Holomorphic implies complex-analytic

This article gives a proof/explanation of the equivalence of multiple definitions for the term holomorphic function

View a complete list of pages giving proofs of equivalence of definitions

Statement

Suppose $U$ is a domain in $\mathbb{C}$, and $f:U\to \mathbb{C}$ is holomorphic: it is complex-differentiable, and its differential is a continuous function. Then, $f$ is a complex-analytic function: for every point in $U$, we can find a disc about that point, and a power series that agrees with $f$ on that disc.

In fact, something stronger is true: given any point $z_0\in U$ and any disc of radius $r$ about $z_0$ contained completely inside $U$, we can find a power series that converges on that disc, and agrees with $f$ on the disc. In other words, we can make the radius of convergence of the power series as large as we can make a disc completely contained inside the open subset.

Facts used

1. Holomorphic function on open disk admits globally convergent power series

Converse

The converse is also true.

Further information: Complex-analytic implies holomorphic

Proof

The proof follows directly from fact (1).