Im reading a proof of G.Polya and G.Szego in my complex analysis book to the following theorem:
Suppose f is nonconstant and analytic on the closed disc D and assumes its maximum modulus at the boundary point $z_0$. Then $ f'\left(z_{0}\right)\neq0 $
I wanted to ask about a little detail in the beggining of the proof.
The proof starts as the following:
Assume that $ f'(z_0) =0 $. For any complex number $\xi $ of sufficiently small modulus we have
$ f\left(z_{0}+\xi\right)=f\left(z_{0}\right)+\frac{f^{(k)}\left(z_{0}\right)}{k!}\xi^{k}+... $
Where $k $ is the least integer with $ f^{(k)}\left(z_{0}\right)\neq0 $ and the omitted terms are all of higher order in $ \xi $ than $ \xi ^k $.
Now, what I dont get is why we can write $ f $ as a power series with center at $ z_0 $ if $ z_0 $ is on the boundary of the region where $ f $ is holomorphic? How can we tell that the convergence radius of this series is not $0$ ?
Actually, the statement is:
Theorem: Suppose $f$ is nonconstant and analytic on the closed disc $D$, and assumes its maximum modulus at the boundary point $z_0$. Then $f′(z_0)\ne0$.
And the author defines analytic function as follows:
Definition: $f$ is analytic on a set $S$ if $f$ is differentiable at all points of some open set containing $S$.
So, if $f$ is analytic on the closed disk $D$, then it is the restriction to $D$ of an analytic function deefined on an open set $A\supset D$. And $z_0\in A$. Therefore, it as a Taylor series with non-zero radius of convergence centered at $z_0$.