let
$$F= \left \{f: \mathbb{C} \to \mathbb{C} \;\;|f \;\;\text{where f is entire function} \right \}\\ \text{such that }{|f'(z)| \leq |f(z)|} \;\;\; \forall z \in \mathbb{C} $$
Then
1). $F$ is an infinite set.
2). $F=\left \{\beta e^{\alpha z}:\beta \in \mathbb{C},|\alpha| \leq 1 \right \}$
The solution I Tried-
As we know that if $f$ s entire then its derivative is also entre so $f'(z)$ is also entire ,now according to question it is given that $$|f'(z) \leq |f(z)|$$ so it implies that $$\frac{|f'(z)|}{ |f(z)|} \leq 1\;\; \forall z \in \mathbb{C}$$but as we know that denominator should not be zero for fraction to be exist thus $f(z)$ should't be zero for any $z \in \mathbb{C}$ $\Rightarrow$ function $f$ should be of the form $e^{\alpha z}$ because exponential function doesn't have zero in complex Plane.Next I am not getting how to proceed .
Please help!
Best Answer
The first part is easy because all constant functions are in $F$.
For the second part, here is a proof.
The zero function is of the form $\beta e^{\alpha z}$ with $\beta=0$.
If $f$ is not the zero function, fix $z_0 \in \mathbb C$. Write $f(z)=(z-z_0)^m g(z)$, where $g(z_0)\ne0$ and $m\ge 0$.
If $m\ge 1$, then $|f'(z) \leq |f(z)|$ at $z=z_0$ gives $|mg(z_0)|=0$, a contradiction.
Thus, $f$ has no zeros. By Liouville's theorem $f'/f$ is a constant function, and so $f'(z)=\alpha f(z)$.
Let $E(z)=e^{-\alpha z}$. Then $(fE)'=0$ and so $fE=\beta$, a constant. Thus, $f(z)=\beta e^{\alpha z}$.
Finally, $|\alpha| = |f'(z)/f(z)| \le 1$.