- 2074 Bhadra · 1+4 marks
Pole of order m; ∮cot z dz on |z|=1
Short answer
A pole of order m has (z−a)^m in denominator with analytic non-zero numerator and highest negative Laurent power (z−a)^−m. cot z has a simple pole at z=0 inside |z|=1 with residue 1. By Cauchy's residue theorem, ∮_C cot z dz = 2πi.
Define pole of order m. Using Cauchy's residue theorem evaluate ∮_C cot z dz; where C is |z| = 1.
Pole of order m
If f(z) = g(z)/(z − a)ᵐ, where g(z) is analytic at a and g(a) ≠ 0, then z = a is a pole of order m. In the Laurent series about a, the highest negative power is (z − a)⁻ᵐ. A pole of order 1 is a simple pole.
Evaluation
cot z = cos z / sin z has poles where sin z = 0, i.e. z = nπ (n = 0, ±1, ±2, …). Only z = 0 lies inside |z| = 1 (π ≈ 3.14 > 1).
At z = 0, sin z has a simple zero and cos 0 = 1 ≠ 0, so z = 0 is a simple pole.
Residue = lim (z → 0) z cot z = lim z cos z / sin z = 1 · 1 = 1.
(Or by the formula Res = p(a)/q′(a) = cos 0 / cos 0 = 1.)
By the residue theorem:
Answer: ∮_C cot z dz = 2πi × 1 = 2πi
✓ Answer checked by Claude Opus 5.5 (high effort)
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