Let $f$ be an entire function such that $f(1)=2f(0)$. Prove that $\forall\epsilon>0, \exists z\in\mathbb{C}$ such that $|f(z)|<\epsilon$ I am asked to prove this:

Let $f$ be an entire function such that $f(1)=2f(0)$. Prove that $\forall\epsilon>0, \exists z\in\mathbb{C}$ such that $|f(z)|<\epsilon$

I considered a function $g(z)=f(z+1)-2f(z)$, which is also entire and has a zero at $z=0$, but I am not sure this is going to help me solve the problem.
 A: Hint Assume by contradiction that this is not true. Show that $g(z)=\frac{1}{f(z)}$ is entire and bounded.
A: I was just giving my answer a final polish edit when N.S. posted; our proofs are essentially the same, though I have fleshed out a few details.
Perhaps someone will find it useful.
If
$\forall \epsilon > 0 \exists z \in \Bbb C, \; \vert f(z) \vert < \epsilon \tag 1$
is not true, then 
$\exists \epsilon > 0 \forall z \in \Bbb C, \; \vert f(z) \vert \ge \epsilon \tag 2$
is true.
Thus 
$g(z) = \dfrac{1}{f(z)} \tag 3$
is a well-defined entire function, and
$\forall z \in \Bbb C, \; \vert g(z) \vert = \dfrac{1}{\vert f(z) \vert} \le \dfrac{1}{\epsilon}; \tag 4$
thus $g(z)$ is a bounded entire function, hence constant by Liouville's theorem; hence $f(z)$ must be constant as well; in fact via (1),
$\forall \epsilon > 0 \forall z \in \Bbb C, \; \vert f(z) \vert < \epsilon \Longrightarrow \forall z \in \Bbb C, \; f(z) = 0; \tag 5$
that is, $f(z)$ is identically zero.  
The hypothesis $f(1) = 2f(0)$ is not necessary to attain this result.
A: If $f(0)=0$ the result is immediate. If $f(0)\ne 0$, we have that $f(1)\ne f(0)$ and so $f$ is not constant. According to Picard's little theorem, a non-constant entire function takes every complex value, with one possible exception. Even if that possible exception turns out to be zero, all other complex values are attained by $f$ and the result follows.
