Integration of the complex function $f\left(z\right)=\frac{z^2}{z^2-1}$ in the counterclockwise direction, around |z – 1| = 1, is

Concept:

Cauchy’s Theorem:

If f(z) is an analytic function and f’(z) is continuous at each point within and on a closed curve C, then

$\underset{C}{\oint }f\left(z\right)dz=0$

Cauchy’s Integral Formula:

If f(z) is an analytic function within a closed curve and if a is any point within C, then

$f\left(a\right)=\frac{1}{2\pi i}\underset{C}{\oint }\frac{f\left(z\right)}{z-a}dz$

$f^n\left(a\right)=\frac{n!}{2\pi i}\underset{C}{\oint }\frac{f\left(z\right)}{{\left(z-a\right)}^{n+1}}dz$

Residue Theorem:

If f(z) is analytic in a closed curve C except at a finite number of singular points within C, then

$\underset{C}{\int }f\left(z\right)dz=2\pi i\times \left[\mathrm{s}\mathrm{u}\mathrm{m}\mathrm{o}\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{i}\mathrm{d}\mathrm{u}\mathrm{e}\mathrm{s}\mathrm{a}\mathrm{t}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{u}\mathrm{a}\mathrm{l}\mathrm{r}\mathrm{p}\mathrm{o}\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{s}\mathrm{w}\mathrm{i}\mathrm{t}\mathrm{h}\mathrm{i}\mathrm{n}\mathrm{C}\right]$

Formula to find residue:

1. If f(z) has a simple pole at z = a, then

$Resf\left(a\right)=\underset{z\to a}{lim}\left[\left(z-a\right)f\left(z\right)\right]$

2. If f(z) has a pole of order n at z = a, then

$Resf\left(a\right)=\frac{1}{\left(n-1\right)!}{\left\{\frac{d^{n-1}}{d{z}^{n-1}}\left[{\left(z-a\right)}^{n}f\left(z\right)\right]\right\}}_{z=a}$

Application:

Given function is $f\left(z\right)=\frac{z^2}{z^2-1}$

Poles: z = 1, -1

|z – 1| = 1

⇒ |x – 1 + iy| = 1

$\Rightarrow \sqrt{{\left(x-1\right)}^2+y^2}=1$

The given region is a circle with the centre at (1, 0) and the radius is 1.

Only pole z = 1, lies within the given region.

Residue at z = 1 is, $\underset{z\to 1}{lim}\frac{z^2}{\left(z+1\right)}=0.5$

The value of the integral = 2πi × 0.5 = πi