# Entire Functions with no zeros are equal

I am trying to prove the following problem:

Assume that $f(z)$ and $g(z)$ are entire functions without any zeros such that $\lim_{|z| \rightarrow \infty} \frac{f(z)}{g(z)} = 1$. Then $f(z) =g(z)$.

My first instinct was to define a new function $h(z) := \frac{f(z)}{g(z)}$, which is entire since both $f$ and $g$ are and they have no zeros. I think that I should show that this is bounded and then invoke Liouville's theorem to show that $h$ is constant. Moreover, the condition that $\lim_{|z| \rightarrow \infty} \frac{f(z)}{g(z)}$ would seem to suggest that if $h$ is constant then $h(z) = 1$, thereby establishing the result.

However it does not seem apparent to me that $h$ is bounded since the limit condition does not exclude the possibility that $h$ becomes arbitrarily large for small values of $z$.

Any hints?

Hint: Say you know that $h$ is bounded outside some closed disk $D$ of radius $r$. The set $D$ is compact; what does that tell you about the image of $D$ under $h$?
• $h(D)$ would be compact as well and, in particular, bounded. But how do I know that such a disk exists? – Nala Ball Feb 22 '18 at 5:16
• Well, you're going to need to use the fact that $\lim_{|z| \rightarrow \infty} f(z) / g(z)=1$ at some point... – Eric Wofsey Feb 22 '18 at 5:26
• This is exactly what I had in mind. +$1$ – Clayton Feb 22 '18 at 5:37
There exists $R$ such that $|h(z)| <2$ for $|z| > R$. $h$ is a continuous function on $\{z:|z|\leq R\}$, hence it is bounded there. If $|h(z)| <M$ for $|z| \leq R$ then $|h(z)| < max \{2,M\}$ for all z. Apply Liouville,s Theorem.