Cauchy's Integral Formula and Derivatives of Every Order

Cauchy's integral formula. Let γ be a simple closed piecewise C1 curve traversed once counterclockwise, f holomorphic on an open set containing γ and its filled interior, and a strictly inside. Then the loop integral of f(z)/(z - a) dz is 2πi f(a), and f(a) is that integral divided by 2πi. Worked example: for f(z) = z² with a = 0.5 strictly inside the unit circle, the integral is 2πi × 0.25 = πi/2. Clockwise traversal reverses the sign; traversing k times multiplies by k, the signed turn count.

Why the formula holds. The quotient f(z)/(z - a) is holomorphic on the region between γ and a small circle C_ρ around a, so deformation preserves the integral I. Parametrising a + ρe^(it) gives I = i times the integral from 0 to 2π of f(a + ρe^(it)) dt. |I - 2πi f(a)| ≤ 2π times the supremum of |f(z) - f(a)| on C_ρ. Differentiability at a gives f(a + h) - f(a) = ((f(a + h) - f(a))/h) h, which tends to 0, so the supremum tends to 0 as ρ shrinks. Since I does not depend on ρ, I = 2πi f(a).

Derivatives of every order. Differentiating under the integral sign n times gives f^(n)(a) = (n!/(2πi)) I_n, where I_n is the loop integral of f(z)/(z - a)^(n+1) dz. Solving for the integral, I_n = 2πi f^(n)(a)/n!, with n a nonnegative integer and 0! = 1. Worked inverse: for f(z) = z³, a = 1 and n = 2, f''(1) = 6, so I_2 = 2πi × 6/2 = 6πi on any circle centred at 1 traversed once counterclockwise.

Why differentiation is justified. Fix the contour γ and let d be the distance from a to γ, which is positive. For any w with |w - a| < d/2, every contour point z has |z - w| ≥ d/2, so each finite-order kernel derivative is uniformly bounded on γ. Repeated differentiation under the integral sign is therefore legitimate. This needs f holomorphic on an open neighbourhood of a. The function |z|² is complex differentiable only at 0, with no neighbourhood, and gets no such formula.

The Cauchy estimate. If f is holomorphic on a neighbourhood of the closed disk |z - a| ≤ R and |f| ≤ M on its boundary circle, then |f^(n)(a)| ≤ n! M/R^n. The factorial comes from differentiating the kernel n times, and the radius power from the contour length 2πR against the denominator R^(n+1). Worked estimate: with M = 4, R = 2 and n = 1, the bound on |f'(a)| is 1 × 4/2 = 2.

Liouville's theorem. A function is entire when holomorphic on the whole complex plane. If one bound M holds for |f(z)| at every z, then at any centre a the estimate gives |f'(a)| ≤ M/R. Letting R grow without bound forces f'(a) = 0 at every a, so f is constant. Per-circle bounds alone do not imply this: a bound like M(R) = R on each circle of radius R is consistent with f(z) = z, which is not constant.

Fundamental theorem of algebra. Let p be a nonconstant polynomial with no zero. Then 1/p is entire. Outside a large disk the leading term dominates, so |1/p| is small there; inside the fixed disk, continuity on the compact set with no zeros gives a finite bound. One global bound makes 1/p constant by Liouville, contradicting nonconstancy. So p has a root. Factoring out each root and repeating, counting multiplicity, gives exactly deg p roots.