Homework question using Liouville theorem and Cauchy estimate. The statement is if $f$ is an entire function that satisfies 
$$ |f(z)| \leq \pi e^{2Rez} $$
Then $\forall z \in \mathbb{C}$ , there is a complex number $a \in \mathbb{C}$ such that:
$$f(z) = ae^{2z}$$
I tried applying cauchy’s estimate or using an auxiliary function to apply Liouville’s theorem, but it yielded nothing. 
If anyone can give me a hint towards where should I start, I really appreciate that. 
 A: With
$z = x + iy \tag 1$
we observe that
$0 < \pi e^{2x} = \pi e^{2\Re(z)} \in \Bbb R, \; \forall z \in \Bbb C; \tag 2$
furthermore, 
$\vert e^{2i\Im(z)} \vert = \vert e^{2iy} \vert = 1, \forall z \in \Bbb C; \tag 3$
it follows that
$\vert e^{2z} \vert = \vert e^{2\Re(z) + 2i\Im(z)} \vert = \vert e^{2\Re(z)} \vert \vert e^{2i\Im(z)} \vert = \vert e^{2\Re(z)} \vert = e^{2\Re(z)} = e^{2x} \ne 0, \; \forall z \in \Bbb C; \tag 4$
now we are given
$\vert f(z) \vert \le \pi e^{2\Re(z)}; \tag 5$
in the light of (4) this yields
$\vert f(z) \vert \le \pi \vert e^{2z} \vert, \tag 6$
which since
$e^{2z} \ne 0 \tag 7$
implies
$\left \vert \dfrac{f(z)}{e^{2z}} \right \vert = \dfrac{\vert f(z) \vert}{\vert e^{2z} \vert} \le \pi; \tag 8$
since $f(z)$ and $e^{2z}$ are both entire, so is $f(z) / e^{2z}$ also entire, and it is bounded according to (8); but a bounded entire function is constant by Liouville's theorem; thus 
$\dfrac{f(z)}{e^{2z}} = a \in \Bbb C, \tag 9$
whence
$f(z) = ae^{2z}, \; \forall z \in \Bbb C, \tag{10}$
as was to be shown.  $OE\Delta$.
A: Hint: Show that $g(z) = e^{-2z} f(z)$ is bounded, and therefore constant. Note that $|e^w| = e^{\operatorname{Re} w}$ for all complex numbers $w$.
Don't be misled by the constant $\pi$ – the actual value of that constant is not relevant.
