Entire function with some restriction

Question

Consider the set $F=\{f:\mathbb C\to\mathbb C\mid f\text{ is an entire function, }|f'(z)|\leq|f(z)|\text{ for all }z\in\mathbb C\}$.

Then which of the following are true?

1. $F$ is a finite set

2. $F$ is an infinite set

3. $F=\{\beta e^{\alpha z}\mid\beta\in\mathbb C,\;\alpha\in\mathbb C\}$

4. $F=\{\beta e^{\alpha z}\mid\beta\in\mathbb C,\;|\alpha|\leq1\}$

My attempt: I take $f(z)=2$ then option $2$ is discarded and when I take $f(z)=\exp(z)$ then option $1$ is also discarded. But I'm not know how to solve it by a general method. Please help me.

Answer 1

Since $(\forall z\in\mathbb C):\left\lvert\frac{f'(z)}{f(z)}\right\rvert\leqslant1$, you know, by Liouville's theorem, that $\frac{f'}f$ is constant. If $\alpha$ is that constant, then $\lvert\alpha\rvert\leqslant1$ and $\frac{f'}f=\alpha$ and therefore $f(z)=\beta e^{\alpha z}$ for some $\beta$. It is now easy to deduce that the options 2. and 4. are the ones which are true.

Answer 2

If $f$ has a zero of order $n$ at $c$ then $f'$ has a zero of order at least $n$ at $c$. [I will will add some details if necessary]. This makes $\frac {f'} f$ a bounded entire function, hence a constant. But $f'(z)=\alpha f(z)$ implies $f(z)=\beta e^{\alpha z}$. This function satisfies the given inequality iff $|\alpha| \leq 1$. It should now be clear that 1) is false, 2) is true, 3) is false and 4) is true.