---
title: "Residue theorem: ∮sin z / z⁶ on |z|=1"
subject: "Applied Mathematics"
chapter: 1
key: "semester_4/applied_mathematics:residue-sinz-over-z6"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/residue-sinz-over-z6
site: BCT Engineering
---

# Residue theorem: ∮sin z / z⁶ on |z|=1

Applied Mathematics · Chapter 1: Complex Analysis

- Asked in: 2073 Magh, 1+4 marks; 2068 Bhadra (asked 2 times)
- Video: [Residue, Residue Theorem (Part 1) | Engineering Mathematics | S Chand Academy](https://www.youtube.com/watch?v=Xk50mL9HqIY) — S Chand Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/residue-sinz-over-z6
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-1

## Question

State Cauchy's residue theorem. Using it evaluate ∮_C (sin z)/z⁶ dz where C: |z| = 1.

## Short answer

Cauchy’s residue theorem: ∮_C f(z)dz = 2πi Σ Res(f, z_k) inside C. For ∮ sin z/z⁶ dz over |z|=1, the residue at z=0 is 1/5! = 1/120. Hence the integral equals 2πi·(1/120) = πi/60.

## Answer

#### Cauchy's residue theorem
If f(z) is analytic inside and on a simple closed contour C except for a finite number of singular points z₁, z₂, …, zₖ inside C, then

∮_C f(z) dz = 2πi Σ Res[f(z), zₖ],

that is, 2πi times the sum of the residues at the singularities inside C.

#### Solution
f(z) = sin z / z⁶ has its only singularity at z = 0, which lies inside C: |z| = 1.

The residue is the coefficient of 1/z in the Laurent series about z = 0. Use

sin z = z − z³/3! + z⁵/5! − z⁷/7! + …

```text
sin z / z⁶ = 1/z⁵ − 1/(3! z³) + 1/(5! z) − z/7! + …
```

The coefficient of 1/z is 1/5! = 1/120. So Res(f, 0) = 1/120.

(z = 0 is a pole of order 5 because sin z has a simple zero.)

By the residue theorem:

∮_C sin z / z⁶ dz = 2πi · (1/120) = πi/60

**Answer: πi/60 ≈ 0.0524 i**

## Check yourself

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

1. The residue theorem says ∮_C f dz =
   A. 2πi × sum of residues inside C
   B. 2π × sum of residues
   C. sum of residues inside C
   D. 0
   Answer: A. Integral = 2πi Σ Res. (Key point: Residue theorem)

2. The residue at z=0 is:
   A. The coefficient of 1/z in the Laurent series
   B. The constant term
   C. The coefficient of z
   D. The value f(0)
   Answer: A. By definition. (Key point: Residue meaning)

3. The 1/z term in sin z/z⁶ comes from:
   A. z in sin z
   B. z⁵/5! in sin z
   C. z³/3! in sin z
   D. z⁷/7! in sin z
   Answer: B. z⁵/z⁶ = 1/z, coefficient 1/120. (Key point: Series)

4. The integral is:
   A. 2πi/5
   B. 0
   C. πi/120
   D. πi/60
   Answer: D. 2πi × 1/120 = πi/60. (Key point: Result)

## Close variants set in other exams

- [Residue theorem: ∮z²/(3+4z+z²) on |z|=2](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Aresidue-z2-over-3-4z-z2) — Applied Mathematics, chapter 1, asked 3×

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