Suppose that $f$ and $g$ are two holomorphic functions on all of $\mathbb{C}$ and that $g$ never vanishes. If $|f(z)|\leqslant|g(z)|$ for all $z \in \mathbb{C}$, then show that there is a constant $C$ so that $f(z) = Cg(z)$ for all $z \in \mathbb{C}$.
So this is what I have gotten:
Liouville's Theorem: every bounded entire function must be constant, i.e. every holomorphic function for which there exists a positive number $M$ such that, $|h(z)|\leqslant M$, for all $z$ in $C$, is constant.
Given that $f$ and $g$ are holomorphic on $C$ (i.e. entire) and $g$ is non-zero in $C$, $|f(z)|\leqslant|g(z)|$ for all $z$ in $C$.
Consider, $h(z)=f(z)/g(z)$
Then, $h$ is defined on $C$ and $h$ is holomor[hic on $C$ (since $f$ and $g$ are holomorphic on $C$)
Hence, $h$ is entire.
Also, $|h(z)|=|f(z)/g(z)|=|f(z)|/|g(z)|\leqslant 1$ (since, $|f(z)|\leqslant |g(z)|$ for all $z$ in $C$)
Hence, $|h(z)|\leqslant 1$ implies $h$ is bounded.
Also, $h$ is entire.
Hence, by Liouville's theorem, $h$ is constant.
Therefore, there exists $C$ such that, $h(z)=C$ for $z$ in $C$.
i.e. $f(z)/g(z)=C$
i.e. $f(Z)=Cg(z)$ for all $z$ in $C$.
Comment: $g$ vanishes nowhere by assumption. So the proof should be valid. (There ist not enough reputation to make comments. Thus I have to answer.)