GATE Electrical Engineering (EE) 2021 Solved Papers

Vertices of rectangle are $\left(-1,-1\right),\left(3,-1\right),\left(3,1\right)$ and $\left(-1,1\right)$
Let $\mathit{f}\left(\mathit{z}\right)=\frac{1}{{\mathit{z}}^2\left(\mathit{z}-4\right)}$
Poles of $\mathit{f}\left(\mathit{z}\right)$ are $\mathit{z}=4$

$\therefore \mathit{Z}=0$ lies inside ' $\mathit{c}$ '
$\therefore \prime \mathit{z}=0$ ' is a singular point of index $\left(\mathit{n}+1\right)=\mathit{z}$
Residue of $\mathit{f}\left(\mathit{z}\right)$ 'at $\mathit{z}=\mathit{a}$ ' $=\frac{1}{\mathit{n}!}\underset{\mathit{z}\to \mathit{a}}{\lim }\frac{{\mathit{d}}^{\mathit{n}}}{\mathit{d}{\mathit{z}}^{\mathit{n}}}{\mathit{z}}^{\mathit{n}+1\mathit{f}\left(\mathit{z}\right)}$
$\mathit{n}=1,\mathit{a}=0$
$\mathit{R}=\frac{1}{1!}\underset{\mathit{z}\to 0}{\lim }\frac{\mathit{d}}{\mathit{d}\mathit{z}}{\mathit{z}}^2\frac{1}{{\left(\mathit{z}-4\right)}^2}$
$=\underset{\mathit{z}\to 0}{\lim }\frac{\left(-1\right)}{{\left(\mathit{z}-4\right)}^2}=\frac{-1}{16}$
By Cauchy's Residue theorem
${\oint }_{\mathit{C}}\mathit{f}\left(\mathit{z}\right)\mathit{d}\mathit{z}=2\mathit{\pi }\mathit{j}[$ sum of residues $]$
${\oint }_{\mathit{C}}\frac{\mathit{d}\mathit{z}}{{\mathit{z}}^2\left(\mathit{z}-4\right)}=2\mathit{\pi }\mathit{j}\mathit{R}$
$=2\mathit{\pi }\mathit{j}\left(\frac{-1}{16}\right)$
$=\frac{-\mathit{j}\mathit{\pi }}{8}$