Suppose $f$ and $g$ are entire functions, and $|f(z)| \leq |g(z)|$ for all $z \in \mathbb{C}$, Prove that $f(z)=cg(z)$. Suppose $f$ and $g$ are entire functions, and $|f(z)| \leq |g(z)|$ for all $z \in \mathbb{C}$, Prove that $f(z)=cg(z)$. 

My try :

I consider $h(z)=\frac{f(z)}{g(z)}$. If I prove that $h(z)$ is entire, using the fact that $|h(z)|\leq 1$, the result follows immediately from Liouville's theorem.

To prove that $h(z)$ is entire, I have to prove that $h(z)$ has removable singularities at possible zeros of $g(z)$.

Suppose $g(z_0)=0 \Rightarrow |g(z_0)|=0 \Rightarrow |f(z_0)|=0$, BUT what if $z_0$ be a root of of order $k$ of $g(z)$ (i.e. $f^{(n)}(z_0) = 0 $ for all $ 0 \leq n \leq k-1$) and be a zero of order $m$ of $f(z)$ and $k&gt;m$ ?

And if that's not the case, how to rigorously show that $z_0$ is a removable singularity ?

Thank you all in advance.
Suppose that $g$ has a zero at $a$. If $f$ is identically zero, we are done, as $0=0g$, so we may assume this is not the case, which implies that $g$ is not identically zero either, and therefore we may assume instead that it has a zero of order precisely $k$ at $a$, so $g(z)=(z-a)^kh(z)$, where $h$ is analytic in a neighborhood of $a$, and $h(a)\ne 0$. 

Since $|f|\le|g|$, then $f(a)=0$, so $f(z)=(z-a)^mj(z)$ where $j(a)\ne0$ and $j$ is analytic in a neighborhood of $a$. For $z\ne a$ near $a$, $|f|\le|g|$ means that $$\displaystyle\left|\frac{(z-a)^{m-k}j(z)}{h(z)}\right|\le 1.$$ 

Since $j$ and $h$ are not zero near $a$, by continuity there is a constant $K$ such that $|j/h|\ge 1/K$ in a neighborhood of $a$, so $$\left|\frac{(z-a)^{m-k}j(z)}{h(z)}\right|\ge\frac{|z-a|^{m-k}}K,$$ or $|z-a|^{m-k}\le K$ for $z\ne a$ near $a$. But this is impossible if $m&lt;k$.
Let $z_0$ be a zero of $g(z)$.Unless $g$ is identically zero, it can not be zero at points arbitrarily close to $z_0$. (By "identity principle".) And in that case $\displaystyle \lim_{z \to z_0} (z-z_0)\frac{f(z)}{g(z)} = 0$, (Since $|f(z)| \le |g(z)|$ ) which tells you that the singularity indeed is removable.