# Residues - Complex Function

The coefficient of (z-a)^-1 in the expansion of f(z) around an isolated singularity is called the residue of f(z) at that point. Thus is the Laurent's series expansion of f(z) around z=a i.e $$f(z)=a_{0}+a_{1}(z-a)+a_{2}(z-a)^{2}+\ldots+a_{-1}(z-a)^{-1}+a_{-2}(z-a)^{-2}+\ldots$$ $$, the residue of f(z) at z=a is a_{-1}$$

Since,

$$a_{n} =\frac{1}{2 \pi i} \int \frac{f(z)}{(z-a)^{n+1}} d z$$

$$a_{-1} =\operatorname{Res} f(a)=\frac{1}{2 \pi i} \int_{C} f(z) d z$$

$$\int_{C} f(z) d z =2 \pi i \operatorname{Res} f(a)$$

## Residue Theorem

If f(z) is analytic in a closed curve C except at a finite number of singular point within C, then ∫f(z)dz= 2πi x (sum of residues at the singular points within C)

Let us surround each of the singular points a₁, a₂,..an by a small circle such that it encloses no other singular point. Then these circles C₁, C₂, ..., Cn together with C, from a multiply connected region in which f(z) is analytic.
∴ Applying Cauchy's Theorem, we have
$$\int_{C} f(z) d z =\int_{C_{1}} f(z) d z+\int_{C_{2}} f(z) d z+\ldots \int_{C_{n}} f(z) d z$$
$$=2 \pi\left[\operatorname{Res} f\left(a_{1}\right)+\operatorname{Res} f\left(a_{2}\right)++\operatorname{Res} f\left(a_{n}\right)\right]$$

## Calculation of Residues

1. If f(z) has a simple pole at z=a, then
$$\text { Res } f(a)=\operatorname{Lt}_{z \rightarrow a}[(z-a) f(z)]$$
Laurent's series in this case is
$$f(z)=c_{0}+c_{1}(z-a)+c_{2}(z-a)^{2}+c_{-1}(z-a)^{-1}$$
Multiplying throughout by z-a, we have
$$(z-a) f(z)=c_{0}(z-a)+c_{1}(z-a)^{2}+...+_{-1}$$
Taking limits as z→a, we get
$$\operatorname{Li}_{z \rightarrow a}[(z-a) f(z)]=c_{-1}=\operatorname{Res} f(a)$$

2 Another formula for Res f(a):
Let
f(z)=ф(z)/Ψ(z) , where Y(z)=(z-a) F(z), F(a)≠ 0
Then
$$\operatorname{Li}_{z \rightarrow a}[(z-a) \phi(z) \psi(z)]$$
$$=\operatorname{L}_{z \rightarrow a} \frac{(z-a)\left[\phi (a)+\left(z-a) \phi^{\prime}(a)+\right....]\right.}{\psi(a)+(z-a) \psi^{\prime}(a)+....}$$
$$=\operatorname{L}_{z \rightarrow a} \frac{(z-a)\left[\phi (a)+\left(z-a) \phi^{\prime}(a)+\right....]\right.}{\psi^{\prime}(a)+(z-a) \psiˊˊ(a)+....}$$, since 𝜸(a)=0
Thus,
$${Res} f(a)=\frac{ф(a)}{Ψ^{\prime}(a)}$$
3. If f(z) has a pole of order n at z=a, then
$$Res f(a)=\frac{1}{(n-1)!}\left \{ \frac{d^{n-1}}{dz^{n-1}}\left [ (z-a)^{n}f(z) \right ] \right \}_{z=a}$$
Observation: In many cases, the residue of a pole (z=a) can be found, by putting z= a+t in f(z) and expanding it in powers of t where |t| is quite small.

Example 1
The residues of a complex function X(z)= (1-2z)/z(z-1)(z-2) at its poles are ?
Solution:
X(z)= (1-2z)/z(z-1)(z-2)
Poles are z=0, z=1 and z=2
Residue at z=0
residue=value of (1-2z)/z(z-1)(z-2) at z=0
= (1 - 2 x 0)/(0-1)(0-2)
=1/2

Residue at z=1
residue=value of (1-2z)/z(z-1)(z-2) at z=1
=(1 - 2 x 1)/1(1-2)
=1

Residue at z=2
residue=value of (1-2z)/z(z-1)(z-2) at z=2
= (1 - 2 x 2)/2(2 - 1)
=-3/2
Therefore, the residues at its poles are 1/2, 1, -3/2.

Example 2
For f(z)= sin(z)/z², the residue of the pole at z=0 is ?
Solution:
Residue of sin z/z²= Coefficient of 1/z in

$$\left \{ \frac{z-\frac{z^{3}}{3!}+\frac{z^{5}}{5!}- - -}{z^{2}} \right \}$$
=Coefficient of 1/z in {1/z - z/3! - z³/5! - - -}
=1

Example 3
The residue of f(z)= z³/(z-1)⁴(z-2)(z-3) at z=3 is ?
Solution:
Residue at z=3 is
lim z-->3
[(z-3) (z³/(z-1)⁴(z-2)(z-3))]
3³/(3-1)⁴(3-2)
27/16