← 2024 Paper 2

UPSC 2024 Maths Optional Paper 2 Q3a — Step-by-Step Solution

15 marks · Section A

Residues: computation at poles of various orders · Complex Analysis · asked 3× in 14 yrs · Read the full method →

Question

Locate the poles and their order for $f(z)=\dfrac{1}{z(\sin\pi z)(z+\tfrac{1}{2})}$. Also find the residue at each pole.

Technique

Order-2 pole at $z=0$ (both $z$ and $\sin\pi z$ vanish); simple poles elsewhere; use Taylor of $\sin\pi z$ near $0$ for the residue; $1/g'(z_0)$ formula for simple poles.

Solution

Step 1 — Locate poles

• $z=0$: both $z$ and $\sin\pi z$ vanish to order 1 each → pole of order 2.
• $z=-1/2$: only $z+1/2$ vanishes; $\sin\pi(-1/2)=-1\ne 0$ → simple pole.
• $z=n$ for $n\in\mathbb Z\setminus\{0\}$: only $\sin\pi z$ vanishes; $n\ne 0$ and $n+1/2\ne 0$ → simple pole.

Step 2 — Residue at $z=0$ (order 2)

Near $z=0$: $z/\sin\pi z=1/\pi+O(z^2)$, so $g(z):=z/[(\sin\pi z)(z+1/2)]=\dfrac{1}{\pi(z+1/2)}+O(z)$.

Step 3 — Residue at $z=-1/2$ (simple pole)

Step 4 — Residue at $z=n\ne 0$ (simple pole)

Writing $f=\dfrac{1/[z(z+1/2)]}{\sin\pi z}$ and using $(\sin\pi z)'|_{z=n}=\pi\cos\pi n=\pi(-1)^n$:

Answer