---
title: "Pole of order m; ∮cot z dz on |z|=1"
subject: "Applied Mathematics"
chapter: 1
key: "semester_4/applied_mathematics:residue-cot-z"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/residue-cot-z
site: BCT Engineering
---

# Pole of order m; ∮cot z dz on |z|=1

Applied Mathematics · Chapter 1: Complex Analysis

- Asked in: 2074 Bhadra, 1+4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/residue-cot-z
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-1

## Question

Define pole of order m. Using Cauchy's residue theorem evaluate ∮_C cot z dz; where C is |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.

## Answer

#### 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**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. When is z = a a pole of order m?
   A. f = g/(z-a)^m with g analytic and g(a) ≠ 0
   B. f = g(z-a)^m with g(a) ≠ 0
   C. f(a) = 0 to order m
   D. f is unbounded for z ≠ a
   Answer: A. g(a) ≠ 0 ensures the order is exactly m. (Key point: Pole of order m)

2. Which poles of cot z lie inside |z| = 1?
   A. z = 0 and ±π
   B. Only z = 0
   C. z = ±π/2
   D. None
   Answer: B. Poles are at nπ and π ≈ 3.14 > 1. (Key point: Poles inside)

3. What is the residue of cot z at 0?
   A. 0
   B. π
   C. 1
   D. -1
   Answer: C. lim z cos z / sin z = 1, or cos 0 / cos 0 = 1. (Key point: Residue)

4. What is the integral?
   A. 0
   B. πi
   C. -2πi
   D. 2πi
   Answer: D. 2πi × 1 = 2πi. (Key point: Result)

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