State and prove Cauchy's integral formula

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. ∎