Let $f,g:\mathbb C\to\mathbb C$ be holomorphic with $|f(z)|\leq|g(z)|$ for all $z\in\mathbb C$. Show that there is a $c\in\mathbb C$ with $|c|\leq 1$ such that $f=cg$.
If $g\equiv 0$ then we trivially have $c=0$. Otherwise, we can write $$|h(z)|:=\frac{|f(z)|}{|g(z)|}\leq 1\,\forall z\in\mathbb C,\,g(z)\neq 0.$$ Since $h$ is bounded it follows from Liouville's theorem that it must be constant with $h(z)=c$ as well as $|h(z)|=|c|\leq 1$ which shows the claim.
However, if $g$ has an isolated singularity $z_0$ then Riemann's theorem on removable singularities implies that it can be removed since $h$ is bounded. Therefore there exists an analytic continuation $\hat{h}$ with $$\left.h\right|_{\mathbb C\setminus\{z_0\}} = \left.\hat h\right|_{\mathbb C\setminus\{z_0\}}.$$ Furthermore $\hat h$ is bounded as well and once again Liouville's theorem implies $\hat h(z)=c$ and thus $|\hat h(z)|=|c|\leq 1$ and therefore the claim follows.
