenience. Addition and subtraction of complex variables

are easier in the Cartesian representation. Multiplication, division,

powers, roots are easier to handle in polar form,

1/ 22 2

1tan /

r z x y

y x

212121

ierrzz

212121 // ierrzz innn erz

)()(

)()(

2221212121

212121

yxyxiyyxxzz

yyixxzz

4

From z, complex functions f(z) may be constructed. They can be written

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

in which v and u are real functions.

For example if , we have

The relationship between z and f(z) is best pictured as a mapping operation, we address it in detail later.

2121 zzzz

xyiyxzf 222

Using the polar form,

)arg()arg()arg( 2121 zzzz

2)( zzf

5

Function: Mapping operation

x

y Z-plane

u

v

The function w(x,y)=u(x,y)+iv(x,y) maps points in the xy plane into pointsin the uv plane.

nin

i

ie

ie

)sin(cos

sincos

We get a not so obvious formula

Since

ninin )sin(cossincos

6

Complex Conjugation: replacing i by –i, which is denoted by (*),

We then have

Hence

Note:

ln z is a multi-valued function. To avoid ambiguity, we usually set n=0

and limit the phase to an interval of length of 2. The value of lnz with

n=0 is called the principal value of lnz.

iyxz *

222* ryxzz

21*zzz Special features: single-valued function of a real variable ---- multi-valued function

irez

nire 2

irlnzln

nirz 2lnln

7

Another possibility

even and 1|cos||,sin|possibly however,

x;real afor 1|cos||,sin|

zz

xx

Question:

yx

yx

yxiyxiyx

yxiyxiyxi

ee iziz

222

222

iziz

sinhcos|cosz |

sinhsin|sinz | (b)

sinhsincoshcos)cos(

sinhcoscoshsin)sin( (a) show to

2

esinz ;

2

ecosz

:identities theUsing

8

Cauchy – Riemann conditions

Having established complex functions, we now proceed to

differentiate them. The derivative of f(z), like that of a real function, is

defined by

provided that the limit is independent of the particular approach to the

point z. For real variable, we require that

Now, with z (or zo) some point in a plane, our requirement that the

limit be independent of the direction of approach is very restrictive.

Consider

zfdz

df

z

zf

z

zfzzf

zz

00limlim

oxxxx

xfxfxfoo

limlim

yixz viuf

,

yix

viu

z

f

9

Let us take limit by the two different approaches as in the figure. First,

with y = 0, we let x0,

Assuming the partial derivatives exist. For a second approach, we set

x = 0 and then let y 0. This leads to

If we have a derivative, the above two results must be identical. So,

x

vi

x

u

z

f

xz

00limlim

x

vi

x

u

y

v

y

ui

z

f

z

0lim

y

v

x

u

,

x

v

y

u

10

These are the famous Cauchy-Riemann conditions. These Cauchy-

Riemann conditions are necessary for the existence of a derivative, that

is, if exists, the C-R conditions must hold.

Conversely, if the C-R conditions are satisfied and the partial

derivatives of u(x,y) and v(x,y) are continuous, exists. (see the proof

in the text book).

xf

zf

11

Analytic functions

If f(z) is differentiable at and in some small region around ,we say that f(z) is analytic at

Differentiable: Cauthy-Riemann conditions are satisfied the partial derivatives of u and v are continuous

Analytic function:

Property 1:

Property 2: established a relation between u and v

022 vu

Example:

xeyxvb

xyxyxua

yxivyxuzw

y sin),( )(

3),( )( if

),(),()( functions analytic theFind23

0zz

0zz 0z

12

Cauchy’s integral Theorem

We now turn to integration. in close analogy to the integral of a real function The contour is divided into n intervals .Let wiith for j. Then 01 jjj zzz

0

01

lim

z

z

n

jjj

ndzzfzf

'00 zz n

The right-hand side of the above equation is called the contour (path) integral of f(z)

. and

bewteen curve on thepoint a is where

, and points thechoosing

of details theoft independen

is and existslimit that theprovided

1

j

j

jj

j

zz

z

13

As an alternative, the contour may be defined by

with the path C specified. This reduces the complex integral to the

complex sum of real integrals. It’s somewhat analogous to the case of

the vector integral.

An important example

22

11

2

1

,,

yx

yxc

z

zc

idydxyxivyxudzzf

22

11

22

11

yx

yx

yx

yxcc

udyvdxivdyudx

c

ndzz

where C is a circle of radius r>0 around the origin z=0 in the direction of counterclockwise.

14

In polar coordinates, we parameterize

and , and have

which is independent of r.

Cauchy’s integral theorem– If a function f(z) is analytical (therefore single-valued) [and its

partial derivatives are continuous] through some simply connected region R, for every closed path C in R,

irez diredz i

2

0

1

1exp22

1dni

rdzz

i

n

c

n

1- n for 1

-1n for 0{

0 dzzf

c

15

Stokes’s theorem proof

Proof: (under relatively restrictive condition: the partial derivative of u, v

are continuous, which are actually not required but usually

satisfied in physical problems)

These two line integrals can be converted to surface integrals by

Stokes’s theorem

Using and

We have

c c c

udyvdxivdyudxdzzf

c s

sdAldA

yAxAA yx zdxdyds

c c s

yx sdAldAdyAdxA

dxdy

y

A

x

Axy

16

For the real part, If we let u = Ax, and v = -Ay, then

=0 [since C-R conditions ]

For the imaginary part, setting u = Ay and v = Ax, we have

As for a proof without using the continuity condition, see the text book.

The consequence of the theorem is that for analytic functions the line

integral is a function only of its end points, independent of the path of

integration,

dxdyy

u

x

vvdyudx

c

y

u

x

v

0dxdyy

v

x

uudyvdx

0dzzf

1

2

2

1

12

z

z

z

z

dzzfzFzFdzzf

17

•Multiply connected regions

The original statement of our theorem demanded a simply connected

region. This restriction may easily be relaxed by the creation of a

barrier, a contour line. Consider the multiply connected region of

Fig.1.6 In which f(z) is not defined for the interior R

Cauchy’s integral theorem is not valid for the contour C, but we can

construct a C for which the theorem holds. If line segments DE and

GA arbitrarily close together, then

1.6 Fig.

E

D

A

G

dzzfdzzf

18

dzzfdzzf

EFGGADEABDABDEFGAC

0dzzfEFGABD

21 CC

dzzfdzzf

'2

'1 CEFGCABD

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Cauchy’s Integral Formula

Cauchy’s integral formula: If f(z) is analytic on and within a closed contour C then

in which z0 is some point in the interior region bounded by C. Note that

here z-z0 0 and the integral is well defined.

Although f(z) is assumed analytic, the integrand (f(z)/z-z0) is not

analytic at z=z0 unless f(z0)=0. If the contour is deformed as in Fig.1.8

Cauchy’s integral theorem applies.

So we have

00

2 zifzz

dzzf

C

C C

dzzz

zf

zz

dzzf

2

000

20

Let , here r is small and will eventually be made to

approach zero

(r0)

Here is a remarkable result. The value of an analytic function is given at

an interior point at z=z0 once the values on the boundary C are

specified.

What happens if z0 is exterior to C?

In this case the entire integral is analytic on and within C, so the

integral vanishes.

i0 rezz

drie

re

rezfdz

zz

dzzf i

C Ci

i

2 2

0

0

00 2

2

zifdzif

C

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Derivatives

Cauchy’s integral formula may be used to obtain an expression for

the derivation of f(z)

Moreover, for the n-th order of derivative

C

zf

zz

dzzf

i exterior z ,0

interiorz ,

2

1

0

00

0

0

0 0

1

2

f z dzdf z

dz i z z

2

000 2

11

2

1

zz

dzzf

izzdz

ddzzf

i

1

00 2

!n

n

zz

dzzf

i

nzf

22

We now see that, the requirement that f(z) be analytic not only guarantees a first derivative but derivatives of all orders as well! The derivatives of f(z) are automatically analytic. Here, it is worth to indicate that the converse of Cauchy’s integral theorem holds as well

book). text the(see R throught analytic

is f(z) then R, within C closedevery for 0)( and

Rregion connectedsimply ain continuous is f(z)function a If

C dzzf

Morera’s theorem:

23

. find origin, about the circle a

within andon analytic is a)( If 1.

Examples

0n

n

n

n

a

zzf

jn

jnnj

j zaajzf

1

!

jj ajf !0

12

1

!

0n

n

nz

dzzf

in

fa

24

2.In the above case, on a circle of radius r about the origin,

then (Cauchy’s inequality)

Proof:

where

3. Liouville’s theorem: If f(z) is analytic and bounded in the complex

plane, it is a constant.

Proof: For any z0, construct a circle of radius R around z0,

Mzf

Mra nn

nn

rznn

r

M

r

rrM

z

dzzfa

11 2

2

2

1

rfMaxrM rz

22

00

2

22

1

R

RM

zz

dzzf

izf

R

R

M

25

Since R is arbitrary, let , we have

Conversely, the slightest deviation of an analytic function from a

constant value implies that there must be at least one singularity

somewhere in the infinite complex plane. Apart from the trivial constant

functions, then, singularities are a fact of life, and we must learn to live

with them, and to use them further.

R

.const)z(f,e.i,0zf

26

Laurent Expansion

Taylor Expansion

Suppose we are trying to expand f(z) about z=z0, i.e.,

and we have z=z1 as the nearest point for which f(z) is not analytic. We

construct a circle C centered at z=z0 with radius

From the Cauchy integral formula,

0n

n0n zzazf

010 zzzz

C 00C zzzz

zdzf

i2

1

zz

zdzf

i2

1zf

C 000 zzzz1zz

zdzf

i2

1

27

Here z is a point on C and z is any point interior to C. For |t| <1, we

note the identity

So we may write

which is our desired Taylor expansion, just as for real variable power

series, this expansion is unique for a given z0.

0

211

1

n

ntttt

C nn

n

zz

zdzfzz

izf

01

0

0

2

1

01

002

1

n Cn

n

zz

zdzfzz

i

0

00 !

n

nn

n

zfzz

28

Schwarz reflection principle

From the binomial expansion of for integer n (as anassignment), it is easy to see, for real x0

Schwarz reflection principle:

If a function f(z) is (1) analytic over some region including the real axis

and (2) real when z is real, then

We expand f(z) about some point (nonsingular) point x0 on the real axis

because f(z) is analytic at z=x0.

Since f(z) is real when z is real, the n-th derivate must be real.

n0xzzg

*n

0**n

0* zgxzxzzg

** zfzf

0

00 !

n

nn

n

xfxzzf

*

0

00

**

!zf

n

xfxzzf

n

nn

29

Laurent Series

We frequently encounter functions that are analytic in annular region

30

Drawing an imaginary contour line to convert our region into a simply

connected region, we apply Cauchy’s integral formula for C2 and C1,

with radii r2 and r1, and obtain

We let r2 r and r1 R, so for C1, while for C2, .

We expand two denominators as we did before

(Laurent Series)

zz

zdzf

izf

CC

21

2

1

00 zzzz 00 zzzz

21000000 112

1

CCzzzzzz

zdzf

zzzzzz

zdzf

izf

zdzfzzzzizz

zdzfzz

i

n

n Cn

n Cn

n

001

001

00

21

1

2

1

2

1

n

nn zzazf 0

31

where

Here C may be any contour with the annular region r < |z-z0| < R encircling z0 once in a counterclockwise sense.

Laurent Series need not to come from evaluation of contour integrals. Other techniques such as ordinary series expansion may provide the coefficients.

Numerous examples of Laurent series appear in the next chapter.

Cnn

zz

zdzf

ia

10

2

1

32

01,22

2

1

mmni

i

1- n for 0

-1nfor 1a n

1

3211

1

1

n

nzzzzzzz

The Laurent expansion becomes

0

222

1

mmnimn

i

ner

drie

ia

021 2

1

1

1

2

1

mn

mnn

z

zdz

izz

zd

zia

11zzzf

Example: (1) Find Taylor expansion ln(1+z) at point z

(2) find Laurent series of the function

If we employ the polar form

1

1)1()1ln(n

nn

n

zz

33

For example

which has a simple pole at z = -1 and is analytic

elsewhere. For |z| < 1, the geometric series expansion f1, while expanding it about z=i leads to f2,

0

20

1 1

1 ;)( ;

1

1)(

n

n

n

n

iz

iz

ifzf

zzf

z11zf

Analytic continuation

34

Suppose we expand it about z = i, so that

converges for (Fig.1.10)

The above three equations are different representations of the same

function. Each representation has its own domain of convergence.

0

211

1

n

nzzzz

i1iz1i1

1

izi1

1zf

2

2

i1

iz

i1

iz1

i1

1

2i1iz

A beautiful theory:

If two analytic functions coincide in any region, such as the overlap of s1 and s2,of coincide on any line segment, they are the same function in the sense that theywill coincide everywhere as long as they are well-defined.