---
title: "Zero of order m; poles and residues"
subject: "Applied Mathematics"
chapter: 1
key: "semester_4/applied_mathematics:zero-order-m-poles-residue"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/zero-order-m-poles-residue
site: BCT Engineering
---

# Zero of order m; poles and residues

Applied Mathematics · Chapter 1: Complex Analysis

- Asked in: 2073 Bhadra, 1+4 marks
- Video: [Zeros, Poles, & Removable Singularities Explained | Complex Analysis #7](https://www.youtube.com/watch?v=UhZ5UjR5NPE) — TheMathCoach
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/zero-order-m-poles-residue
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-1

## Question

Define zero of order m of function of complex variable. Determine the poles and residue at poles of the function f(z) = (1 + z)/((z + 2)(1 - z)²).

## Short answer

Zero of order m: f(z)=(z−a)^m g(z) with g analytic and g(a)≠0. f(z)=(1+z)/((z+2)(1−z)²) has a simple pole at z=−2 with residue −1/9 and a double pole at z=1 with residue 1/9.

## Answer

#### Zero of order m
A point z = a is a **zero of order m** of an analytic function f(z) if f(a) = f′(a) = … = f⁽ᵐ⁻¹⁾(a) = 0 but f⁽ᵐ⁾(a) ≠ 0. Equivalently f(z) = (z − a)ᵐ g(z) where g is analytic and g(a) ≠ 0. A zero of order 1 is a simple zero.

#### Poles and residues of f(z) = (1 + z)/((z + 2)(1 − z)²)
The denominator vanishes at z = −2 and z = 1; the numerator is non-zero at both (−1 and 2). So:

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

**Residue at z = −2 (simple pole):**

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

**Residue at z = 1 (double pole):**

Res = d/dz [ (z − 1)² f(z) ] at z = 1 = d/dz [ (1 + z)/(z + 2) ] at z = 1
= [ (z + 2) − (1 + z) ] / (z + 2)² = 1/(z + 2)²
= 1/9

**Answer:** simple pole at z = −2 with residue −1/9; double pole at z = 1 with residue 1/9. (Their sum is 0, as expected since f(z) ~ 1/z² at infinity.)

## Check yourself

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

1. What defines a zero of order m at a?
   A. f^(m)(a) = 0 only
   B. f(a) ≠ 0 and f^(m)(a) = 0
   C. f(a) = 0 and f'(a) ≠ 0 only for m=1, m>1 undefined
   D. f(a) = ... = f^(m-1)(a) = 0 but f^(m)(a) ≠ 0
   Answer: D. Order m means the first m derivatives vanish and the m-th does not. (Key point: Zero of order m)

2. What are the poles of (1+z)/((z+2)(1-z)²)?
   A. Simple pole at z = -2, double pole at z = 1
   B. Double pole at z = -2, simple at z = 1
   C. Simple poles at -2 and 1
   D. Simple pole at -1
   Answer: A. (1-z)² = (z-1)² gives order 2 at z=1; numerator is non-zero at both. (Key point: Poles)

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

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

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