Complex-analytic implies holomorphic

Statement

Suppose $U\subset \mathbb{C}$ is a domain and $f:U\to \mathbb{C}$ is a complex-analytic function: for every point $z_0\in U$, there exists a real number $r>0$ and a power series $\sum _{n=0}^{\mathrm{\infty }}{a}_n(z-z_0{)}^n$ such that the power series converges and agrees with $f$ in the ball of radius $r$.

Then, $f$ is a holomorphic function: it is complex-differentiable, and the complex differential is a continuous function. In fact, $f$ is differentiable infinitely often.

Proof

We will show the following somewhat stronger fact: if $f$ is a complex-analytic function at $z_0$ with a power series:

$f(z):=\sum _{n=0}^{\mathrm{\infty }}{a}_n(z-z_0{)}^n$

By basic facts of power series, $f$ converges absolutely, and uniformly on every disc of radius strictly smaller than its radius of convergence. We'll show that $f^{\prime }$ exists on the open disc of the radius of convergence, and is given by the following absolutely convergent power series:

$f^{\prime }(z):=\sum _{n=0}^{\mathrm{\infty }}n{a}_n(z-z_0{)}^{n-1}$

Note that the power series on the right clearly has the same radius of convergence as the power series for $f$, so we can concentrate on showing that it equals $f$. Observe that if $s_n$ is the partial sum till the $n^{th}$ term of $f$, then ${s_{n^{\prime }}}^{\prime }(z)$ indeed equals the sum, till the $n^{th}$ term, for $f^{\prime }(z)$. Thus, what we really need to show is that the error term goes to zero.