This is a question from a previous complex analysis qualifying exam that I'm trying to work through in order to study for my own upcoming qual. I'm not sure where to start with this one, to be honest.
Problem:
If $f(z)$ is a one-to-one analytic function on the unit disk and $f(0)=0$, show that there is an analytic function $g(z)$ such that $g(z)^2 = f(z^2)$ on the unit disk.
Current progress:
The only place I can think to start is with the definition of a one-to-one function, which is both
Injective: If $f(z_1) = f(z_2)$, then $z_1 = z_2$.
Surjective: If $z_2 \in \mathbb{D}$, where $\mathbb{D}$ is the unit disk, then there exists a $z_1 \in \mathbb{D}$ such that $f(z_1) = z_2$.
Also, if $f(0)=0$ and $f$ is one-to-one, then $f$ has no other zeros and there is no other fixed point such that $f(z_1) = z_1$.
Any help would be appreciated.
Best Answer
If $f$ is not identically $0$ it has a zero of some finite order at $0$. So we can write $f(z)=z^{N} F(z)$ where $F$ is analytic and has no zeros in the unit disk. [$F(z)=z^{-N}f(z) \neq 0$ for $0<|z|<1$ since $f$ has no zeros other than $z=0$].
Now $f(z^{2})=z^{2N} F(z^{2})$. Since the unit disk is simply connected any analytic fucniton on it with no zeros has an anayltic square root. So $F(z^{2})=(h(z))^{2}$ for some analytic function $h$. It follows that $f(z^{2})= [z^{N}h(z)]^{2}$.