Show that $f=\lambda g$

Question

Let $f$ and $g$ be holomorphic on $\mathbb{C}$ and $|f|\le C|g|$.

Show that there exists $\lambda\in\mathbb{C}$ such that $f=\lambda g$.

I know I should use Liouville theorem but I don't know how. Any input?

Answer 1

Define $$h(x)=\frac {f(x)} {g(x)}$$ so it's given that $|h(x)|\le C$ means it's bounded and $h$ is also entire (since $f,g$ are holomorphic) so by Louville's theorem: $$\exists \lambda\in\mathbb{C}:\lambda=h(x)=\frac{f(x)}{g(x)}\Rightarrow f(x)=\lambda g(x)$$as required.