Let $f:\mathbb C \to \mathbb C$ be an entire function such that $$\lim _{z\to 0}\left|f\left(\frac{1}{z}\right)\right|=\infty. $$ Then which of the following is true?
$f$ is constant
$f$ can have infinitely many zeros
$f$ can have at most finitely many zeros
$f$ is necessarily nowhere vanishing
My Try:- Let $f(z)=z$, I could eliminate 1,2,4. So, Answer is 3. How do I solve formally? $$\lim _{z\to 0}\left|f\left(\frac{1}{z}\right)\right|=\infty.\iff $$ for $\epsilon>0, \exists \delta: |z|<\delta \implies |f(1/z)|>\epsilon.\iff $ Put $w=1/z.$ for $\epsilon>0, \exists \delta: |1/w|<\delta \implies |f(w)|>\epsilon \iff $ for $\epsilon>0, \exists \delta: |w|>1/ (\delta) \implies |f(w)|>\epsilon$. So, the function has no zero out side the closed disk of radius $\delta$. So, Zeroes may or may not be lie inside the disk of radius $\delta$. How do I proceed further?
If the zeros lie in a bounded set and there are infinitely many of them, does that imply the zero set has a limit point? If it does, what does that imply?