---
title: "Residue theorem: ∮tan z dz on |z|=2"
subject: "Applied Mathematics"
chapter: 1
key: "semester_4/applied_mathematics:residue-tan-z"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/residue-tan-z
site: BCT Engineering
---

# Residue theorem: ∮tan z dz on |z|=2

Applied Mathematics · Chapter 1: Complex Analysis

- Asked in: 2069 Bhadra
- Video: [Evaluate the complex integral∮tanz dz within a circle |z|=2](https://www.youtube.com/watch?v=nYFHfbem3xI) — Ayan Sarkar
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/residue-tan-z
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-1

## Question

Evaluate ∮_C tan z dz where C is a circle |z| = 2 by Cauchy's residue theorem.

## Short answer

tan z = sin z/cos z has simple poles at z=±π/2 inside |z|=2. Residue at each pole is −1, sum of residues = −2. Integral = 2πi(−2) = −4πi.

## Answer

tan z = sin z / cos z has simple poles where cos z = 0, i.e. z = (2n + 1)π/2.

For C: |z| = 2, the poles inside are z = π/2 ≈ 1.57 and z = −π/2 ≈ −1.57. The next poles at ±3π/2 ≈ ±4.71 are outside.

Residue at a simple pole z = a of p(z)/q(z) is p(a)/q′(a). Here p = sin z and q = cos z, q′ = −sin z, so

Res = sin a / (−sin a) = −1 at every pole.

- At z = π/2: Res = −1
- At z = −π/2: Res = −1

Sum of residues inside C = −2. By Cauchy's residue theorem:

**Answer: ∮_C tan z dz = 2πi(−2) = −4πi**

## Check yourself

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

1. Which poles of tan z lie inside |z| = 2?
   A. z = ±π/2
   B. z = 0 and ±π
   C. z = ±π/2 and ±3π/2
   D. z = 0 only
   Answer: A. Poles where cos z = 0; ±π/2 ≈ ±1.57 inside, ±3π/2 ≈ 4.71 outside. (Key point: Poles inside)

2. How is the residue at a simple pole of p/q computed?
   A. p'(a)/q(a)
   B. p(a)/q'(a), here sin a/(-sin a) = -1
   C. p(a) q'(a)
   D. p(a)/q'(a), here sin a/cos a
   Answer: B. q = cos z, q' = -sin z, so the residue is -1 at every pole. (Key point: Residue formula)

3. What is the integral?
   A. 4πi
   B. -2πi
   C. -4πi
   D. 0
   Answer: C. Sum of residues is -2, so 2πi(-2) = -4πi. (Key point: Result)

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