← 2022 Paper 2

UPSC 2022 Maths Optional Paper 2 Q3a — Step-by-Step Solution

15 marks · Section A

Cauchy's residue theorem · Complex Analysis · asked 5× in 14 yrs · Read the full method →

Question

Evaluate $\displaystyle\int_C\dfrac{z+4}{z^2+2z+5}\,dz$, where $C$ is $|z+1-i|=2$.

Technique

Factor denominator → find poles; check which are inside the contour; residue theorem with single pole.

Solution

Setup. Find roots of denominator: $z^2+2z+5=0\Rightarrow z=\dfrac{-2\pm\sqrt{4-20}}{2}=-1\pm 2i$.

So poles at $z_1=-1+2i$ and $z_2=-1-2i$. Both simple poles.

Step 1 — Identify which poles lie inside $C$

$C$ is circle centred at $z_0=-1+i$ with radius $2$.

Distance from $z_0$ to $z_1=-1+2i$: $|(-1+2i)-(-1+i)|=|i|=1<2$. So $z_1$ is inside $C$.

Distance from $z_0$ to $z_2=-1-2i$: $|(-1-2i)-(-1+i)|=|-3i|=3>2$. So $z_2$ is outside $C$.

Step 2 — Residue at $z=-1+2i$

$f(z)=\dfrac{z+4}{(z-z_1)(z-z_2)}$ with $z_1=-1+2i$, $z_2=-1-2i$.

$\text{Res}_{z_1}f=\dfrac{z_1+4}{z_1-z_2}=\dfrac{(-1+2i)+4}{(-1+2i)-(-1-2i)}=\dfrac{3+2i}{4i}$.

Simplify: $\dfrac{3+2i}{4i}=\dfrac{(3+2i)\cdot(-i)}{4i\cdot(-i)}=\dfrac{-3i+2}{4}=\dfrac{2-3i}{4}$.

Step 3 — Apply residue theorem

Answer