How do I show that if $f$ and $g$ are entire and $|f|\ge |g|$, then there is some $\beta$ such that $f(z) = \beta g(z)$ for all $z$?

Question

I was wondering if you could help me with a question:

Suppose that $ f $ and $ g $ are entire functions, and that $ |f(z)| \leq |g(z)| ,\forall z \in C $. Prove that there $ \exists \beta \in C $ such that $f(z) = \beta g(z), \forall z ∈ C$.

I tried to show $f(z)/g(z) $ was constant by Liouville theorem however we don't know if $ f(z)/g(z)$ is entire as $g(z)$ might be equal to $0$. So I couldn't use the fact that it is entire and bounded to use Liouville theorem. Do you have an idea? thank you in advance

Answer 1

If $g(z)$ is nonvanishing Liouville's theorem applies nicely, so assume $g$ has a nontrivial set $Z$ of zeros.

If $z_0 \in Z$ then the function $h(z) = \dfrac{f(z)}{g(z)}$ is analytic in some punctured disk centered at $z_0$, has an isolated singularity at $z_0$, and is bounded. This rules out the possibility of an essential singularity, so either the singularity at $z_0$ is removable or is a pole of finite order.

If $h$ has a pole of finite order $m$ at $z_0$ then there is a function $w$ that is analytic in a disk centered at $z_0$ that satisfies $w(z_0) \not= 0$ and $w(z) = (z - z_0)^m h(z)$. This in turn leads to $g(z) w(z) = (z - z_0)^m f(z)$, and upon taking modulus it follows that $|w(z)| \le |z - z_0|^m$. This contradicts $w(z_0) \not= 0$.

It follows that $h(z)$ has a removable singularity at $z_0$ and may be modified to be analytic at $z_0$. Consequently $h$ is a bounded entire function, hence constant.

Answer 2

Using little Picard, if a meromorphic function on the entire complex plane misses three points it must be constant. Can you see how to use this?