← 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 13 yrs · Read the full method →
Question
Locate the poles and their order for f(z)=z(sinπz)(z+21)1. Also find the residue at each pole.
Technique
Order-2 pole at z=0 (both z and sinπz vanish); simple poles elsewhere; use Taylor of sinπz near 0 for the residue; 1/g′(z0) formula for simple poles.
Solution
Step 1 — Locate poles
- z=0: both z and sinπz vanish to order 1 each → pole of order 2.
- z=−1/2: only z+1/2 vanishes; sinπ(−1/2)=−1=0 → simple pole.
- z=n for n∈Z∖{0}: only sinπz vanishes; n=0 and n+1/2=0 → simple pole.
Step 2 — Residue at z=0 (order 2)
Resz=0f=z→0limdzd[z2f(z)]=z→0limdzd[(sinπz)(z+1/2)z].
Near z=0: z/sinπz=1/π+O(z2), so g(z):=z/[(sinπz)(z+1/2)]=π(z+1/2)1+O(z).
g′(z)z=0=−π(z+1/2)21z=0=−π4.
Resz=0f=−π4.
Step 3 — Residue at z=−1/2 (simple pole)
Resz=−1/2f=z→−1/2lim(z+1/2)f(z)=(−1/2)sin(−π/2)1=(−1/2)(−1)1=2.
Step 4 — Residue at z=n=0 (simple pole)
Writing f=sinπz1/[z(z+1/2)] and using (sinπz)′∣z=n=πcosπn=π(−1)n:
Resz=nf=n(n+1/2)⋅π(−1)n1=πn(2n+1)(−1)n⋅2.
Answer
Resz=0=−π4,Resz=−1/2=2,Resz=n=πn(2n+1)2(−1)n(n∈Z∖{0}).