← 2021 Paper 2

UPSC 2021 Maths Optional Paper 2 Q1d — Step-by-Step Solution

10 marks · Section A

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

Question

$c(t)=e^{4\pi i t}$, $0\le t\le 1$. Evaluate $\int_c\dfrac{dz}{2z^2-5z+2}$.

Technique

Recognise the curve winds twice; residue theorem with winding number.

Solution

Setup. $c(t)=e^{4\pi i t}$ traces the unit circle $|z|=1$ twice (as $t$ goes from 0 to 1, the angle goes $0\to 4\pi$).

Step 1 — Factor denominator

$2z^2-5z+2=(2z-1)(z-2)$. Roots: $z=1/2$ and $z=2$.

Only $z=1/2$ is inside $|z|=1$.

Step 2 — Residue at $z=1/2$

$\dfrac{1}{2z^2-5z+2}=\dfrac{1}{(2z-1)(z-2)}$.

$\text{Res}_{z=1/2}=\lim_{z\to 1/2}\dfrac{z-1/2}{(2z-1)(z-2)}=\lim_{z\to 1/2}\dfrac{z-1/2}{2(z-1/2)(z-2)}=\dfrac{1}{2(1/2-2)}=\dfrac{1}{2(-3/2)}=-\dfrac{1}{3}$.

Step 3 — Apply residue theorem (with winding number)

The curve $c$ winds around $z=1/2$ twice (winding number $n=2$).

$\int_c f\,dz=2\pi i\cdot n\cdot\text{Res}=2\pi i\cdot 2\cdot(-1/3)=-\dfrac{4\pi i}{3}$.

Answer