Holomorphic functions are analytic

Statement

Let $U \subseteq \mathbb{C}$ be an open set and let $f : U \to \mathbb{C}$ be holomorphic on $U$. Let $c \in U$ and $r > 0$ be such that the open disc $D = \{z : |z - c| < r\}$ is contained in $U$. Then there are complex numbers $a_0, a_1, a_2, \dots$ such that the power series $\sum_{n \ge 0} a_n (z - c)^n$ converges to $f(z)$ for each $z \in D$. For each $\rho$ with $0 < \rho < r$, the coefficients are $$a_n = \frac{1}{2 \pi i} \oint_{|z - c| = \rho} \frac{f(z)}{(z - c)^{n+1}} \, dz.$$ In particular, $f$ is analytic on $U$: near each point of $U$, $f$ is the sum of a convergent power series.

Remarks

Proof idea: in Cauchy's integral formula, expand $\frac{1}{z - a}$ as a geometric series in $\frac{a - c}{z - c}$, and integrate term by term. The radius of convergence of the Taylor series at $c$ is at least the distance from $c$ to the boundary of $U$. For example, the Taylor series of $1/(1 - z)$ at $0$ has radius $1$, the distance from $0$ to the point $1$ where the function is not defined. The converse is also true: power series are holomorphic in their disc of convergence.

Explanations

L1 human

A function with one complex derivative everywhere on an open set is automatically equal to its Taylor series near each point. This is very different from real functions, where a differentiable function can fail to have a second derivative.