# 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 $|z|<|\alpha|$. [duplicate]

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.

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 $$f(z)\not=0$$.
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.
• $\phi_a$ is actually called Blaschke factor, which appears in the proof of Jensen's formula and other theorems concerning the log of certain function's modulus. Commented Aug 2, 2022 at 4:07