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. Example: A phase difference on opposite sides of the cut line.
Consider
Two branch points: z=-1 and z=1

5. 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) ?

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

8. Consider the path integral indicated in the figure
Consider the path integral indicated in the figure. The Cauchy theorem leads to
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
Residue theorem: Zm Z2 C2

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. Infinite semicircle
In both cases, we have (Cauchy principle value)

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

14. From this
Our integral becomes a contour integral of the unit circle
Example

15. Suppose that our definite integral has the above form and
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. Example
Where are the poles ?
Consider the above definite integral with a real and positive (This is a Fourier transform). We assume the two conditions:

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

21. C2 C1

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

25. Attempt to do the following exercises!
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.
Attempt to do the following exercises!

26. 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

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

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