---
title: "Residue theorem: ∮(1-cos z)/z³ dz"
subject: "Applied Mathematics"
chapter: 1
key: "semester_4/applied_mathematics:residue-1-cosz-over-z3"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/residue-1-cosz-over-z3
site: BCT Engineering
---

# Residue theorem: ∮(1-cos z)/z³ dz

Applied Mathematics · Chapter 1: Complex Analysis

- Asked in: 2070 Magh, 5 marks
- Video: [23. Residue Theorem | Problem#1 | Complete Concept](https://www.youtube.com/watch?v=ZyOIZq__zxE) — MKS TUTORIALS by Manoj Sir
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/residue-1-cosz-over-z3
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-1

## Question

Evaluate the integral by residue theorem ∮_C (1 - cos z)/z³ dz.

## Short answer

The integrand (1−cos z)/z³ has a simple pole at z=0 with Laurent coefficient 1/2. Residue = 1/2. ∮_ C (1−cos z)/z³ dz = 2πi × 1/2 = πi.

## Answer

Here C is taken as any simple closed contour enclosing z = 0 (for example |z| = 1), since the only singularity of f is at the origin.

f(z) = (1 − cos z)/z³

**Nature of the singularity:** the numerator 1 − cos z = z²/2 − z⁴/24 + … has a zero of order 2 at z = 0, the denominator has a zero of order 3. So f(z) = 1/(2z) − z/24 + … and z = 0 is a **simple pole**.

**Residue:**

Res at 0 = lim (z → 0) z f(z) = lim (1 − cos z)/z² = 1/2

(Laurent series: f(z) = 1/(2z) − z/24 + z³/720 − …; coefficient of 1/z is 1/2.)

By the residue theorem:

**Answer: ∮_C (1 − cos z)/z³ dz = 2πi × 1/2 = πi**

## Check yourself

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

1. Why is z=0 a simple pole of (1-cos z)/z³?
   A. Numerator is non-zero at 0
   B. Both have zeros of order 3
   C. Numerator has a zero of order 4
   D. Numerator has a zero of order 2, denominator of order 3, leaving z^-1
   Answer: D. 1 - cos z ~ z²/2, so f ~ 1/(2z). (Key point: Nature of singularity)

2. How is the residue obtained?
   A. lim z f(z) = lim (1-cos z)/z² = 1/2
   B. lim f(z) = 1/2
   C. lim (1-cos z)/z³ = 1/6
   D. d/dz of z f(z) at 0 = 0
   Answer: A. For a simple pole, residue = lim z f(z). (Key point: Residue computation)

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

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