---
title: "Residues of a function"
subject: "Mathematics III"
chapter: 2
key: "semester_3/mathematics_iii:residues-of-a-function"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/residues-of-a-function
site: BCT Engineering
---

# Residues of a function

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

- Asked in: 2082 Chaitra (new course), 2 marks
- Video: [How to find the Residues of a Complex Function](https://www.youtube.com/watch?v=sSj7z-pz-yY) — Faculty of Khan
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/residues-of-a-function
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

Find the residue of f(z) = (2z + 1)/((z − 3)(z + 5)) at each of the singularities.

## Short answer

At simple pole z=3: Res = (2·3+1)/(3+5)=7/8. At z=−5: Res = (2·(−5)+1)/(−5−3)=9/8.

## Answer

f(z) = (2z + 1)/((z − 3)(z + 5)) has two simple poles, z = 3 and z = −5. For a simple pole at z = a, Res = lim (z − a) f(z).

```text
At z = 3:   Res = (2z + 1)/(z + 5) at z = 3 = 7/8
At z = −5:  Res = (2z + 1)/(z − 3) at z = −5 = (−9)/(−8) = 9/8
```

**Answer:** residue = 7/8 at z = 3 and 9/8 at z = −5. (Check: the sum of the residues, 7/8 + 9/8 = 2, equals the limit of z·f(z) as z → ∞, which is 2, as it must.)

## Check yourself

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

1. The singularities z=3 and z=-5 are:
   A. Double poles
   B. Essential singularities
   C. Removable singularities
   D. Simple poles
   Answer: D. Each appears with power 1 in the denominator. (Key point: Type)

2. Residue at simple pole a is:
   A. lim (z-a) f(z)
   B. f(a)
   C. lim f(z)/(z-a)
   D. derivative of f at a
   Answer: A. Multiply by (z-a) and take limit. (Key point: Method)

3. Residue at z=3 is:
   A. 9/8
   B. 7/8
   C. 7/5
   D. -7/8
   Answer: B. (2*3+1)/(3+5) = 7/8. (Key point: Residue at 3)

4. Residue at z=-5 is:
   A. 7/8
   B. -9/8
   C. 9/8
   D. 9/2
   Answer: C. (-9)/(-8) = 9/8. (Key point: Residue at -5)

## Close variants set in other exams

- [Residues of a function](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Aresidues-of-a-function-two) — Mathematics III, chapter 2, asked 3×
- [Residues of a function](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Aresidues-of-a-function-four) — Mathematics III, chapter 2, asked 3×

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