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 $N > 0$ such that $|f(z)/g(z) - 1| < 1$ for all $|z| > N$. (Note that I am taking $\epsilon = 1$ for concreteness.)
Then, $|f(z)/g(z) - 1| \geq ||f(z)/g(z)| - 1|$ $\implies ||f(z)/g(z)| - 1| < 1$ $\implies |f(z)/g(z)| - 1 < 1$ or $1 - |f(z)/g(z)| < 1$ $\implies 0 < |f(z)/g(z)| < 2$.
That is, $|f(z)/g(z)|$ is bounded above by $2$ for $|z| > N$. 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?