GATE 2024 EC Q.43: solution and answer

Official answer: DSource: Official answer key, GATE 2024 EC (IISc Bengaluru)

1. Find the singularities. The denominator $z^2(z-2)$ is zero at $z=0$ and $z=2$. Both points satisfy $|z|<3$, so both lie inside $C$.

2. Classify $z=0$. $\sin (\pi z)=\pi z-\frac{(\pi z{)}^3}{6}+\dots$ has a simple zero at $z=0$, which cancels one factor of $z^2$: $$f(z)=\frac{\sin (\pi z)}{z}\cdot \frac{1}{z(z-2)}$$ The first factor is analytic and equals $\pi$ at $z=0$, so $z=0$ is a simple pole, not a double pole.

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

4. Classify $z=2$. The numerator is $\sin (2\pi )=0$. $\sin (\pi z)$ has a simple zero at $z=2$, which cancels the simple zero of $(z-2)$, so $z=2$ is a removable singularity and its residue is $0$. (Treating it as a simple pole gives the same value: ${\lim }_{z\to 2}(z-2)f(z)=\frac{\sin 2\pi }{2^2}=0$.)

5. Cauchy's residue theorem: $${\oint }_{C}f(z)dz=2\pi j\sum \text{Res}=2\pi j\left(-\frac{\pi }{2}+0\right)=-{\pi }^{2}j$$

6. This is option (D).