---
title: "State and prove Cauchy's integral formula"
subject: "Applied Mathematics"
chapter: 1
key: "semester_4/applied_mathematics:cauchy-integral-formula-prove"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/cauchy-integral-formula-prove
site: BCT Engineering
---

# State and prove Cauchy's integral formula

Applied Mathematics · Chapter 1: Complex Analysis

- Asked in: 2073 Bhadra, 5 marks; 2071 Magh, 5 marks; 2070 Bhadra, 5 marks; 2071 Bhadra; 2068 Bhadra (asked 5 times)
- Video: [Cauchy's Integral Formula (Proof)](https://www.youtube.com/watch?v=ZQCtNuIr55k) — IGNITED MINDS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/cauchy-integral-formula-prove
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-1

## Question

State and prove Cauchy's integral formula.

## Short answer

If f is analytic inside and on a simple closed contour C and a is inside, then f(a)=1/(2πi)∮ f(z)/(z−a) dz. Proof: reduce C to a small circle around a, set z−a=ρe^{iθ}, then let ρ→0 to get ∮=2πi f(a). Derivatives satisfy f^(n)(a)=n!/(2πi)∮ f(z)/(z−a)^(n+1) dz.

## Answer

**Statement.** If f(z) is analytic inside and on a simple closed contour C, and *a* is any point inside C, then

f(a) = (1/2πi) ∮_C f(z)/(z − a) dz

(C is traversed counter-clockwise.) Also, f′(a) = (1/2πi) ∮_C f(z)/(z − a)² dz and, in general, f⁽ⁿ⁾(a) = (n!/2πi) ∮_C f(z)/(z − a)ⁿ⁺¹ dz.

#### Proof
Let g(z) = f(z)/(z − a). It is analytic everywhere in and on C except at z = a.

Draw a small circle C₁: |z − a| = ρ, lying entirely inside C. In the region between C and C₁ (a doubly connected region) g(z) is analytic, so by Cauchy's theorem for multiply connected regions

∮_C f(z)/(z − a) dz = ∮_C₁ f(z)/(z − a) dz.

On C₁, put z − a = ρe^{iθ}, so dz = iρe^{iθ} dθ, and 0 ≤ θ ≤ 2π. Then

∮_C₁ f(z)/(z − a) dz = ∫₀^{2π} f(a + ρe^{iθ}) · iρe^{iθ}/(ρe^{iθ}) dθ = i ∫₀^{2π} f(a + ρe^{iθ}) dθ.

The left side does not depend on ρ, so we may let ρ → 0. Since f is continuous at *a*, f(a + ρe^{iθ}) → f(a), so

∮_C f(z)/(z − a) dz = i ∫₀^{2π} f(a) dθ = 2πi f(a).

(Rigorous check: |∮_C₁ [f(z) − f(a)]/(z − a) dz| ≤ 2πρ · ε/ρ = 2πε, where |f(z) − f(a)| < ε for small ρ; so this part tends to 0 and the remaining part is f(a) ∮_C₁ dz/(z − a) = 2πi f(a).)

Therefore

f(a) = (1/2πi) ∮_C f(z)/(z − a) dz. ∎

## Check yourself

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

1. According to Cauchy's integral formula, what is f(a) for a inside C?
   A. (1/2πi) ∮ f(z)/(z − a) dz
   B. 2πi ∮ f(z)/(z − a) dz
   C. (1/2πi) ∮ f(z)/(z − a)^2 dz
   D. (1/2πi) ∮ f(z) dz
   Answer: A. f(a) = (1/2πi) ∮_C f(z)/(z − a) dz, with C counter-clockwise. (Key point: Statement)

2. Why is a small circle C1: |z − a| = ρ drawn inside C in the proof?
   A. It makes the integral over C zero
   B. It removes the need for f to be continuous
   C. It makes f(z) constant on C
   D. g(z)=f(z)/(z − a) is analytic between C and C1, so the integral over C equals that over C1
   Answer: D. The region between C and C1 has no singularity of g, so Cauchy's theorem for multiply connected regions equates the two integrals. (Key point: Small circle)

3. On C1, with z − a = ρe^{iθ}, the integral of f(z)/(z − a) dz becomes:
   A. ∫₀^{2π} f(a + ρe^{iθ}) dθ / i
   B. ρ ∫₀^{2π} f(a + ρe^{iθ}) dθ
   C. i ρ ∫₀^{2π} f(a + ρe^{iθ}) e^{iθ} dθ
   D. i ∫₀^{2π} f(a + ρe^{iθ}) dθ
   Answer: D. dz = iρe^{iθ}dθ cancels the denominator ρe^{iθ}, leaving i f dθ. (Key point: Parametrisation)

4. Why can we let ρ → 0 to get 2πi f(a)?
   A. Because dθ tends to zero
   B. The left side is independent of ρ and f is continuous at a, so f(a + ρe^{iθ}) → f(a)
   C. Because f(a) = 0
   D. Because the circle shrinks the contour C
   Answer: B. The integral over C is fixed, and continuity gives f → f(a), so i·2π·f(a) = 2πi f(a). (Key point: Limit ρ→0)

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