Holomorphic Function is Continuously Differentiable

Theorem
Let $U \subseteq \C$ be an open set.

Let $f : U \to \C$ be a holomorphic function.

Then $f$ is continuously differentiable in $U$.

Proof
Let $z_0 \in U$.

By definition of open set, there exists an $\epsilon$-neighborhood $\map {N_\epsilon}{z_0}$ of $z_0$ for some $\epsilon \in \R_{>0}$.

From Holomorphic Function is Analytic, it follows that $f$ can be written as a complex power series:


 * $\map f z = \ds \sum_{n \mathop = 0}^\infty c_n \paren {z - z_0}^n$

for all $z \in \map {N_\epsilon}{z_0} $.

From Power Series is Taylor Series, it follows that $f$ is of differentiability class $C^\infty$.

Specifically, both the derivatives $f'$ and $f''$ exist.

It follows that $f'$ is complex-differentiable.

Complex-Differentiable Function is Continuous shows that $f'$ is continuous.

Proof 2
Let $z \in U$.

$\map {f ' '} z$ exists by Cauchy's integral formula.

Therefore $f'$ is continuous at $z$ by Complex-Differentiable Function is Continuous.