Bounded analytic function is constant

Question

I'm trying to prove this :
Let $f(z)= \sum_{n=0}^{\infty}a_n(z-z_0)^n$ be an analytic function on $D(z_0,R)$. Suppose $\lvert f(z) \rvert \leq \vert f(z_0) \rvert = \lvert a_0 \rvert$ on $D(z_0,R)$. Show $f$ is contant.
My idea : If $f$ is not constant, let $a_k, k \geq 1$ be the first non zero coefficient. Then, on $\overline{D}(z_0,r)$ as $r \to 0^{+}$, $\sum_{n=k+1}^{\infty}a_n(z-z_0)^n$ becomes negligeable compared to $a_k(z-z_0)^k$, so $\lvert f(z) \rvert \approx \lvert a_0+ a_k(z-z_0)^k \rvert$, where then I can find $z= r \frac{( \overline{a_k}(sign(\Re(a_0)) + isign(\Im(a_0)))^{1/k} }{ \lvert\overline{a_k}(sign(\Re(a_0)) +i sign(\Im(a_0)) \rvert^{1/k}} + z_0 \in \partial D(z_0,r)$ (one of the $k^{th}$-roots) to arrive at a contradiction. I just don't know how to find $r$ such that all of this works (where the tail of the series becomes small enough).

Answer 1

Let $0<r<R$ where $r$ is small enough that $\forall z (|z-z_0|=r\implies |f(z)-a_0-a_k(z-z_0)^k|<\frac 1 2 |a_kr^k|)$. There exists $z_1$ with $|z_1-z_0|=r$, such that $\frac {a_0}{a_k(z_1-z_0)^k} $is a non-negative real. It follows by elementary algebra that $|f(z_1)|>|a_0|=|f(z_0)|$.