Theorem of liouville Consider two entire functions with no zeroes and having a ratio equal to unity at infinity. Use Liouville’s Theorem to show that they are in fact the same function.

My attempt
Consider $h(z) = f(z)/g(z)$.
First of all, $h$ is entire, since $f$ and $g$ are entire, and $g(z)$ is nonzero for all $z$ in $\mathbb{C}$.

The fact that $\lim_{z→\infty} h(z) = \lim_{z→\infty} f(z)/g(z) = 1$ suggests that $h(z)$ is bounded as well.

Why: Since $\lim_{z→\infty} f(z)/g(z) = 1$, there exists $N &gt; 0$ such that $|f(z)/g(z) - 1| &lt; 1$ for all $|z| &gt; N$.
(Note that I am taking $\epsilon = 1$ for concreteness.)

Then, $|f(z)/g(z) - 1| \geq ||f(z)/g(z)| - 1|$
$\implies ||f(z)/g(z)| - 1| &lt; 1$
$\implies |f(z)/g(z)| - 1 &lt; 1$ or $1 - |f(z)/g(z)| &lt; 1$
$\implies 0 &lt; |f(z)/g(z)| &lt; 2$.

That is, $|f(z)/g(z)|$ is bounded above by $2$ for $|z| &gt; N$.
Moreover, we know that $|f(z)/g(z)|$ has a maximum $M$ on $|z| \leq N$ by maximum modulus principle (or simply from $|z| \leq N$ being compact). Hence, $|f(z)/g(z)|$ is bounded above by $\max \{2, M\}$ for all z in $\mathbb{C}$.

Hence, $h(z)$ is constant by Liouville's Theorem, i.e. $h(z) = f(z)/g(z) = c$ for some constant $c$.

Since this is true for all $z$ in $\mathbb{C}$, taking the limit as $z\to \infty$ yields $c = 1$.
Hence $f(z) = g(z)$ for all $z$ in $\mathbb{C}$.

Is correct my work?
Since both are entire functions without zeros, $h(z) = \frac{f(z)}{g(z)}$ is an entire function. $\lim_{z \rightarrow \infty} \frac{f(z)}{g(z)} \rightarrow 1$ where $\infty$ can be interpreted as the point at infinity, implies that $h$ is a bounded. A bounded entire function is constant $c$. $h(z) = c$ implies that $f(z) = c g(z)$. However $\lim_{z \rightarrow \infty} h(z) = 1$ implies that $c = 1$.
Since both are entire functions without zeros, $h(z) = \frac{f(z)}{g(z)}$ is an entire function. $\lim_{z \rightarrow \infty} \frac{f(z)}{g(z)} \rightarrow 1$ where $\infty$ can be interpreted as the point at infinity, implies that $h$ is a bounded. A bounded entire function is constant $c$. $h(z) = c$ implies that $f(z) = c g(z)$. However $\lim_{z \rightarrow \infty} h(z) = 1$ implies that $c = 1$.