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

Hint Assume by contradiction that this is not true. Show that $g(z)=\frac{1}{f(z)}$ is entire and bounded.

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.