---
title: "Singular points, poles, residue of z²/((z-2)(z²+1))"
subject: "Applied Mathematics"
chapter: 1
key: "semester_4/applied_mathematics:poles-residues-z2-over-z-2-z2-1"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/poles-residues-z2-over-z-2-z2-1
site: BCT Engineering
---

# Singular points, poles, residue of z²/((z-2)(z²+1))

Applied Mathematics · Chapter 1: Complex Analysis

- Asked in: 2069 Poush
- Video: [To find poles and corresponding Residues of the function-complex Analysis](https://www.youtube.com/watch?v=-C6ETPbrfH0) — Btech Maths Hub
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/poles-residues-z2-over-z-2-z2-1
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-1

## Question

Define singular points and poles. Compute the residue of f(z) = z²/((z - 2)(z² + 1)) at its pole(s).

## Short answer

A singular point is where f is not analytic; a pole is an isolated singularity where f(z)=g(z)/(z−a)^m with g(a)≠0. f(z)=z²/(( z−2)(z²+1)) has simple poles at z=2 (residue 4/5), z=i (residue (1−2i)/10), z=−i (residue (1+2i)/10).

## Answer

#### Definitions
- **Singular point:** a point z = a where f(z) is not analytic (it fails to be differentiable there), although every neighbourhood of a contains points where f is analytic. An **isolated singular point** has a punctured neighbourhood in which f is analytic.
- **Pole:** an isolated singular point z = a where f(z) = g(z)/(z − a)ᵐ with g analytic and g(a) ≠ 0. The integer m is the order of the pole (m = 1: simple pole). Near a pole, |f(z)| → ∞.

#### Residues of f(z) = z²/((z − 2)(z² + 1))
Denominator = (z − 2)(z − i)(z + i). The numerator z² does not vanish at 2, i or −i, so there are three **simple poles**: z = 2, z = i, z = −i.

For a simple pole, Res at a = lim (z → a) (z − a) f(z).

- At z = 2: Res = z²/(z² + 1) = 4/5
- At z = i: Res = z²/((z − 2)(z + i)) = i²/((i − 2)(2i)) = −1/(−2 − 4i) = 1/(2 + 4i) = (2 − 4i)/20 = **(1 − 2i)/10**
- At z = −i: Res = z²/((z − 2)(z − i)) = (−1)/((−i − 2)(−2i)) = −1/(−2 + 4i) = 1/(2 − 4i) = **(1 + 2i)/10**

**Check:** sum = 4/5 + (1 − 2i)/10 + (1 + 2i)/10 = 4/5 + 1/5 = 1, which equals the coefficient of 1/z in f at infinity (f ~ 1/z), as it should.

**Answer:** Res(2) = 4/5, Res(i) = (1 − 2i)/10, Res(−i) = (1 + 2i)/10.

## Check yourself

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

1. What is a pole?
   A. Any point where f is zero
   B. Any point where f is continuous
   C. Isolated singular point where f = g/(z-a)^m with g analytic and g(a) ≠ 0
   D. Point where f is analytic but its derivative is zero
   Answer: C. Near a pole |f| → ∞. (Key point: Pole definition)

2. What poles does z²/((z-2)(z²+1)) have?
   A. Simple pole at 2 only
   B. Double pole at i
   C. Poles at 2 and 1
   D. Simple poles at 2, i, -i
   Answer: D. Denominator = (z-2)(z-i)(z+i) and the numerator is non-zero at each. (Key point: Pole list)

3. What is the residue at z = 2?
   A. 4/5
   B. 4/3
   C. 1/5
   D. 2/5
   Answer: A. z²/(z²+1) at 2 = 4/5. (Key point: Residue at 2)

4. What check confirms the three residues?
   A. Their sum is 0
   B. Their sum is 1, since f ~ 1/z at infinity
   C. Their product is 1
   D. Each is real
   Answer: B. 4/5 + (1-2i)/10 + (1+2i)/10 = 1. (Key point: Check)

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