LEt $f$ be an entire function such that $|f'(z)|\leq |f(z)|$ for all $z$. Show that $f(z)=ae^{cz}$
My proof:
Consider $g(z)=\frac{f'(z)}{f(z)}$ This function is bounded by 1 per our inequality. It is holomorphic everywhere except at zeroes of $f(z)$ however, these must be removable singularities as $g(z)$ is bounded around them. Thus $g(z)$ is entire and bounded (or can be extended to be such). Thus by louvile theorem it is a constant funciton. Hence $f'(z)=af(z)$ for all $z$ that are not zeroes of $f$, but it also works for zeroes as our inequality implies that $f(z)=0 \implies f'(z)=0$. Thus we get that $f^{(n)}(z)=a^nf(z)$ Hence looking the the taylors series we find that it is of the form $\Sigma \frac{f(0)((\frac{z}{a})^n}{n!}=f(0)e^{z/a}$ and this conclued our proofs.
Lauds to our OP 2132123 for providing an elegant argument to show that we may take
$f'(z) = cf(z), \tag 1$
for some
$c \in \Bbb C. \tag 2$
However, it is not necessary to invoke Taylor series to show that
$f(z) = ae^{cz} \tag 3$
is the unique solution to (1); indeed, from (1) it follows that
$f'(z)e^{-cz} = ce^{-cz}f(z), \tag 4$
or
$f'(z)e^{-cz} - ce^{-cz}f(z) = 0; \tag 5$
we next note that
$(f(z)e^{-cz})' = f'(z)e^{-cz} - ce^{-cz}f(z), \tag 6$
and thus combining (5) and (6) yields
$(f(z)e^{-cz})' = 0, \tag 7$
which implies that
$f(z)e^{-cz} = a \in \Bbb C \; \text{a constant}, \tag 8$
and thus
$f(z) = ae^{cz}; \tag 9$
taking
$z = 0 \tag{10}$
yields
$a = ae^0 = f(0), \tag{11}$
and thus we may write
$f(z) = f(0)e^{cz}. \tag{12}$
It will be observed that the above argument shows that (12) is the only possible solution to (1).