GATE 2018 EC Q.51: solution and answer

Official answer: 2Source: Official answer key, GATE 2018 EC (IIT Guwahati)

1. Singularities. $$\frac{1}{z^2-1}=\frac{1}{(z-1)(z+1)}$$ has simple poles at $z=1$ and $z=-1$ and is analytic everywhere else. That includes the crossing point $z=0$, where the integrand equals $-1$, so the self-crossing of $C$ causes no difficulty. In the figure, $z=-1$ is inside the left lobe, $z=1$ is inside the right lobe, and neither pole lies on $C$.

2. Direction of each lobe, read from the arrows. On top of the left lobe the arrow points right (east, toward the origin). A loop whose top is travelled eastward goes clockwise, so the left lobe goes clockwise round $z=-1$. On top of the right lobe the arrow points left (west). A loop whose top is travelled westward goes counter-clockwise, so the right lobe goes counter-clockwise round $z=1$. Both readings fit one continuous path: along the top of the left lobe, through the origin, east along the bottom of the right lobe, back west along its top, through the origin again, west along the bottom of the left lobe, and back up.

3. Winding numbers. From step 2, $n(C,1)=+1$ and $n(C,-1)=-1$. A clockwise loop round a pole contributes $-2\pi j$ times that pole's residue, not $+2\pi j$ times it.

4. Residue at each simple pole, using ${\text{Res}}_{z=a}f={\lim }_{z\to a}(z-a)f(z)$: $${\text{Res}}_{z=1}=\frac{1}{1+1}=\frac{1}{2},{\text{Res}}_{z=-1}=\frac{1}{-1-1}=-\frac{1}{2}.$$

5. Generalised residue theorem, ${\oint }_{C}fdz=2\pi j{\sum }_{k}n(C,a_k){\text{Res}}_{a_k}f$: $${\oint }_C\frac{dz}{z^2-1}=2\pi j\left[(+1)\left(\frac{1}{2}\right)+(-1)\left(-\frac{1}{2}\right)\right]=2\pi j\left[\frac{1}{2}+\frac{1}{2}\right]=2\pi j.$$

6. Scale by the prefactor: $$\frac{1}{\pi j}\cdot 2\pi j=2.$$