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.