1

Chp. 2. Functions of A Complex Variable II

2.1 Singularities

Isolated singular point:

We define z0 as an isolated singular point of the function f(z) if it is not analytic at z = z0 but is analytic at neighboring points.

Poles

In the Laurent expansion

A pole of order m: If an = 0 for n < -m < 0 and a-m 0, we say that z0 is a pole of order m. A simple pole: if we have a pole of order one, i.e., m = 1, often called a simple pole. Essential singularity: the summation continues to n = - , the z0 is a pole of infinite order

2

One point of fundamental difference between a pole of finite order and an essential singularity:

a pole of order m can be removed by multiplying f(z) by This obviously cannot be done for an essential singularity.

The behavior of f(z) as z is defined in terms of the behavior of f(1/t) as t 0. Consider the function

As z , we replace the z by 1/t to obtain

Clearly, from the definition, sinz has an essential singularity at . This result could be expected from the following analysis.

3

Branch points (optional reading) A branch point may be informally thought of as a point Z0 at which a

multi-valued function changes values when one winds once around z0

Consider

4

A phase difference on opposite sides of the cut line.

Example:

Consider

Two branch points: z=-1 and z=1

5

2/)( is f(z) of phase the1 1 ii ezrez

Remarks:(1) The phase at points 5 and 6 is not the sameas the phase at 2 and 3

(2) The phase at 7 exceeds that at 1 by 2andF(z) is therefore single-valued for the contour.

How about if a path cross with the cut line, forexample (1,2,6,7) ?

6

Generalizing from this example, for a function

the phase is the algebraic sum of the phase of its individual factors:

The phase of an individual factor may be taken as the arctangent of the

ratio of its imaginary part to its real part,

For the case of a factor of the

Z0

7

2.2 Calculus of Residues

Residue Theorem

Since

if C encircles one isolated singular point z0 of f(z), we have

n

nn zzazf )()( 0

A set of isolated singularities can be handled by deforming our contour.

8

Zm

Z2

Z1

C1

C2

Cm

C

Consider the path integral indicated in the figure. The Cauchy theorem leads to

C

CCircles

linesparallelanti

dzzfdzzfdzzfzf )()()()(0'

Where C’ is the union of all the contours, and the minus sign on the first integral is due to the clockwise direction.

iC

iziadzzf )(2)( 1

Residue theorem:

9

The problem of evaluating one or more contour integrals is replaced by the algebraic problem of computing residues at the enclosed singular points.

10

Example:

11

Cauchy Principal Value

Occasionally an isolated first-order pole will be directly on the contour of integration. In this case we may deform the contour to include or exclude the residue as desired by including a semicircular detour of infinitesimal radius,

12

residues enclosed2

)()()()()(

semicircleinfinite

00

0

i

dzzfdxxfdzzfdxxfdzzfxC

x

x

clockwise if )(

clockwise-counter if )(

0

110

1

2

110

1

0 0

0 0

aidiadaxz

adazf

aidiadaxz

adazf

x x

x x

C C

C C

deidzezz ii ,set ,semicircle On the 0

residues enclosedother 2)()()( 1

0

0

iaidxxfdxxfdxxfPx

x

In both cases, we have (Cauchy principle value)

0xC

Infinite semicircle

13

Evaluation of Definite Integrals

Definite integrals appears frequently in problems of mathematical

physics as well as in pure mathematics. We here introduce several

techniques to evaluate them.、

We consider integrals of the form

2

0

cos,sin dfI

14

From this

Our integral becomes a contour integral of the unit circle

Example

15

The denominator has roots

Suppose that our definite integral has the above form and

satisfies the two conditions:

16

With these conditions, we may take as a path integral along the real

axis and a semicircle in the upper half-plane as shown in the Fig. We

let the radius R of the semicircle become infinitely large. Then

From the 2nd condition, the 2nd integral vanishes and

-R R

17

21 x

dxI

Consider the above definite integral with a real and positive (This is a Fourier transform). We assume the two conditions:

Example

plane)-half(upper residues2 iI

izizz

11

1

12

have we,|z| as 1/z )z1/(1f(z) Since 22

Where are the poles ?

i)f(z Res i2I

)2/(1|i)f(z)-(zi)f(z Res So

plane halfupper in the iz and i,zat poles simple twohave )(

iz i

zf

18

We employ the path shown in Fig2.5. The application of the calculus

of it is the same as the one just considered, but here it is a little harder

to show that the integral over the (infinite) semicircle goes to zero

(please see the text book, it is called Jordan's lemma).

19

Jordan’s Lemma

1

0)(

thatstates Lemma sJordan'then

R as 1/zn faster tha 0|)(| 0,a if 2.

R as 0|)(| ,0 if 1.

where)()( form theoffunction aConsider

limR C

iaz

dzzf

zg

zga

zgezf

UHP]in the Res[f(z)i2)(

have we theorem,residue theusingBy

)()()(

Lemma sJordan' ofn Applicatio

21

dxxf

dxxfdzzfdzzff(z)dzCCC

20

21

C2C1

1C

izixiz

z

dze

x

dxeP

z

dze

22

(4) Exponential form:

With exponential or hyperbolic functions present in the integrand, life

gets somewhat more complicated than before. Instead of a general

overall prescription, the contour must be chosen to fit the specific

integral.

As an example, we consider an integral that will be quite useful in

developing a relation between z! and (-z)!

Example Factorial Function

We wish to evaluate

23

The limits on a are necessary (and sufficient) to prevent the integral

from diverging as x . This integral may be handled by integrate

around the contour shown in Fig.2.7

24

25

Using the beta function (we shall study it later), we can show that the

integral to be (a-1)!(-a)!. This results in the interesting and useful factorial function relation

Although the integral result holds for real a, 0 < a < 1, the above

equation may be extended by analytical continuation to all values of a,

real and complex, excluding only real integer values.

)sin(1)!()!1(

adx

e

eaa

x

ax

Attempt to do the following exercises!

.||sin

that Show )2(

36)4)(1(

x that Show )1(

22

-222

2

k

tk

kx

dxeP

xx

dx

itx

26

2.3 Dispersion Relation

The name dispersion comes from optical dispersion.

Dispersion relation describe the ways that thewave propagation varies with the wavelength orfrequency of a wave.

The index of refraction n

ir innn

material he through tpropagates wave when theloss absorption :n

velocityphase the:

i

rn

27

We consider f(z) that is analytic in the upper half plane and on the real axis. We also require that

in order that the integral over an infinite semicircle will vanish. By the Cauchy integral formula,

An important result: (Kronig and Kramers in 1926-1927 )

]]1[Re[~]1[Im ]]1[Im[~]1Re[ 2222 nfnnfn

Generalizing this, any pair of equations giving the integral relation between real part and its imaginary part

28

29

Splitting the above equation into real and imaginary parts yields

These are the dispersion relations ( Kronig and Kramers relations).

30

Symmetry Relations

Suppose f(-x) = f*(x)

Then u(-x) +i v(-x) = u(x) – i v(x)

The real part is even and the imaginary part is odd. In quantum

mechanical scattering problems these relations are called crossing

conditions.

To exploit the crossing condition, we rewrite

Letting x -x in the first integral on the RHS and substituting v(-x) = -

v(x), we obtain

31

Similarly,

Note: the present case is of considerable physical importance because

the variable x might represent a frequency and only zero and positive

frequencies are available for physical measurements.

Optical Dispersion

From Maxwell's equations and Ohm's law, one has

32