GATE 2022 IN Q.61: solution and answer

Official answer: 0.49 to 0.51Source: Official answer key, GATE 2022 IN (IIT Kharagpur)

1. The factors $(z+1)$, $(z+2)$ and $(z+3)$ each appear once in the denominator and the numerator is $1$, which is never zero. So $z=-1$ is a simple pole of $f(z)$. The poles at $-2$ and $-3$ are not needed here.

2. Residue at a simple pole: ${\mathrm{Res}}_{z=z_0}f(z)={\lim }_{z\to z_0}(z-z_0)f(z)$.

3. Take $z_0=-1$. The factor $(z+1)$ cancels: $${\mathrm{Res}}_{z=-1}f(z)=\underset{z\to -1}{\lim }(z+1)\cdot \frac{1}{(z+1)(z+2)(z+3)}=\underset{z\to -1}{\lim }\frac{1}{(z+2)(z+3)}$$

4. Put $z=-1$: $$\frac{1}{(-1+2)(-1+3)}=\frac{1}{(1)(2)}=\frac{1}{2}$$

5. Cross-check with partial fractions (cover-up rule): $f(z)=\frac{1/2}{z+1}-\frac{1}{z+2}+\frac{1/2}{z+3}$. The coefficients are $\frac{1}{(1)(2)}=\frac{1}{2}$ at $z=-1$, $\frac{1}{(-1)(1)}=-1$ at $z=-2$ and $\frac{1}{(-2)(-1)}=\frac{1}{2}$ at $z=-3$. The coefficient of $\frac{1}{z+1}$ is $\frac{1}{2}$, and that is the residue at $z=-1$.

6. Residue $=0.5$