---
title: "Laurent series expansion"
subject: "Mathematics III"
chapter: 2
key: "semester_3/mathematics_iii:laurent-series-expansion"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/laurent-series-expansion
site: BCT Engineering
---

# Laurent series expansion

Mathematics III · Chapter 2: Line, Surface and Volume Integrals

- Asked in: 2082 Chaitra (new course), 4 marks
- Video: [Laurent Series Explained | How to Determine Laurent Series | Complex Analysis #9](https://www.youtube.com/watch?v=RC15R-ktnUI) — TheMathCoach
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/laurent-series-expansion
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

Express the function 1/(z² − 3z + 2) in Laurent series form in the region 1 < |z| < 2.

## Short answer

Partial fractions: 1/(z-2)-1/(z-1). In 1<|z|<2, 1/(z-2) = -∑_{n=0}^∞ z^n/2^(n+1) and 1/(z-1) = ∑_{n=0}^∞ z^{-n-1}. Thus f(z) = -∑_{n=0}^∞ z^n/2^(n+1) - ∑_{n=1}^∞ z^{-n}.

## Answer

Resolve into partial fractions: z² − 3z + 2 = (z − 1)(z − 2).

```text
1/((z − 1)(z − 2)) = 1/(z − 2) − 1/(z − 1)
```

In the annulus 1 < |z| < 2 we have |z/2| < 1 and |1/z| < 1. Expand each term accordingly.

```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/2^(n+1)

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

Subtracting:

```text
f(z) = − Σ_{n=0}^∞ zⁿ/2^(n+1) − Σ_{n=1}^∞ 1/zⁿ
     = ... − 1/z³ − 1/z² − 1/z − 1/2 − z/4 − z²/8 − ...
```

**Answer:** f(z) = − Σ_{n=0}^∞ zⁿ/2^(n+1) − Σ_{n=1}^∞ z^(−n), valid for 1 < |z| < 2. The principal part has infinitely many negative powers because z = 1 is a singularity inside the inner circle.

## Check yourself

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

1. 1/((z-1)(z-2)) =
   A. 1/(z-1) - 1/(z-2)
   B. 1/(z-1) + 1/(z-2)
   C. 1/(z-2) - 1/(z-1)
   D. 1/(z-1)(z-2)
   Answer: C. Check: (z-1) - (z-2) = 1. (Key point: Partial fractions)

2. In 1<|z|<2 expand 1/(z-2) as:
   A. sum 1/z^(n+1)
   B. sum z^n/2^n
   C. -sum 2^n/z^(n+1)
   D. -sum z^n/2^(n+1) (since |z/2|<1)
   Answer: D. Expand in z/2. (Key point: Expansion of 1/(z-2))

3. 1/(z-1) is expanded as:
   A. sum 1/z^(n+1) (since |1/z|<1)
   B. -sum z^n
   C. sum z^n/2^(n+1)
   D. sum z^(n+1)
   Answer: A. Factor out z and expand in 1/z. (Key point: Expansion of 1/(z-1))

4. Why does the series have infinitely many negative powers?
   A. z = 2 is inside the inner circle
   B. z = 1 is a singularity inside the inner circle
   C. The region is |z|<1
   D. Because of the factor 1/2
   Answer: B. The 1/(z-1) term gives the negative powers. (Key point: Principal part)

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