GATE 2020 IN Q.15: solution and answer

Official answer: BSource: Official answer key, GATE 2020 IN (IIT Delhi)

1. Singularity. $f(z)=\frac{1}{z+a}$ has exactly one singularity: a simple pole at $z=-a$, which is the point $(-a,0)$ in the complex plane.

2. Where it lies. The circle C is centred at $(-a,0)$, so the pole sits at its centre. That means the pole is inside C for every radius $R>0$, however small or large.

3. Residue at the simple pole. ${\mathrm{Res}}_{z=-a}f={\lim }_{z\to -a}(z+a)\cdot \frac{1}{z+a}=1$.

4. Cauchy's residue theorem (the contour is traversed anti-clockwise): $${\oint }_C\frac{dz}{z+a}=2\pi i\times 1=2\pi i$$

5. Check by parametrising. Put $z=-a+R{e}^{i\theta }$ with $0\le \theta \le 2\pi$. Then $dz=iR{e}^{i\theta }d\theta$ and $${\oint }_C\frac{dz}{z+a}={\int }_0^{2\pi }\frac{iR{e}^{i\theta }}{R{e}^{i\theta }}d\theta ={\int }_0^{2\pi }id\theta =2\pi i,$$ which does not depend on $R$. The answer is option (B).