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
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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.
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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
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A phase difference on opposite sides of the cut line.
Example:
Consider
Two branch points: z=-1 and z=1
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z  1  re
i
z  1  e
i
the phase of f(z) is (   ) / 2
Remarks:
(1) The phase at points 5 and 6 is not the same
as the phase at 2 and 3
(2) The phase at 7 exceeds that at 1 by 2p and
F(z) is therefore single-valued for the contour.
How about if a path cross with the cut line, for
example (1,2,6,7) ?
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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
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2.2 Calculus of Residues
Residue Theorem
Since

f (z) 
a
n
( z  z0 )
n
n  
if C encircles one isolated singular point z0 of f(z), we have
A set of isolated singularities can be handled by deforming our contour.
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Consider the path integral indicated in the figure. The Cauchy theorem leads to
0


f (z)  
C'
Circles

f ( z ) dz 
f ( z ) dz 
anti  parallel
lines

f ( z ) dz
C
Where C’ is the union of all the contours, and the minus sign on the
first integral is due to the clockwise direction.
C

C1
Cm
Z1
Zm
Z2

f ( z ) dz  2 p ia 1 ( z i )
Ci
Residue theorem:
C2
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The problem of evaluating one or more contour
integrals is replaced by the algebraic problem of
computing residues at the enclosed singular points.
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Example:
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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,
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Infinite semicircle
C x0

f ( z ) dz 

x0 
f ( x ) dx 


 2 p i  enclosed
Cx

Cx
f ( z ) da 
f ( z ) da 


Cx
0

x0 
f ( x ) dx 

f ( z ) dz
infinite
semicircle
residues
i
Cx
0
0
i
, set z  z 0   e , dz  i  e d 
On the semicircle

f ( z ) dz 
Cx

a 1
0
da ia 1 
z  x0
p
a 1
0
2p
d   i p a 1 if counter
- clockwise
0
da ia 1  d    i p a 1 if clockwise
z  x0
p
In both cases, we have (Cauchy principle value)
P


f ( x ) dx  
x0 

f ( x ) dx 


x0 
f ( x ) dx  i p a 1  2 p i  other enclosed
residues12
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
2p
I 
 f sin  , cos  d 
0
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From this
Our integral becomes a contour integral of the unit circle
Example
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The denominator has roots
Suppose that our definite integral has the above form and
satisfies the two conditions:
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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
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Example

I 


dx
1 x
2
Since f(z)  1/(1  z )  1/z as | z |  , we have
2
2
I  2 p i  residues
(upper half - plane)
1
Where are the poles ?
z 1
2
f ( z ) have two simple

1

1
zi zi
poles at z   i, and z  i in the upper half plane
So Res f(z  i)  (z - i)f(z) | z  i  1 /( 2 i )
I  2 p i Res f(z  i)  p
Consider the above definite integral with a real and positive (This is a
Fourier transform). We assume the two conditions:
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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).
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Jordan’s Lemma
Consider
a function
f (z)  e
of the form
iaz
g ( z ) where
1. if a  0 , | g ( z ) | 0 as R  
2. if a  0, | g ( z ) | 0 faster tha n 1/z as R  
then Jordan' s Lemma
lim 
R
C1
states that
f ( z ) dz  0
Applicatio
n of Jordan' s Lemma


C
f(z)dz 

C1
f ( z ) dz 

f ( z ) dz 
C2
By using the residue theorem,

f ( x ) dx

we have


f ( x ) dx  2 p i  Res[f(z)
in the UHP]

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iz

e dz
z
e dx

x
 P
C1
ix

iz


C1
e dz
z
C2
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(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
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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
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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
( a  1)! (  a )! 
ax

e

1 e

x
dx 
p
sin( p a )
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.
Attempt to do the following exercises!

(1) Show that
2
 (x
x dx
2
-

( 2 ) Show that P

 1)( x  4 )
2
itx
sin k | t |
2
e dx
x

p
2
k
2
 p
36
.
k
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2.3 Dispersion Relation
The name dispersion comes from optical dispersion.
Dispersion relation describe the ways that the
wave propagation varies with the wavelength or
frequency of a wave.
The index of refraction n
n  n r  in i
n r : the phase velocity
ni :
absorption
loss when the
wave propagates
through t
he material
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An important result: (Kronig and Kramers in 1926-1927 )
Re[ n  1] ~
2

f [Im[ n  1]]
2
Im [ n  1] ~
2

f [Re[ n  1]]
2
Generalizing this, any pair of equations giving the integral relation between
real part and its imaginary part
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,
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Splitting the above equation into real and imaginary parts
yields
These are the dispersion relations ( Kronig and Kramers
relations).
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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
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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
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