Let $f : \Bbb C \longrightarrow \Bbb C$ be an entire function such that $$\lim\limits_{z \to 0} \left \lvert f \left (\dfrac {1} {z} \right ) \right \rvert = \infty.$$ Then show that $f$ can have at most finitely many zeros.
What I can observe is that if $f$ has infinitely many zeros and zero set of $f$ is bounded then by Bolzano-Weierstrass theorem it has a limit point. Then by identity theorem it follows that $f \equiv 0$ on $\Bbb C.$ But then the above limit becomes $0,$ a contradiction. Hence if the zero set of $f$ has infinite cardinality then the set has to be unbounded. Now how do I proceed? Any help in this regard will be appreciated.
Thanks in advance.
Best Answer
There exists $\delta >0 $ such that $|z| <\delta $ implies $|f(\frac 1 z)| >1$. In particular $|z| >\frac 1 {\delta} $ implies $f(z) \neq 0$. So all the zeros are in $\{z: |z| \leq \frac1 {\delta}\}$. If there are infinitely many zeros in this compact set there would be a limit point for the zeros which leads to a contradiction. .