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

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

Applied Mathematics · Chapter 1: Complex Analysis

- Asked in: 2072 Asoj, 5 marks; 2068 Bhadra (asked 2 times)
- Video: [19. Taylor's and Laurent's Series | Problem#1 | Most Important | Complete Concept](https://www.youtube.com/watch?v=OZAZImytcXE) — MKS TUTORIALS by Manoj Sir
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/laurent-1-over-z2-3z-2
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-1

## Question

Write the statement of Taylor's theorem. Find the Laurent series for the function f(z) = 1/(z² - 3z + 2) in the region 1 < |z| < 2.

## Short answer

Taylor’s theorem states that an analytic function inside a circle can be expanded as a power series Σ aₙ(z−a)ⁿ with aₙ=f⁽ⁿ⁾(a)/n!. For f(z)=1/(z²−3z+2) in 1<|z|<2, partial fractions yield f(z)=−Σ_{n=0}^∞ zⁿ/2ⁿ⁺¹ − Σ_{n=1}^∞ 1/zⁿ.

## Answer

#### Taylor's theorem
If f(z) is analytic inside and on the circle C: |z − a| = R, then for every z inside C,

f(z) = Σ_{n=0}^{∞} aₙ (z − a)ⁿ, where aₙ = f⁽ⁿ⁾(a)/n! = (1/2πi) ∮_C f(ζ)/(ζ − a)ⁿ⁺¹ dζ.

#### Laurent series in 1 < |z| < 2
f(z) = 1/(z² − 3z + 2) = 1/((z − 1)(z − 2)).

Partial fractions: 1/((z − 1)(z − 2)) = 1/(z − 2) − 1/(z − 1).

The poles are at z = 1 and z = 2; in the ring 1 < |z| < 2 we have |z| > 1 and |z| < 2, so expand each fraction in the matching way.

**Term 1: 1/(z − 2).** Since |z/2| < 1:

```text
1/(z − 2) = −1/(2 − z) = −(1/2) · 1/(1 − z/2)
          = −(1/2) Σ_{n=0}^{∞} (z/2)ⁿ = −Σ_{n=0}^{∞} zⁿ/2ⁿ⁺¹
```

**Term 2: 1/(z − 1).** Since |1/z| < 1:

```text
1/(z − 1) = (1/z) · 1/(1 − 1/z) = (1/z) Σ_{n=0}^{∞} z⁻ⁿ = Σ_{n=1}^{∞} z⁻ⁿ
```

#### Result
f(z) = 1/(z − 2) − 1/(z − 1)

**f(z) = − Σ_{n=0}^{∞} zⁿ/2ⁿ⁺¹ − Σ_{n=1}^{∞} 1/zⁿ,  for 1 < |z| < 2.**

Written out:

f(z) = … − 1/z³ − 1/z² − 1/z − 1/2 − z/4 − z²/8 − …

The principal part (negative powers) comes from the pole at z = 1, and the analytic part (non-negative powers) from the pole at z = 2.

## Check yourself

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

1. In Taylor's theorem, aₙ equals:
   A. fⁿ(a)·n!
   B. f(a)/n!
   C. fⁿ(a)/n!
   D. fⁿ(a)
   Answer: C. aₙ = fⁿ(a)/n!. (Key point: Taylor's coefficient)

2. In 1 < |z| < 2, the fraction 1/(z−2) is expanded in:
   A. Powers of 1/z
   B. Powers of z/2
   C. Powers of 2/z
   D. Powers of z − 1
   Answer: B. |z/2| < 1. (Key point: Region)

3. 1/(z−1) in 1<|z|<2 is expanded in:
   A. A single term
   B. Powers of z/2
   C. Powers of z
   D. Powers of 1/z, as Σ z^{-n} from n=1
   Answer: D. |1/z| < 1. (Key point: Pole at 1)

4. The Laurent series is:
   A. −Σ_{n≥0} zⁿ/2ⁿ⁺¹ + Σ_{n≥1} z⁻ⁿ
   B. −Σ_{n≥0} zⁿ/2ⁿ⁺¹ − Σ_{n≥1} z⁻ⁿ
   C. Σ_{n≥0} zⁿ − Σ_{n≥1} z⁻ⁿ
   D. Σ_{n≥0} zⁿ/2ⁿ⁺¹ + Σ_{n≥1} z⁻ⁿ
   Answer: B. 1/(z−2) = −Σ zⁿ/2ⁿ⁺¹ and −1/(z−1) = −Σ z⁻ⁿ. (Key point: Result)

## Related in Applied Mathematics

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