GATE 2020 EE Q.5: solution and answer

Official answer: CSource: Official answer key, GATE 2020 EE (IIT Delhi)

1. Factor the denominator: $z^2-2z=z(z-2)$, so $f(z)=\frac{z^2+1}{z(z-2)}$. The numerator is non-zero at $z=0$ (value $1$) and at $z=2$ (value $5$), so both are simple poles.

2. Locate the poles relative to $C$, the circle $|z|=1$: $z=0$ lies inside ($|0|<1$), while $z=2$ lies outside ($|2|=2>1$). Neither pole is on the contour, and only $z=0$ contributes.

3. Residue at the simple pole $z=0$: $${\text{Res}}_{z=0}f=\underset{z\to 0}{\lim }zf(z)=\underset{z\to 0}{\lim }\frac{z^2+1}{z-2}=\frac{0+1}{0-2}=-\frac{1}{2}$$

4. Cauchy's residue theorem for a closed contour traversed counterclockwise: $${\int }_{C}f(z)dz=2\pi i\sum {\text{Res}}_{\text{inside}}=2\pi i\left(-\frac{1}{2}\right)=-\pi i$$

5. Hence the integral equals $-\pi i$, which is option (C).