---
title: "Laurent series of z/((z+1)(z+2)), 1<|z|<2"
subject: "Applied Mathematics"
chapter: 1
key: "semester_4/applied_mathematics:laurent-z-over-z1-z2"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/laurent-z-over-z1-z2
site: BCT Engineering
---

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

Applied Mathematics · Chapter 1: Complex Analysis

- Asked in: 2073 Bhadra, 1+4 marks
- Video: [find the Laurent series 7z-2/(z+1)(z)(z-2) important question](https://www.youtube.com/watch?v=U62lzpuxVLk) — MB Uday
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/laurent-z-over-z1-z2
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-1

## Question

State Taylor's theorem. Find the Laurent's series representation of the function f(z) = z/((z + 1)(z + 2)) in the annular region between |z| = 1 and |z| = 2.

## Short answer

Taylor’s theorem: if f is analytic on a disc |z−a|<R, f(z) = Σ aₙ(z−a)ⁿ with aₙ = f⁽ⁿ⁾(a)/n!. For 1<|z|<2, the Laurent series of z/((z+1)(z+2)) is Σ_{n=0}^∞ (−1)ⁿ zⁿ/2ⁿ + Σ_{n=1}^∞ (−1)ⁿ z⁻ⁿ.

## Answer

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

f(z) = Σ (n = 0 to ∞) aₙ (z − a)ⁿ,  where aₙ = f⁽ⁿ⁾(a)/n! = (1/2πi) ∮_C f(w)/(w − a)ⁿ⁺¹ dw.

#### Laurent series of f(z) = z/((z + 1)(z + 2)) for 1 < |z| < 2
Partial fractions: z/((z + 1)(z + 2)) = −1/(z + 1) + 2/(z + 2)  (check: −(z + 2) + 2(z + 1) = z).

In the region 1 < |z| < 2 we have |1/z| < 1 and |z/2| < 1, so expand each term in the convergent form.

**First term** (|z| > 1, use powers of 1/z):

−1/(z + 1) = −(1/z) · 1/(1 + 1/z) = −(1/z)[1 − 1/z + 1/z² − …]
= −1/z + 1/z² − 1/z³ + … = Σ (n = 1 to ∞) (−1)ⁿ z⁻ⁿ

**Second term** (|z| < 2, use powers of z/2):

2/(z + 2) = 1/(1 + z/2) = 1 − z/2 + z²/4 − z³/8 + … = Σ (n = 0 to ∞) (−1)ⁿ zⁿ/2ⁿ

**Answer:**

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

= … − 1/z³ + 1/z² − 1/z + 1 − z/2 + z²/4 − z³/8 + …

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

## Check yourself

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

1. Which formula gives the Taylor coefficients in Taylor's theorem?
   A. a_n = f(a)/n!
   B. a_n = (1/2πi)∮ f(w)(w-a)^n dw
   C. a_n = f^(n)(a)/n! = (1/2πi)∮ f(w)/(w-a)^{n+1} dw
   D. a_n = f^(n)(a) n!
   Answer: C. The coefficients are f^(n)(a)/n!, equal to the Cauchy integral formula. (Key point: Taylor's theorem)

2. In 1 < |z| < 2, how is 1/(z+1) expanded?
   A. In powers of z, since |z| < 1
   B. In powers of z/2
   C. In powers of (z+1)
   D. In powers of 1/z, since |1/z| < 1
   Answer: D. |z|>1 so 1/(z+1) = (1/z)/(1+1/z) is expanded in 1/z. (Key point: Choice of expansion)

3. What are the partial fractions of z/((z+1)(z+2))?
   A. -1/(z+1) + 2/(z+2)
   B. 1/(z+1) + 2/(z+2)
   C. -1/(z+1) - 2/(z+2)
   D. 2/(z+1) - 1/(z+2)
   Answer: A. Check: -(z+2) + 2(z+1) = z. (Key point: Partial fractions)

4. Which pole produces the principal (negative power) part?
   A. The pole at z = -2
   B. The pole at z = -1 (inner boundary)
   C. Both poles equally
   D. Neither; there is no principal part
   Answer: B. The 1/(z+1) term is expanded in 1/z, giving negative powers; the z=-2 term gives the analytic part. (Key point: Principal part)

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