I am trying to solve a problem, that might seem easy to you, but i am really stuck with it and would be great if somebody could help me or give me a hint how to show that proof.
Given are two holomorphic functions on the whole $\mathbb{C}$: $f$ and $g$ and $\forall z\in \mathbb{C}: f(g(z))=0$. To show is that $f\equiv 0$ or $g$ is constant.
Okay, what i know is that a composition of two holomorphic functions is also holomorphic. I know also what the theorem of Liouville says: every bounded entire function is constant. $g$ is entire, this is clear, but how to show that it's bounded? Or do i have to show that $\mid g\mid $ could have a local minimum at some point $z_{0}$, so it's $g(z_{0})=0$ or constant?
Thank you in advance!
Best Answer
Suppose $g: \mathbb{C} \to \mathbb{C}$ is not constant. Then there are $a, b \in \mathbb{C}$ such that $g(a) \ne g(b)$. Consider $g(\mathbb{C})$. As $\mathbb{C}$ is connected, $g(\mathbb{C})$ is also connected, and since it contains two different points $g(a) \ne g(b)$, it necessarily has an accumulation point (otherwise it would be discrete, and thus disconnected). Since $f(z) = 0$ for $z \in g(\mathbb{C})$, $f$ takes the same values as zero function on a set with an accumulation point $g(\mathbb{C})$. Thus, by the identity theorem for holomorphic functions, $f$ must be zero function.
If you know the open mapping theorem for holomorphic function, you can say conclude that $g(\mathbb{C})$ is actually open, and so $f$ is equal to zero function on an open set, so it's zero everywhere.