Prove that if analytic function $f$ is such $f(0) \neq 0$, it has no zeroes in a certain disk

Question

Problem. Assume $\displaystyle f(z)=\sum_{n=0}^\infty a_n z^n$ is analytic in $\overline{U}=\{|z|\leqslant R\}$ and $a_0\ne 0$。Prove: $f$ has no zeroes in the circular disk $\left \{|z|< \dfrac{|a_0|R}{|a_0|+M}\right\}$, where $M = \max_{z\in \partial \overline U} |f(z)|$.

My attempt: I tried to use proof by contradiction, that is to assume $f$ has zeroes in the disk, but however I was not able to deduce the contradiction. So I guess probably it is not the right approach for this problem. Also I guess it may be related to the maximum modulus principle, which I failed to establish connections with.

Any help is appreciated.

Answer 1

The function $$ z \in D \longmapsto \frac{f(Rz)-a_0}{M+|a_0|} $$ is an analytic function from $D$ to itself that is zero at $0$. So by Schwarz lemma $$ |f(z)-a_0| \leq \frac{M+|a_0|}{R}|z|. $$

So if $f(z)=0$, then $|a_0| \leq \frac{M+|a_0|}{R}|z|$, QED.