TITLE: Multiple choice question: Let $f$ be an entire function such that $\lim_{|z|\rightarrow\infty}|f(z)|$ = $\infty$.
QUESTION [9 upvotes]: Let $\displaystyle f$ be an entire function such that $$\lim_{|z|\rightarrow \infty} |f(z)| = \infty .$$ Then, 

$f(\frac {1}{z})$ has an essential singularity at 0.
$f$ cannot be a polynomial.
$f$ has finitely many zeros.
$f(\frac {1}{z})$ has a pole at 0.

Please suggest which of the options seem correct. 
I am thinking that $f$ can be a polynomial and so option (2) does not hold.
Further, if $f(z) = \sin z $ then it has infinitely many zeros... which rules out (3) while for $f(z) = z$ indicates that it has a simple pole at $0$ and option (4) seems correct.

REPLY [5 votes]: You are correct about $2$, and given that, you should be able to determine whether $1$ is true or not--consider your example $f(z)=z$.
Your example $f(z)=\sin z$ does not meet the given criteria. Note that if there are infinitely many zeros, then the set of zeros is necessarily unbounded, for if not, it has a limit point, and so the function is identically zero, contradicting our assumption that $\lim_{|z|\to\infty}|f(z)|=\infty$. But then we have a sequence $\{z_n\}$ such that $|z_n|\to\infty$ but $f(z_n)=0$ for all $n$, so that once again contradicts our assumption. That takes care of $3$.
For $4$, note that since $|1/z|\to\infty$ as $z\to 0$, then by assumption, $\lim_{|z|\to 0}|f(1/z)|=\infty$, which means that $f(1/z)$ has a pole at $z=0$. (H/T to J.J. for reminding me of that characteristic of poles.)