---
title: "Evaluate ∮e^{2z}/((z-1)(z-2)) by CIF"
subject: "Applied Mathematics"
chapter: 1
key: "semester_4/applied_mathematics:cif-e2z-z-1-z-2"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/cif-e2z-z-1-z-2
site: BCT Engineering
---

# Evaluate ∮e^{2z}/((z-1)(z-2)) by CIF

Applied Mathematics · Chapter 1: Complex Analysis

- Asked in: 2073 Magh, 5 marks; 2071 Magh, 5 marks; 2070 Bhadra, 5 marks (asked 3 times)
- Video: [Complex Analysis | Unit 2 | Lecture 13 | Example of Cauchy's Integral Formula](https://www.youtube.com/watch?v=wUOFqFBnQDk) — Ranjan Khatu
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/cif-e2z-z-1-z-2
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-1

## Question

Evaluate ∮_C e^{2z}/((z-1)(z-2)) dz where C is the circle |z| = 3, using Cauchy's integral formula.

## Short answer

Both poles z=1 and z=2 lie inside |z|=3. Partial fractions give integrand = e^{2z}/(z−2) − e^{2z}/(z−1); by Cauchy's integral formula, I=2πi e⁴ − 2πi e² = 2πi(e⁴−e²).

## Answer

**Cauchy's integral formula (CIF):** if f(z) is analytic inside and on a simple closed curve C and *a* lies inside C, then ∮_C f(z)/(z − a) dz = 2πi f(a).

#### Solution
The poles z = 1 and z = 2 both lie inside C: |z| = 3, since |1| < 3 and |2| < 3.

Partial fractions: 1/((z − 1)(z − 2)) = 1/(z − 2) − 1/(z − 1).

So I = ∮_C e^{2z}/(z − 2) dz − ∮_C e^{2z}/(z − 1) dz.

Take f(z) = e^{2z}, which is analytic everywhere. By CIF:

```text
∮_C e^{2z}/(z − 2) dz = 2πi f(2) = 2πi e⁴
∮_C e^{2z}/(z − 1) dz = 2πi f(1) = 2πi e²
```

I = 2πi e⁴ − 2πi e² = 2πi (e⁴ − e²)

**Answer: ∮_C e^{2z}/((z − 1)(z − 2)) dz = 2πi (e⁴ − e²) ≈ 296.62 i**

## Check yourself

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

1. For C: |z| = 3, which poles of e^{2z}/((z−1)(z−2)) lie inside?
   A. Only z = 2
   B. Both z = 1 and z = 2
   C. None
   D. Only z = 1
   Answer: B. |1| < 3 and |2| < 3. (Key point: Poles inside)

2. 1/((z−1)(z−2)) equals:
   A. (1/2)[1/(z−2) − 1/(z−1)]
   B. 1/(z−2) − 1/(z−1)
   C. 1/(z−1) − 1/(z−2)
   D. 1/(z−1) + 1/(z−2)
   Answer: B. 1/(z−2) − 1/(z−1) = 1/((z−1)(z−2)). (Key point: Partial fractions)

3. With f(z) = e^{2z}, ∮ e^{2z}/(z−2) dz equals:
   A. 2πi e^{2}/2
   B. 2πi e^2
   C. 2πi e^4
   D. πi e^4
   Answer: C. CIF gives 2πi f(2) = 2πi e^4. (Key point: Apply CIF)

4. The value of the integral is:
   A. 2πi(e^4 − e^2)
   B. 2πi e^6
   C. 2πi(e^4 + e^2)
   D. 2πi(e^2 − e^4)
   Answer: A. Subtract: 2πi e^4 − 2πi e^2. (Key point: Result)

## Close variants set in other exams

- [Evaluate ∮(4-3z)/(z(z-1)(z-2)) on |z|=3/2](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Acif-4-3z-over-z-z-1-z-2) — Applied Mathematics, chapter 1, asked 6×
- [Evaluate ∮z/((z-1)(z-3)) on |z|=3/2](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Acif-z-over-z-1-z-3) — Applied Mathematics, chapter 1, asked 6×
- [Evaluate ∮e^z/((z-1)(z-3)) on |z|=2](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Acif-e-z-z-1-z-3) — Applied Mathematics, chapter 1, asked 4×
- [Cauchy theorem vs formula; evaluate integral](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Acif-distinguish-e-z-z-1-z-2) — Applied Mathematics, chapter 1, asked 4×

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