---
title: "Isolated pole; ∮(z-1)/((z+1)²(z-2))"
subject: "Applied Mathematics"
chapter: 1
key: "semester_4/applied_mathematics:residue-z-1-over-z1sq-z-2"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/residue-z-1-over-z1sq-z-2
site: BCT Engineering
---

# Isolated pole; ∮(z-1)/((z+1)²(z-2))

Applied Mathematics · Chapter 1: Complex Analysis

- Asked in: 2075 Baisakh, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/residue-z-1-over-z1sq-z-2
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-1

## Question

Define an isolated pole. Using Cauchy's residue theorem evaluate ∮_C (z - 1)/((z + 1)²(z - 2)) dz where C is the circle |z - i| = 2.

## Short answer

An isolated pole is a singularity where the Laurent series has finitely many negative powers. For the integral around |z−i|=2, only the double pole at z=−1 lies inside C. Its residue is −1/9, so ∮_ C f(z) dz = −2πi/9.

## Answer

#### Isolated pole
A point z = a is an **isolated singularity** of f(z) if f is analytic in a punctured neighbourhood 0 < |z − a| < δ but not at a. It is an **isolated pole** (of order m) if the Laurent series about a has only finitely many negative powers:

f(z) = b_m/(z − a)ᵐ + … + b₁/(z − a) + a₀ + a₁(z − a) + …, with b_m ≠ 0.

Equivalently f(z) = g(z)/(z − a)ᵐ with g analytic and g(a) ≠ 0.

#### Evaluation
f(z) = (z − 1)/((z + 1)²(z − 2)); C: |z − i| = 2 (centre i, radius 2).

Poles: z = −1 (double) and z = 2 (simple).

- Distance from i to −1 = |−1 − i| = √2 ≈ 1.41 < 2, so z = −1 is **inside** C.
- Distance from i to 2 = |2 − i| = √5 ≈ 2.24 > 2, so z = 2 is **outside** C.

Residue at the double pole z = −1:

Res = d/dz [ (z − 1)/(z − 2) ] at z = −1
= [ (z − 2) − (z − 1) ] / (z − 2)² = −1/(z − 2)²
= −1/(−3)² = −1/9

By Cauchy's residue theorem, ∮_C f dz = 2πi × (sum of residues inside C):

**Answer: ∮_C f(z) dz = 2πi(−1/9) = −2πi/9**

## Check yourself

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

1. What defines an isolated pole of order m at a?
   A. Laurent series about a has finitely many negative powers, lowest (z-a)^-m
   B. f is analytic at a with f(a)=0
   C. Laurent series has infinitely many negative powers
   D. f is bounded near a
   Answer: A. Equivalent to f = g/(z-a)^m with g analytic and g(a) ≠ 0. (Key point: Isolated pole)

2. For C: |z-i| = 2, which poles are inside?
   A. z = 2 only
   B. z = -1 only (distance √2); z = 2 is at √5 > 2
   C. Both
   D. Neither
   Answer: B. |-1-i| = √2 < 2 and |2-i| = √5 > 2. (Key point: Poles inside)

3. How is the residue at the double pole z = -1 computed?
   A. lim (z+1)f(z) = 2/3
   B. d/dz[(z-1)/(z-2)] at z=-1 = 1/9
   C. d/dz[(z-1)/(z-2)] at z = -1 = -1/9
   D. (z-1)/(z-2) at z=-1 = 2/3
   Answer: C. For order 2, differentiate (z+1)² f(z); the derivative is -1/(z-2)² = -1/9. (Key point: Double-pole residue)

4. What is the value of the integral?
   A. 2πi/9
   B. -2πi/3
   C. 0
   D. -2πi/9
   Answer: D. 2πi × (-1/9) = -2πi/9. (Key point: Result)

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