1
2. Functions of A Complex Variable II
2.1 Singularities
We have known that the Laurent expansion represents a generalization of the Taylor
series in the presence of singularities. 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
n 
f(z) =

an (z - z0)n,
n  
If an = 0 for n < -m < 0 and a-m  0, we say that z0 is a pole of order m. For instance, if m
= 1, that is a-1/(z-z0) is the first non-vanishing term, we have a pole of order one, often
called a simple pole.
If, on the other hand, the summation continues to n = - , the z0 is a pole of infinite
order and is called an essential singularity. One point of fundamental difference between
a pole of finite order and an essential singularity is that a pole of order m can be removed
by multiplying f(z) by (z-z0)m. 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

 1n z 2n1
sin z  
2n  1!
n 0

As z  , we replace the z by 1/t to obtain
sin 1 t  
 1n
 2n  1!t 2n1
n 0

Clearly, from the definition, sin z has an essential singularity at . This result could be
expected from the following analysis.


When x = 0, sin z = sin iy = e i iy  e i iy 2i  = i sinh y,
which approaches  exponentially as y  . Thus, the absolute value of sin z is not
bounded.
2
• Branch points
There is another sort of singularity. Consider
f(z) = z a,
in which a is not an integer, As z moves around the unit circle from e 0i to e 2i,
f(z)  e 2ai  e 0i.
We have a branch point at the origin and another at infinity. The points e
the z -plane coincide but they lead to different values of f(z).
0i
and e 2i in
The problem is resolved by constructing a cut line joining both branch points so that f(z)
will be uniquely specified for a given point in the z-plane.
Note that a function with a branch point and a required cut line will not be continuous
across the cut line. In general, there will be a phase difference on opposite sides of this
cut line. Hence line integrals on opposite sides of this cut line will not generally cancel
each other.
The contour line used to convert a multiply connected region into a simply connected
region (Section 1.3) is completely different. The function is continuous across this
contour line, and no phase difference exists.
Example:
Consider
f(z) = (z2 - 1) 1/2 = (z + 1)1/2(z - 1)1/2
The first factor on the right hand side(RHS), (z+1)1/2, has a branch point at z = -1. The
second factor has one at z = 1. At infinity f(z) has a simple pole. The cut line has to
connect both branch points. To check on the possibility of taking the line segment
jointing z = +1 and -1 as a cut line, let us follow the phases of these two factors as we
move along the contour shown in Fig2.1.
Fig.2.1
For convenience, let z + 1 =r e i and z - 1 =  e i. Then the phase of f(z) is ()/2. At
point 1,  = 0; 1 to 2, , but  - unchanged; then  stays constant until 6; 6 to 7,
3
= .  increases by 2 as we move from 3 to 5. The phase of f(z) is tabulated in
the final column of table2.1.
Table2.1 Phase Angle
(1). The phase at points 5 and 6 is not the same as the phase at 2 and 3. This behavior can
be expected at a branch point cut line.
(2). The phase at 7 exceeds that at 1 by 2 and f(z) = (z -1)1/2 is therefore single-valued
for the contour shown, encircling both branch points.
If we take the x-axis -1 ≤ x ≤ 1 as a cut line, f(z) is uniquely specified. Alternatively, the
positive x-axis x > 1 and the negative x -axis for x < 1 may be taken as cut lines. The
branch points cannot be encircled and the function remains single-valued.
Generalizing from this example, for a function
f(z) = f1(z)∙f2(z)∙f3(z) ...,
the phase is the algebraic sum of the phase of its individual factors:
arg f(z) = arg f1(z) + arg f2(z) + arg f3(z) + ...
The phase of an individual factor may be taken as the arctangent of the ratio of its
imaginary part to its real part,
arg fi(z) = tan-1 (vi/ui).
For the case of a factor of the form
fi(z) = (z - z0i)
The phase corresponds to the phase angle of a two-dimensional vector from +z0 to z, the
phase increasing by 2 as the point z0 is encircled. Conversely, the traversal of any closed
loop not encircling z0 does not change the phase of z-z0.
4
2.2 Calculus of Residues
• Residue Theorem
Since
0
n  1
 2i
n  -1

n


z

z
dz


0

C
if C encircles one isolated singular point z0 of f(z), we have
1
2i

C

  z  z0 n dz  a1
1
f z dz 
an
2i n
C
The a-1, the coefficient of (z-z0)-1 in the Laurent expansion, is called the residue of f(z) at
z = z 0.
A set of isolated singularities can be handled by deforming our contour as in Fig.2.2
Fig.2.2
Cauchy's integral theorem leads to
 f z dz   f z dz   f z dz   f z dz    0
C
C0
C1
C2
The circular integral around any given singular point is given by
 f z dz  2ia 1 zi 
Ci
assuming a Laurent expansion about the singular point zi.
So
 f z dz  2ia1z0  a1z1  a1z 2  
C
5
= 2i (sum of enclosed residues)
This is the residue theorem. The problem of evaluating one or more contour integrals is
replaced by the algebraic problem of computing residues at the enclosed singular points.
How to calculate the residue of f(z0) ?
If z0 is a simple pole, we can write
f(z) = a-1/(z-z0) + a0 + a1(z-z0) + ... +
a-1 = lim (z-z0) f(z)
If z0 is a pole of order m, then
a 1 


1
d m1
 z  z 0  m f  z  z  z0
m

1
m  1! dz
Example
Find Res of f(z) = 1 / (z2+1) at z =i.
1

z 1
2

1 1
z i z i
z = i is a simple pole
a-1 = lim (z - i) f(z) = 1/(2i) .
• 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, (Fig. 2.3)
Fig.2.3
On the semicircle, set z - x0 =  e i, dz = i e id,

2

dz
 i d  i 
z  z0

i.e., i a-1
if counterclockwise
6

0

dz
 i d  i
z  z0
i.e., -i a-1
if clockwise.

This contribution, + or -, appears on the LHS of the equation of Residue theorem. If our
detour were clockwise, the residue would not be enclosed and there would be no
corresponding term on the RHS of the equation. However, if our detour were
counterclockwise, this residue would be enclosed by the contour C and a term 2ia-1
would appear on the RHS. The net result for either clockwise or counterclockwise detour
is that a simple pole on the contour is counted as one half what it would be if it were
within the contour. This corresponds to the Cauchy principal value.
For example, let us suppose that f(z) with a simple pole at z = x0 is integrated over the
entire real axis. (Fig. 2.4)
Fig. 2.4
Then
x0 

 f z dz   f xdx   f z dz   f xdx 

C x0
x 0 

Cinfinite semicircle
=2 Res f(z)
If Cx0 include x0, its contribution appears twice as a-1 in

of the LHS and as 2ia-1
Cx0
in the RHS --- for a net contribution of ia-1. If Cx0 does not include Cx0, the only
contribution is from the clockwise integration over Cx0 which yields -ia-1; moving this
to the RHS, we have ia-1 as before.
We therefore define the Cauchy principal value as


 x0 



lim  f  x dx 
f  x dx   P f  x dx
 0 

x0 

 



which presents the proceeding limiting process.
7
Some time, the same limiting technique is applied to the integration limits . We may
define
P

a

a
f  x dx
 f x dx  alim
 
• 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.
2
(1) Form of
 f sin  , cos d
0
We consider integrals of the form
2
I
 f sin  , cos  d
0
where f is finite for all values of . We also require f to be a rational function of sin and
cos so that it will be single-valued. Let
z = e i, dz = i e i d
From this
d = -i dz/z, sin  
z  z 1
z  z 1
, cos  
2i
2
Our integral becomes a contour integral of the unit circle


 z  z 1 z  z 1  dz

I  i f 
,
 2
 z
2


  i 2i Re s
Example

8
2
I
d
 1   cos ,
||<1
0


I  i
dz
unit
circle
 i
2




 

z 1  z  z 1 
 2

dz
2
z  2 z 1
 
The denominator has roots
z  
1


1

1   2 and
z  
1


1

1  2
z+ is within the unit circle; z- is outside. Then
2
1
4
1
I  i 2i lim  z  z  

z  z
z  z  z  z    z   z 

2

1  2

(2) Form of
 f x dx

Suppose that our definite integral has the above form and satisfies the two conditions:
a. f(x) is analytic in the upper half-plane except for a finite number of poles. (If poles on
the real axis, they may be included or excluded as discussed earlier).
b. f(z) vanishes as strongly as 1/z2 for |z|  , 0 < argz < . ( We could use f(z) vanishes
faster than 1/z, but we wish to have f(z) single-valued.)
With these conditions, we may take as a contour of integral the real axis and a semicircle
in the upper half-plane as shown in Fig 2.5. We let the radius R of the semicircle become
infinitely large. Then
R

R
0
i
i
 f z dz  Rlim  f x dx  Rlim  f Re i Re d
= 2 i  residues ( upper half-plane)
9
From the 2nd condition, the 2nd integral vanishes and

 f x dx = 2i  residues (upper half -plane)

Example

I
 1 x2
dx

Since f(z) = 1/(1+z2 ) have two simple poles at z = i, and z = i in the upper half plane,
So, Res f( z=i) = (z-i)f(z)|z=i = 1/(2i)
Consider that f(z)  1/z2 as |z|  , we can use the above formula to obtain
I = 2i Res f(z=i) = .

 f xe
(3) Form of
iax
dx

Consider the above definite integral with a real and positive (This is a Fourier
transform). We assume the two conditions:
a. f(z) is analytic in the upper half-plane except for a finite number of poles.
b. lim f z   0 ,
0  arg z  .
z 
We employ the contour 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 the integral over the
(infinite) semicircle(IR) goes to zero (please see the text book, it is called Jordan's
lemma).



 f z e
 f xe
iax
iaz
dz
dx  lim I R

R 
=2i  residues
Fig. 2.5
10
Since IR  0 as R   ( Jordan's lemma),

 f xe
iax
dx = 2i  residues (upper half -plane)

Example 1: Singularity on contour of integration

I

0
sin x
dx
x
This may be taken as the imaginary part of
Iz  P

i

 ix




e dx
P
x

cos x  i sin x
dx
x
sin x
dx
x
(

cos x
is odd function of x)
x
For f(z) = eiz/z, the only pole is a simple pole at z = 0 and the residue here is a-1 = 1. We
choose the contour shown in Fig.2.6 (1) to avoid the pole, (2) to include the real axis, (3)
to yield a vanishingly small integrand for z = iy, y  . Note that in this case a large
(infinite) semicircle in the lower half-plane would be disastrous. We have

e iz dz

z
 r ix

R
e dx

x
By Jordan's lemma,
R
e iz dz
e ix dx
e iz dz


0
z
x
z

C1

C2


P





0


r
C2
e iz dz
0
z
e ix dx
e iz dz

 i
x
z
sin x
1
dx 
x
2

C1


sin x

dx 
x
2

Fig. 2.6
11
(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

I
e ax
 1  e x dx ,
0<a<1

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
Fig. 2.7
If we let R  , the real axis leads to the integral we want. The return path along y = 2
is chosen to leave the denominate of the integral invariant, at the same time introducing a
constant factor ei2a in the numerator. We have

R
 R e ax
e ax

i 2a
dz

lim

e
R  1  e x
1 ez
1 ex
R
 R
e az




f  z dz

 vertical

sec tion

 1 e
i 2a



e ax
1 ex
Now where are the poles and what are the residues? We have a pole when ez = -1, i.e.,
z = i in the contour. Since


z  i z  i 2
1  e z  1  e z i  z  i 1 

 


2!
3!


12

Res f z   lim z  i 
z i


e ax
 1 ex




,
sin a 
 2ie ia
1  e i 2a

e az
 e ia
z

i



 z  i 1 
 
2!




1 ia
e  e ia
2i

(0<a<1)
Using the beta function (we shall study it later), we can show that the integral to be
(a)!(-a)!. This results in the interesting and useful factorial function relation
a! a !
a
sin 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.
2.3 Dispersion Relation
The name dispersion comes from optical dispersion. The index of refraction n may have
a real part determined by the phase velocity and an imaginary part determined by the
absorption. Kronig and Kramers showed in 1926-1927 that real part of (n2 -1 ) could be
expressed as an integral of the imaginary part. Generalizing this, we shall apply the label
dispersion relations to any pair of equations giving the real part of a function as an
integral of its imaginary part and the imaginary part as an integral of its real part ( See
later).
We consider f(z) that is analytic in the upper half plane and on the real axis. We also
require that
lim f  z   0 , 0  arg z  
z 
in order that the integral over an infinite semicircle will vanish. By the Cauchy integral
formula,
f x0  
1
2i
f z dz
 z  x0
C
Fig. 2.8
13

 x0 

f x dx
f x dx  1
f z dz
1 



 lim I R
2i 
x  x0
x  x0  2i
z  x0 R
x0 
C x0
 




Here Cx0 denotes a small semicircle about x0 in the lower half-plane. Since IR  0, and
f  z dz
 if  x 0 
z  x0

C x0
1
f x0   P
i


f x 
 x  x0 dx

Splitting the above equation into real and imaginary parts yields
f x0   u x0   iv x0 


1


P


vx dx i
 P
x  x0 
u  x0  
1

v x0   




P
1

u x dx
x  x0
vx dx
 x  x0


P


u x dx
x  x0
These are the dispersion relations ( Kronig and Kramers relations). The real and
Imaginary parts are Hilbert transforms of each other.
• 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
14
u  x0  
1


v x dx 1
v x dx
 P
x  x0 
x  x0
0

P


0
Letting x  -x in the first integral on the RHS and substituting v(-x) = -v(x), we obtain
u  x0  


 1
1 
P v x 

dx

x  x0 x  x0 

0

1
2


P
x0 v x 
 x 2  x02 dx
0
Similarly,
v x0   
2


P
x u x 
 x 20  x02 dx
0
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
k2 = 2/c2 ( 1 + i 4)
where k - wave vector,  - electric permittivity, -conductivity,  - angular frequency.
The index of refraction n = ck/ satisfies
n2 =  + i4
We take n2 to be a function of the complex variable  (with  and  depending on ).
However, n2 does not vanish as  infinity but instead approaches unity. So we work
with f() = n2() -1.


Re n  0   1 
2
2


P

0

 d
 Im n 2    1
  0
2
2
15


Im n  0   1  
2
2


P


 d
 0 Re n 2    1
0
 2  0
2
.
• The Parseval Relation
When the functions u(x) and v(x) are Hilbert transform of each other and each is square

integrable (i.e.,
u

2
dx and


 u x 

v
2
dx are finite), the two functions are related by


2
dx 
 v x 
2
dx

This is the Parseval relation (See the text book for the proof).
2.4
The Method of Steepest Decent(Optional Reading)
In analyzing problems, it is often desirable to know the behavior of a function for large
values of the variable, that is, the asymptotic behavior of the function. Specific examples
are furnished by the gamma function and the various Bessel function addressed later. The
method of steepest decent is a method of determining such asymptotic behavior of the
function like
I s   g z e sf  z  dz

C
For the present, let us take s to be real. The contour C is then chosen so that real part of
f(z) approaches minus infinity at both limits and that the integrand will vanish at the
limits, or is chosen as a closed contour. It is further assumed that the factor g(z) in the
integrand is dominated by the exponential in the region of interest.
The integral may be written as
I s   g z e su x, y e isv  x, y  dz

C
As s  infinity, the entire contribution of the integrand to the integral will come from the
region in which the real part of f(z) takes a positive maximum value. Away from this
maximum, the integrand will became negligibly small in comparison. If now, in addition,
16
we impose the condition that the imaginary part of the exponent, iv(x,y), be constant in
this region, i.e., v(x,y) = v(x0,y0) = v0, we may approximate the integral by
I s   e isv0 g z e su x, y  dz

C
Away from the maximum of the real part, the imaginary part may be permitted to
oscillate as it wishes, for the integral is negligibly small and the varying phase factor is
therefore irrelevant.
For a given s, at the maximum, we have
u u

0
x y
and therefor, by the use of Cauchy-Riemann conditions
 u u u
v 


,
  
x 
 x y y
df
0
dz
We proceed to search for the zeros of the derivative.
It is essential to note that the maximum of u(x,y) is the maximum only along a given
contour. In the finite plane neither the real nor the imaginary part of our analytic function
possesses an absolute maximum. This may be seen by recalling that both u and v satisfy
Laplace's equation
 2u  2u

0
x 2 y 2
 2u
 2u
 0 ( minimum along y-axis), and
 0 (maximum along x -axis),
From this, if
y 2
x 2
therefore there is no absolute maximum or minimum for u and v. The vanishing of the
derivative implies that we have a saddle point, which may be a maximum of u for one
contour and minimum for another (Fig.2.9)
Fig2.9
17
Our problem, then is to choose the contour of integration to satisfy two conditions,
(1) The contour must be chosen so that u(x,y) has a maximum at the saddle point;
(2) in such a way that, v(x,y) is constant.
We know that u = constant and v = constant form an orthogonal system. This means that
a curve v = ci, constant, is everywhere tangential to the gradient of u, u. Hence the curve
v = constant, is the curve that gives the line of steepest descent from the saddle point. So,
the above second condition leads to the path of steepest decent and gives the method its
name.
At the saddle point, we expand f(z) in a Taylor series,
f(z) = f(z0) + (1/2)(z-z0)2f''(z0) + ...
The first correction term, (1/2)(z-z0)2f''(z0), is real and negative. It is real, for we have
specified that v = v0, and negative because we are moving down from the saddle point or
mountain pass. Then , assuming that f''(z0)  0, we have
1
t2
f  z   f  z 0   z  z 0 2 f z 0   
2
2s
which serves to define a new variable t. If we set
(z - z0) = e i
(with the phase  held constant), we have
t 2  sf z 0  2 e 2i
Since t is real, it may be written as
t   sf z 0 
12
Substituting the above approximate f(z) into I(s), we obtain
I s  ~ g z 0 e
ef  z0 


e t
2
2
dz dt dt

we have
dz  dt d 


dt  d dz 
1
 sf z 0 

1
i
2e
18
Therefore,
Is  
gz 0 e sf z 0 e i
ef z 0 
12

e
t 2 2
dt

Note that, since the integral is essentially zero when departs appreciably from the origin,
the limits have then been set as minus infinity to positive infinity. The remaining integral
is just Guass error integral equal to (2)1/2, we finally obtain
2 g z 0 e sf  z0 e i
I s  
sf z 0 
12
The phase  was introduced as the phase of the contour as it passed through the saddle
point. It is chosen so that the two conditions given [  = const. Re f(z) = maximum) are
satisfied. It some times happens that the contour passes through two on more saddle
points in succession. If this is the case, we need only add the contribution from each of
the saddle points.
One note of warning, for each new problem, we must check the condition, u(x,y) <<
u(x0,y0), over the contour away from z0.
Example
Asymptotic Form of the Factorial
Function s!
The factorial function may be defined by the integral (in section 5.1)

s! 

s 
e
d
0
Let  = z s,
s! s
s 1

z e
s  zs
dz
0
C
s
s 1

e
s ln z  z 
dz
0
C
As before, we assume that s is real and positive, from which the integrand vanishes at the
limits 0 and infinity.
19
df
d
1
 ln z  z    1
dz dz
z
which shows that z =1 is a saddle point. We let
z-1 =  ei
with  small to describe the contour in the vicinity of the saddle point. Substituting it into
f(z), we have

 

f z   ln 1  e i  1  e i
1
  e i    2 e 2 i    1   e i
2
1
 1   2 e 2i
2
From this, we see that the integral takes on a maximum value e-s at the saddle point if the
contour C follows the real axis. (i.e., =0, f(1) = -1)

s!
2 s s 1e  s
s 1
12
 2s s s e s
The result is the first term in Stirling's expansion of the factorial function.