Prove that a function is a constant multiple of the other

Question

Given $f$ and $g$ be entire functions and: $$|(f(z)| \leq |g(z)|$$ can you show me how to prove that $$f(z)=cg(z)$$ , where c is a (complex) constant? I have been able to prove this for the case $g(z) \neq 0$ by dividing $f$ into $g$: $$ \frac{|f|}{|g|} \leq 1 \rightarrow f(z)=cg(z) \text{by Liouville's theorem} $$ However, I don't know how to prove this result for the more general case in which $g(z)$ is arbitrary.

Answer 1

The zeros of $g$ must be discrete, if not, $g=0$ and hence the result follows. We conclude that $f/g$ has only removable singularities, an entire function $h$ coincides with $f/g$ except for removable singularities, and $h$ is bounded, so $h=c$, and the result follows.

Answer 2

If you know some basic facts about order of zeros of an analytic function (and the fact that if f is analytic at all points except a and if f is bounded near a then it is actually analytic in a neighborhood of a) you can argue as follows: we first show that if g has a zero of order n at a point a then f has a zero of order m at a with $m \geq n$. Indeed, $g(a)=0$ implies $f(a)=0$; let m be the order of zero of f at a. Then we can write $f(z)= (z-a)^{m}f_1 (z)$ and $g(z)= (z-a)^{m}g_1 (z)$ in a disc around a where $f_1,g_1$ are analytic in the disc and do not vanish at a. If $m<n$ divide the inequality $|(z-a)^{m}f_1 (z)| \leq |(z-a)^{n}g_1 (z)|$ by $(z-a)^{m}$ and let $z \to a$ to get a contradiction. Thus $m \geq n$. From this it follows $\frac f g$ can be defined in an obvious way at a whenver $g(a)=0$. This makes $\frac f g$ a bounded entire function, hence a constant by Liouville's Theorem.