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

Cauchy integral formula

Further information: Cauchy integral formula

We use here the version for a circle and its center. The general version, involving winding numbers, is not necessary for this proof.

Converse

The converse is also true.

Further information: Complex-analytic implies holomorphic

Proof

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