GATE 2018 EE Q.13: solution and answer

Official answer: ASource: Official answer key, GATE 2018 EE (IIT Guwahati)

1. The contour is the circle $|z+2|=1$, traversed counter-clockwise (the positive direction).

2. Factor the denominator: $z^2-4=(z-2)(z+2)$. The numerator $z+1$ vanishes at neither $z=2$ nor $z=-2$, so the integrand has simple poles at $z=2$ and $z=-2$ and is analytic everywhere else.

3. Locate the poles: $z=-2$ is the centre of $C$, so it lies inside. For $z=2$, $|2+2|=4>1$, so it lies outside $C$ and contributes nothing.

4. Residue at the simple pole $z=-2$: $${\text{Res}}_{z=-2}=\underset{z\to -2}{\lim }(z+2)\frac{z+1}{(z-2)(z+2)}=\frac{-2+1}{-2-2}=\frac{-1}{-4}=\frac{1}{4}$$

5. Residue theorem, ${\oint }_{C}f(z)dz=2\pi i\sum \text{Res}$ (poles inside $C$): $${\oint }_C\frac{z+1}{z^2-4}dz=2\pi i\cdot \frac{1}{4}=\frac{\pi i}{2}$$

6. Check with Cauchy's integral formula: write the integrand as $\frac{g(z)}{z+2}$ with $g(z)=\frac{z+1}{z-2}$, which is analytic on and inside $C$. Then ${\oint }_C=2\pi ig(-2)=2\pi i\cdot \frac{1}{4}=\frac{\pi i}{2}$, the same result.

7. This matches option (A).