---
title: "Laurent series of (z²-1)/((z+2)(z+3)), 2<|z|<3"
subject: "Applied Mathematics"
chapter: 1
key: "semester_4/applied_mathematics:laurent-z2-1-over-z2-z3"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/laurent-z2-1-over-z2-z3
site: BCT Engineering
---

# Laurent series of (z²-1)/((z+2)(z+3)), 2<|z|<3

Applied Mathematics · Chapter 1: Complex Analysis

- Asked in: 2070 Bhadra, 5 marks
- Video: [Taylor’s and Laurent’s series/Complex Integration(Laurent’s Series) problems in simple explanation](https://www.youtube.com/watch?v=VTJ-zYlR78k) — SK Maths Tutorial
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/laurent-z2-1-over-z2-z3
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-1

## Question

Obtain the Laurent series which represents the function f(z) = (z² - 1)/((z + 2)(z + 3)), 2 < |z| < 3.

## Short answer

After division and partial fractions, in 2<|z|<3: f(z) = 1 + 3 Σ_{n=0}^∞ (−2)ⁿ z^{−n−1} − (8/3) Σ_{n=0}^∞ (−1)ⁿ zⁿ/3ⁿ. The expansion is … + 12/z³ − 6/z² + 3/z − 5/3 + 8z/9 − ….

## Answer

f(z) = (z² − 1)/((z + 2)(z + 3)).

Numerator and denominator have the same degree, so first divide:

(z² − 1)/(z² + 5z + 6) = 1 − (5z + 7)/((z + 2)(z + 3))

Partial fractions: (5z + 7)/((z + 2)(z + 3)) = A/(z + 2) + B/(z + 3)

- A = (5(−2) + 7)/(−2 + 3) = −3
- B = (5(−3) + 7)/(−3 + 2) = 8

so f(z) = 1 + 3/(z + 2) − 8/(z + 3).

In the ring 2 < |z| < 3: |2/z| < 1 and |z/3| < 1.

**Term 3/(z + 2)** (|z| > 2, powers of 1/z):

3/(z + 2) = (3/z) · 1/(1 + 2/z) = 3 Σ (n = 0 to ∞) (−2)ⁿ / zⁿ⁺¹ = 3/z − 6/z² + 12/z³ − …

**Term −8/(z + 3)** (|z| < 3, powers of z/3):

−8/(z + 3) = −(8/3) · 1/(1 + z/3) = −(8/3) Σ (n = 0 to ∞) (−1)ⁿ zⁿ/3ⁿ
= −8/3 + 8z/9 − 8z²/27 + …

**Answer:**

f(z) = 1 + 3 Σ (n = 0 to ∞) (−2)ⁿ z^(−n−1) − (8/3) Σ (n = 0 to ∞) (−1)ⁿ zⁿ/3ⁿ

i.e. f(z) = … + 12/z³ − 6/z² + 3/z − 5/3 + 8z/9 − 8z²/27 + …

(constant term: 1 − 8/3 = −5/3).

## Check yourself

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

1. Why is the division 1 - (5z+7)/((z+2)(z+3)) done first?
   A. The poles are repeated
   B. Numerator and denominator have the same degree
   C. The ring excludes z=0
   D. The numerator vanishes at z=-2
   Answer: B. Equal degrees means a constant (1) must be separated before partial fractions. (Key point: Preliminary division)

2. What is f(z) after partial fractions?
   A. 1 - 3/(z+2) + 8/(z+3)
   B. 1 + 3/(z+2) + 8/(z+3)
   C. 1 + 3/(z+2) - 8/(z+3)
   D. 3/(z+2) - 8/(z+3)
   Answer: C. (5z+7)/((z+2)(z+3)) = -3/(z+2) + 8/(z+3), so f = 1 + 3/(z+2) - 8/(z+3). (Key point: Decomposition)

3. In 2 < |z| < 3, how are the two terms expanded?
   A. Both in powers of 1/z
   B. Both in powers of z
   C. 3/(z+2) in powers of z/2; 8/(z+3) in powers of 3/z
   D. 3/(z+2) in powers of 1/z; 8/(z+3) in powers of z/3
   Answer: D. |2/z|<1 and |z/3|<1 are the convergent conditions. (Key point: Ring choice)

4. What is the constant term of the series?
   A. -5/3 (that is 1 - 8/3)
   B. 1
   C. -8/3
   D. 11/3
   Answer: A. The 1 contributes 1 and the z/3 series contributes -8/3, total -5/3. (Key point: Constant term)

## Close variants set in other exams

- [Laurent series of 1/((1+z²)(z+2)), 1<|z|<2](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Alaurent-1-over-1-z2-z-2) — Applied Mathematics, chapter 1, asked 2×

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