# $f,g \colon \mathbb C \to \mathbb C$ holomorphic with $\vert f \vert \le \vert g \vert$ [duplicate]

Let $f,g \colon \mathbb C \to \mathbb C$ be two holomorphic functions s.t. $$\vert f(z) \vert \le \vert g(z) \vert, \qquad \forall z \in \mathbb C.$$

Is it true that there exists $c \in \mathbb C$ s.t. $f(z)=cg(z)$ for every $z \in \mathbb C$?

My answer is yes: in fact, if $g(z) \ne 0$ then I can divide and I obtain $$\left \vert \frac{f(z)}{g(z)} \right \vert \le 1$$ from which I can conclude that $\frac{f}{g} \equiv c \Leftrightarrow f\equiv c g$, for some $c \in\mathbb C$, by Liouville's theorem.

The equality $f\equiv c g$ still holds for $z$ s.t. $g(z)=0$: from $\vert f(z) \vert \le \vert g(z) \vert$ and $g(z)=0$ I deduce $f(z)=0$. To sum up, we have $f\equiv c g$ for every $z \in \mathbb C$.

Am I right? Thank you in advance.

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## marked as duplicate by Jonas Meyer, Norbert, Davide Giraudo, martini, t.b.Jul 15 '12 at 19:58

This question was marked as an exact duplicate of an existing question.

There is a problem in your proof when $g(z_0)=0$ at some point $z_0$. In this case of course $f(z_0)=0$, but you can't use result from the previous part! This result was proved with assumption that $g(z)\neq 0$ for all $z\in\mathbb{C}$. – Norbert Jul 15 '12 at 18:33
@Norbert Thank you for your comment, you are perfectly right. I derived the first part under the assumption $g(z) \ne 0$ for every $z \in \mathbb C$. I've been so stupid! Do you have any suggestion to conclude correctly the proof? Thank you again for your kindness. – Romeo Jul 15 '12 at 18:39
Hint: $f/g$ is meromorphic and the first part of your argument shows the singularities are removable ... – Matt Jul 15 '12 at 18:40
– Jonas Meyer Jul 15 '12 at 18:45
@JonasMeyer Thanks a lot, I hadn't seen those discussions. Thank you. – Romeo Jul 15 '12 at 18:48