The following is a past qualifying exam question in complex analysis:
Let $f$ and $g$ be two meromorphic functions in $\mathbb{C}$. Assume that $|f(z) + g(z)| \le |g(z)|$ for all $z \in \mathbb{C}$ which are not poles of either $f$ or $g$. Show that there is some constant $c$, with $|c+1| \le 1$, such that $f(z) = cg(z)$.
My intuition would be to use Rouche's Theorem to show that $f/g$ would be entire, and then we would have the result by Louiville's Theorem; but I've been unable to make this a reality. Any hints?
Let $h(z)=\frac{f(z)}{g(z)}=\frac{f(z)+g(z)}{g(z)}-1$. Then $\bigl|h(z)+1\bigr|\leqslant1$ for each $z$ at which $h(z)$ is defined. You can extend $h$ to an entire function, by Riemann's theorem on removable singularities. So, by Liouville's theorem, $h$ is a constant function $c$ and, of course, $c$ is such that $|c+1|\leqslant1$.