Problem of Liouville's theorem Here is my problem:
Let $f(z)$ be an entire function such that  $|f '(z)| < |f(z)|$ for all $z \in \mathbb{C}$,
Show that there exists a constant A such that $|f(z)| < A*e^{|z|}$ for all $z \in \mathbb{C}$.
I am trying to use Liouville's theorem  to prove and try to set $g(z)=|f '(z)|/|f(z)| < 1$ and then g(z) is constant. I am not sure if my thinking is right and how to prove this problem? 
Thanks
 A: Consider $g(z) = \frac{f'(z)}{f(z)}$. Since $|f'(z)| \lt |f(z)|$ we see that $f$ has no zeros, so $g$ is entire. Thus, $g$ is an entire function satisfying $|g(z)| \lt 1$, so by Liouville it is constant. This means that $f'(z) = C f(z)$ for some $C \lt 1$.
Since $f$ is entire we can write $f(z) = \sum_{n=0}^{\infty} a_n z^n$ for all $z \in \mathbb{C}$ and the equation $f'(z) = C f(z)$ leads to
$$f'(z) = \sum_{n=0}^{\infty} (n +1) a_{n+1} z^{n} = C \sum_{n=0}^{\infty} a_n z^n.$$
By comparing coefficients we see that $a_1 = C a_0$, $2 a_2 = C a_1$, $3a_3 = Ca_2$, etc, so that $a_n = \frac{C^{n}}{n!} a_0$. In other words, $f(z) = \sum_{n=0}^{\infty} \frac{a_{0}}{n!} (Cz)^n = a_{0} e^{Cz}$.
Now let us prove the estimate. Since $C \lt 1$ we have for all $z \in \mathbb{C}$ that $$|f(z)| \leq \sum_{n=0}^{\infty} \frac{|a_0|}{n!}\,|Cz|^n \leq \sum_{n=0}^{\infty} \frac{|a_0|}{n!} |z|^n = |a_0| e^{|z|},$$
almost as we wanted. By choosing $A \gt |a_0|$ we get the desired strict inequality $|f(z)| \lt A e^{|z|}$.
