---
title: "Poles and residues of z²/((z-1)²(z+2))"
subject: "Applied Mathematics"
chapter: 1
key: "semester_4/applied_mathematics:poles-residues-z2-z1sq-z2"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/poles-residues-z2-z1sq-z2
site: BCT Engineering
---

# Poles and residues of z²/((z-1)²(z+2))

Applied Mathematics · Chapter 1: Complex Analysis

- Asked in: 2068 Magh, 5 marks
- Video: [ To find poles and corresponding Residues of the function-complex Analysis](https://www.youtube.com/watch?v=v7acr_BYn5k) — Btech Maths Hub
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/poles-residues-z2-z1sq-z2
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-1

## Question

Determine the poles and the residue at each pole of the function f(z) = z²/((z - 1)²(z + 2)).

## Short answer

f(z)=z²/(( z−1)²( z+2)) has a double pole at z=1 with residue 5/9 and a simple pole at z=−2 with residue 4/9.

## Answer

f(z) = z²/((z − 1)²(z + 2)).

The numerator is non-zero at z = 1 and z = −2, so:

- z = 1 is a **pole of order 2** (double pole).
- z = −2 is a **simple pole**.

**Residue at z = −2 (simple):**

Res = lim (z → −2) (z + 2) f(z) = z²/(z − 1)² at z = −2 = 4/9

**Residue at z = 1 (order 2):**

Res = d/dz [ (z − 1)² f(z) ] at z = 1 = d/dz [ z²/(z + 2) ] at z = 1

d/dz [ z²/(z + 2) ] = [ 2z(z + 2) − z² ] / (z + 2)² = (z² + 4z)/(z + 2)²

At z = 1: (1 + 4)/9 = 5/9.

**Answer:** Double pole at z = 1 with residue 5/9; simple pole at z = −2 with residue 4/9.

**Check:** f(z) ~ 1/z for large z, so the sum of all residues must be 1: 5/9 + 4/9 = 1.

## Check yourself

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

1. What are the poles of z²/((z-1)²(z+2))?
   A. Simple pole at 1, double at -2
   B. Double pole at both
   C. Double pole at 1, simple pole at -2
   D. Simple poles at both
   Answer: C. (z-1)² gives order 2 at 1. (Key point: Pole orders)

2. What is the residue at z = -2?
   A. -4/9
   B. 4/3
   C. 1/9
   D. 4/9
   Answer: D. z²/(z-1)² at -2 = 4/9. (Key point: Simple-pole residue)

3. How is the residue at z = 1 found?
   A. d/dz[z²/(z+2)] at 1 = 5/9
   B. z²/(z+2) at 1 = 1/3
   C. d/dz[z²/(z+2)] at 1 = 4/9
   D. d²/dz² of z²/(z+2) at 1
   Answer: A. Derivative (z²+4z)/(z+2)² = 5/9 at z=1. (Key point: Double-pole residue)

4. Why must the residues sum to 1?
   A. f ~ 1/z² for large z
   B. f ~ 1/z for large z
   C. f is analytic at infinity
   D. All residues are equal
   Answer: B. 5/9 + 4/9 = 1. (Key point: Check)

## Related in Applied Mathematics

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