Cauchy's integral formula

Statement

Let $U \subset \mathbb{C}$ be an open domain, let $f : U \to \mathbb{C}$ be holomorphic, and let $\gamma$ be a positively oriented simple closed contour in $U$ whose interior lies entirely in $U$. Then for every point $z_0$ in the interior of $\gamma$, $f(z_0) = \frac{1}{2\pi i} \oint_{\gamma} \frac{f(z)}{z - z_0}\,dz$, and more generally $f^{(n)}(z_0) = \frac{n!}{2\pi i} \oint_{\gamma} \frac{f(z)}{(z - z_0)^{n+1}}\,dz$ for every integer $n \ge 0$.

Why is it true?

A holomorphic function behaves like a taut soap film: its values on a boundary curve $\gamma$ completely lock in its value at every point $z_0$ inside. Dividing by $z - z_0$ creates a single pole at $z_0$ that acts like a magnifying probe; when you integrate around $\gamma$, Cauchy's integral theorem lets you shrink the loop down to an infinitesimal circle around $z_0$, where $f(z)$ is essentially constant at $f(z_0)$ and the winding of $1/(z - z_0)$ contributes exactly $2\pi i$. Even more remarkably, differentiating under the integral sign shows that being complex-differentiable just once automatically forces $f$ to be infinitely differentiable and analytic.

Topics that use this theorem

Related theorems

Step-by-step proofs

No step-by-step proof yet for this theorem.