Suppose $f(z)$ is analytic on $|z|<1$ such that $|f(z)|<1$ for all $|z|<1$ and $f(0) = \alpha \neq 0$. Show $f(z) \neq 0$ for all $|z|<|\alpha|$.

Question

This is a problem from a previous complex analysis qualifying exam that I'm working through to study for my own upcoming exam. I've been playing with it for a while and am stuck.

Problem:

Suppose $f(z)$ is an analytic function on $|z|<1$ such that $|f(z)|<1$ for all $|z|<1$ and $f(0) = \alpha \neq 0$. Show $f(z) \neq 0$ for all $|z|<|\alpha|$.

Thoughts:

Since $f(z)$ is analytic on the unit disk, then it has a power series about zero such that $$ f(z) = \sum_{k=0}^\infty a_k z_k = a_0 + \sum_{k=1}^\infty a_kz^k. $$

If $f(0) = \alpha \neq 0$, then $$ f(0) = a_0 + \sum_{k=1}^\infty a_k0^k = \alpha \quad \implies \quad a_0 = \alpha \quad \implies \quad f(z) = \alpha + \sum_{k=1}^\infty a_kz^k. $$

My thought is to somehow use Rouche's Theorem here, something like letting $g(z) = \sum_{k=1}^\infty a_kz^k$, and $h(z) = \alpha$. Then if $|g(z)|<|h(z)|$ for $|z|<|\alpha|$, then $g+h$ has the same number of zeros as $h$, which is none. I'm guessing it also helps to use that $|f(z)|<1$. But I can't seem to work it out.

Answer 1

By Schwarz–Pick theorem, for all $z_1,z_2\in\mathbb{C}$ such that $|z_1|<1$ and $|z_2|<1$, we have $$\left|\frac{f(z_1)-f(z_2)}{1-\overline{f(z_1)}f(z_2)}\right| \leq \left|\frac{z_1-z_2}{1-\bar z_1 z_2}\right|.$$ Letting $z_1=z$ and $z_2=0$, we get $$\left|\frac{f(z)-\alpha}{1-\overline{f(z)}\alpha}\right| \leq |z|$$ If $f(z)=0$ then we find $|\alpha|\leq |z|$, which means that if $|\alpha|>|z|$ then $f(z)\not=0$.

Answer 2

One idea is to consider the function $$\phi_a(z) = \frac{z-a}{1-\overline{a}z}$$ where $|a|<1$. I claim that this is function is an automorphism of $\Delta$ (biholomorphic map of the unit disc to itself) and $\phi^{-1}_a(z) = \phi_{-a}(z)$. Further it's clear that $\phi_a(a)=0$.

Now try applying the Schwarz lemma to $\phi\circ f$

EDIT: The solution above using Schwarz-Pick is far more clean! That said, I do think there's some value to being familiar with these $\phi_a$ functions-- in some sense they're "recenterings" of the unit disc.