I have been struggling on the following problem: Show that there is no such entire function $f(z)$ such that $|f(z)|>e^{|z|}$ for all $|z|>100$.
I've tried to define at first $g(z)=1/f(z)$ and use liouville theorem but $f$ might have zeros in $\{|z|\leq 100\}$. I also thought about using the argument theorem.
I would really appreciate it if you could help.
Suppose by contradiction that such a function exists. Then all zeros of $f$ lie in the ball of radius 100, and hence $f$ has finitely many zeros. Letting $p$ be a polynomial with the same zeros as $f$, with the same multiplicities, we deduce that $g(z):= p(z)/f(z)$ is an entire function with $$ |g(z)|\leq \frac{|p(z)|}{|f(z)|} \leq \frac{|p(z)|}{e^{|z|}} $$ for $|z|>100$. Since $e^{|z|}$ grows faster than any polynomial, we deduce that $g$ is bounded, and hence constant by Liouville's theorem. Consequently $f$ is polynomial, but it is clear that no polynomial can satisfy the hypothesis.