---
title: "Cauchy theorem vs formula; evaluate integral"
subject: "Applied Mathematics"
chapter: 1
key: "semester_4/applied_mathematics:cif-distinguish-e-z-z-1-z-2"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/cif-distinguish-e-z-z-1-z-2
site: BCT Engineering
---

# Cauchy theorem vs formula; evaluate integral

Applied Mathematics · Chapter 1: Complex Analysis

- Asked in: 2075 Baisakh, 1+4 marks
- 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-distinguish-e-z-z-1-z-2
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-1

## Question

Distinguish between Cauchy integral Theorem and Cauchy integral formula. Using Cauchy integral formula evaluate ∮_C e^z/((z + 1)(z - 2)) dz where C is the circle |z - 1| = 3.

## Short answer

Cauchy integral theorem says ∮_ C f(z)dz=0 when f is analytic inside and on C; integral formula gives ∮_ C f(z)/(z-a)dz = 2πi f(a) for a inside C. For ∮_ C e^z/(( z+1)(z-2)) dz with C:|z-1|=3, both poles z=−1,2 lie inside. Write as (1/3)[ e^z/(z-2) − e^z/(z+1) ]; applying formula gives (2πi/3)(e^2 − e^{-1}).

## Answer

#### Cauchy integral theorem vs Cauchy integral formula
| Point | Cauchy integral theorem | Cauchy integral formula |
|---|---|---|
| Statement | ∮_C f(z) dz = 0 | f(a) = (1/2πi) ∮_C f(z)/(z − a) dz |
| Condition | f analytic inside and on C | f analytic inside and on C, a inside C |
| Integrand | f(z) itself | f(z)/(z − a), has a singularity at a |
| Result | Always zero | Gives the value of f at the interior point a |
| Use | Show integrals vanish, path independence | Evaluate integrals with poles; get derivatives |

#### Solution
C: |z − 1| = 3, a circle with centre 1 and radius 3.

- z = −1: |−1 − 1| = 2 < 3, inside
- z = 2: |2 − 1| = 1 < 3, inside

Both poles are inside C. Partial fractions:

1/((z + 1)(z − 2)) = (1/3) [ 1/(z − 2) − 1/(z + 1) ]

Apply CIF with f(z) = e^z (analytic everywhere):

```text
∮ e^z/(z − 2) dz = 2πi f(2)  = 2πi e²
∮ e^z/(z + 1) dz = 2πi f(−1) = 2πi e⁻¹
```

I = (1/3) [ 2πi e² − 2πi e⁻¹ ] = (2πi/3)(e² − e⁻¹)

**Answer: (2πi/3)(e² − 1/e) ≈ 14.705 i**

## Check yourself

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

1. Cauchy's theorem gives:
   A. none
   B. ∮f = 2πi
   C. ∮f = 0
   D. ∮f = f(a)
   Answer: C. Standard. (Key point: Theorem)

2. The formula yields:
   A. 0
   B. f(a)
   C. 1
   D. infinity
   Answer: B. Value at interior point. (Key point: Contrast)

3. For |z−1|=3 both z=−1 and z=2 are:
   A. Outside
   B. Absent
   C. Inside
   D. On C
   Answer: C. Distances 2 and 1. (Key point: Poles)

4. The answer is:
   A. (2πi)(e² − 1/e)
   B. (2πi/3)(e² − 1/e)
   C. (2πi/3)(e² + 1/e)
   D. (πi/3)(e²−1/e)
   Answer: B. Partial fractions factor 1/3. (Key point: Result)

## Close variants set in other exams

- [Evaluate ∮e^{2z}/((z-1)(z-2)) by CIF](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Acif-e2z-z-1-z-2) — Applied Mathematics, chapter 1, asked 8×

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