This is part (2) of a question that I am working on.
In part(1), I have constructed an entire function $f:=\cosh(\sqrt{z})$ that grows like
$$\lim_{r \to \infty} \frac {\log M(r)}{\sqrt{r}}=1$$
where M(r) = $\max\limits_{|z|=r} |f(z)|$.
Then equivalently, $\cosh(\sqrt{z})$ grows like $\mathrm{e}^{\sqrt{r}}$.
The question is now:
"Does every function $f(z)$ whose growth rate is given above have infinitely many zeros? Give your reasons."
My attempt:
First, we know that analyzing the behavior of such functions $f(z)$ at infinity is equivalent to analyzing $f(\frac{1}{z})$ as $z \to 0$.
Looking at the Laurent series, expanded about the origin, say for $\cosh(\sqrt{z})$, and inverting the argument, we get
$$\sum \frac{z^n}{(2n)!} \to \sum \frac{1}{z^n(2n)!}$$
Here's where I am a little confused. Does this automatically mean that 0 is an essential singularity of $f(1/z)$, so that $\infty$ is an essential singularity for $f(z)$? Just merely writing the series with an inverted argument makes all of the positive power z terms change into negative power terms - and infinitely many negative power terms about $z = 0$ implies, by definition, that $0$ is an essential singularity of $f(1/z)$.
But that seems silly, because then every function, written with the argument inverted, would suggest that it has an essential singularity at $0$, and therefore an essential singularity at $\infty$ for $f(z)$. What is wrong with my thinking?
Moving on, I could then use Big Picard's theorem and argue that near the essential singularity, $f(z)$ picks up every complex value, with an exception of at most one value. Then $f(z) = 0$ is fulfilled infinitely often, as we wanted to show. But, what if $z=0$ were that exceptional point? Then...how can I say anything at all about $f(z)$ attaining $0$...even once?
Any suggestions are welcome.
Thanks,