GATE 2018 EE Q.35: solution and answer

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

1. Factor the denominator: $z^2-3z+2=(z-1)(z-2)$, so $f(z)=\frac{z^2}{(z-1{)}^2(z-2{)}^2}$. The numerator is not zero at $z=1$ or $z=2$, so $f$ has poles of order 2 at $z=1$ and $z=2$. Both lie inside $|z|=4$.

2. For a pole of order 2 at $z=a$, the residue is ${\text{Res}}_{z=a}f={\frac{d}{dz}[(z-a{)}^{2}f(z)\left]\right|}_{z=a}$.

3. At $z=1$: $(z-1{)}^{2}f(z)=\frac{z^2}{(z-2{)}^2}$, and $\frac{d}{dz}\left[\frac{z^2}{(z-2{)}^2}\right]=\frac{2z(z-2)-2z^2}{(z-2{)}^3}=\frac{-4z}{(z-2{)}^3}$. At $z=1$ this is $\frac{-4}{(-1{)}^3}=4$.

4. At $z=2$: $(z-2{)}^{2}f(z)=\frac{z^2}{(z-1{)}^2}$, and $\frac{d}{dz}\left[\frac{z^2}{(z-1{)}^2}\right]=\frac{2z(z-1)-2z^2}{(z-1{)}^3}=\frac{-2z}{(z-1{)}^3}$. At $z=2$ this is $\frac{-4}{1^3}=-4$.

5. By Cauchy's residue theorem, $${\oint }_{C}f(z)dz=2\pi i(4+(-4))=0.$$

6. Quick check: for large $|z|$, $f(z)=\frac{1}{z^2}{\left(1-\frac{3}{z}+\frac{2}{z^2}\right)}^{-2}=\frac{1}{z^2}+\frac{6}{z^3}+\dots$ There is no $\frac{1}{z}$ term, so the residue at infinity is $0$. The residues at all finite poles plus the residue at infinity add up to zero, so the finite residues must add up to $0$. This agrees with $4+(-4)=0$.

7. The answer is $0$, which is option (B).