Showing an analytic function on the unit disk is nonzero in a certain neighborhood Suppose $f(z)$ is analytic for $|z|\le 1$ and $f(0) = a_0 \ne 0$. If $M = \max_{|z|=1} |f(z)|$, then show $f(z)\ne 0$ for all $z$ with $|z| < \frac{|a_0|}{|a_0|+M} =:r$.
I know we can write $f(z) = a_0 + z^kg(z)$, some $k\ge 1$ and $g$ analytic and $g(0)\ne 0$. From here, I've tried various techniques, like contradiction by assuming $f$ has a root in the disk $\{|z| < r\}$, or trying to use Rouche's Theorem on the disk by examining $|f(z)-a_0|$, but I haven't really gotten anywhere. Any hints would be greatly appreciated.

The map $g(z):=f(z)/M$ is a analytic map from the unit disc in itself.
Then by the Schwarz-Pick Theorem, for $|z|< 1$,
$$\left|\frac{g(z)-g(0)}{1-\overline{g(z)}g(0)}\right|\leq |z|.$$
If $f(z)=0$ and $|z|<r<1$ then $g(z)=0$ and we obtain
$$\frac{|a_0|}{M}=|g(0)|\leq |z| <r= \frac{|a_0|}{|a_0|+M}$$
which is a contradiction because $f(0) = a_0 \ne 0$.
