I have the following problem:
Let $f$ be a nonzero analytic function on $B:=\{|z|<1\}$. Show that there exists $c>0$, $r>0$ and $k=0,1,2,...$ such that $$|f(z)|\geq c|z|^k$$when $|z|<r$.
My idea was the following:
Since $f$ is analytic on $B$ and $0\in B$ we know that $f$ is analytic at $0$. By definition $\exists r>0$ such that on $|z-0|=|z|<r$ $$f(z)=\sum_{n=0}^\infty a_n(z-0)^n=\sum_{n=0}^\infty a_n z ^n$$Therefore $$|f(z)|=\left|\sum_{n=0}^\infty a_n z^n\right|$$
I see now some similarities to $c|z|^k$. But I'm somehow thinking about how one could find this $c$.
Does someone have a hint how I can proceed?
Thanks for your help.
If $f(0)\ne0$, take $\delta>0$ such that\begin{align}|z|<\delta&\implies|f(z)-f(0)|<\frac12|f(0)|\\&\implies|f(z)|>\frac12|f(0)|.\end{align}Now, let $r=\delta$, $k=0$, and $c=\frac12|f(0)|$.
And if $f(0)=0$, let $k$ be the order of $0$ as a zero of $f$. So, near $0$ you have$$f(z)=a_kz^k+a_{k+1}z^{k+1}+\cdots,$$with $a_k\ne0$. So, if $z\ne0$,$$\frac{f(z)}{z^k}=a_k+a_{k+1}z+\cdots$$and you can apply the same argument as above to this function; it follows that there are $c>0$ and $r\in(0,1)$ such that, when $|z|\in(0,r)$,$$\left|\frac{f(z)}{z^k}\right|>c\left(\iff|f(z)|>c|z|^k\right).$$