$f,g \in H(\mathbb{C})$ s.t. $|f(z)| \le k|g(z)|$ $\implies$ $f(z) = cg(z)$

Question

Let $f,g \in H(\mathbb{C})$. Suppose that there exists a constant $K>0$ such that $$|f(z)| \le k|g(z)|$$ for all $z \in \mathbb{C}$. How can I prove that there exists $c>0$ such that $$f(z) = cg(z)$$ for all $z \in \mathbb{C}$?

Answer 1

Consider $h=f/g$. This is, a priori, a meromorphic function, but by the Riemann continuation theorem, it is actually holomorphic (and bounded) on the whole complex plane, and hence constant.

Recall Riemann's continuation theorem says that if $f$ is holomorphic in a punctured neighborhood of a point and bounded there, it is holomorphic on the whole neighborhood.