$|f|\leq |g|$ on $\mathbb{C}$, then $f=cg$. Let $f$ and $ g$ be entire functions with $|f(z)|\leq |g(z)|$ for all $z\in \mathbb{C}$. Is it true that $f=cg$ for some $c\in \mathbb{C}$?
My attempt: If $g=0$, we are done. Suppose then that $g$ is not identically zero. Let $h=f/g$. Fix an arbitrary $z_0$. If $g(z_0)\neq 0$, then $g$ is nonzero on a neighborhood of $ z_0$ and $h$ is holomorphic at $z_0$. If $g(z_0)=0$, then $g$ is nonzero on a neighborhood of $z_0$ except at $z_0$ (by analytic continuation). Then $h$ is holomorphic on that neighborhood of $z_0 $ except at $z_0$. As $| h|\leq 1$, $h$ has an avoidable singularity at $z_0$, therefore it can be assumed to be holomorphic at $z_0$. Thus, $h$ is entire and by Liouville constant.

Yes, your solution is correct. Succinctly: consider $h=f/g$ which is mermorphic and bounded, so in particular it has removable singularities. Thus $h$ is in fact holomorphic and bounded, and you can conclude using Liouville.
