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 $k>m$ ?
And if that's not the case, how to rigorously show that $z_0$ is a removable singularity ?
Thank you all in advance. 
 A: If $z_0$ is a singularity of $h$, and $h$ is bounded in a neighbourhood of $z_0$, then $z_0$ is a removable singularity of $h$.  Because if it isn't removable, then it is a pole (and not bounded) or essential (which is also not bounded, but actually this cannot happen).
A: @Shubhodip Mondal, @Andres Caicedo: My proof appears different from those of the above replies. Hope it is not too far off? 
Consider $h:=\dfrac{f}{g},$ so singularties of $h$ are zeroes of $g.$ 
1.) Suppose $g$ has zero of order $n$ at $z_o$ and $f(z_o)\neq 0,$ then $\lim _{\ z\to z_o}h(z)= \infty.$ This contradicts $|h| \leq 1,$ in a neighbourhood of $z_o.$  
2.) Suppose $f,g$ have zero of order $m,n$ at $z_o.$ Then $h= \dfrac{\phi(z)(z-z_o)^m}{\beta(z)(z-z_o)^n},$ for some $\phi, \beta$ analytic at $z_o$ and $\phi(z_o), \beta(z_o) \neq 0.$
If $m < n,$ then $ h(z)= \dfrac{\phi(z)}{\beta(z)(z-z_o)^{\ n-m}}$ and hence $ \lim _{\ z\to z_o}h(z)= \infty.$ (Contradiction)
So, $m \geq n$ and $ \ h(z)=\dfrac{\phi(z)(z-z_o)^{m-n}}{\beta(z)}.$ Hence $z_o$ is removable singularity of $h,$ whence  $h$ can be extended to an entire function. Also, by continuity $, |h(z)| \leq 1,$ where $z$ is removable singularity of $h.$ The result follows from Lioville's theorem.   
A: 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 $m<k$.
A: 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. 
