Sorry, this may be a stupid question, but I am just beginning to learn about this and cannot find the answer anywhere I have looked so far. Clearly if we have any polynomial $P(z)$, then it is easy to show that the order is $0$.
Clearly though not all entire functions that grow at roughly this rate are polynomials as I believe $\sin (z)$ is of order $1$, which would make $f(z) = \frac {\sin (z)}{e^z}$ an entire function of order $0$.
It is obvious that $f$ still has infinitely many $0$'s and is thus not a polynomial.
I was wondering what the characterization of all entire functions of $0$-order look like. I would assume it will just be all functions with $0$'s that are "nicely" spaced. But I do not know how to make this precise exactly as I know very little about this subject material. If anyone wants to aid me in my studies it would be greatly appreciated. Thanks.
EDIT: My friend pointed out that $\frac {\sin z}{e^z}=e^{-z} \sin z$, which grows at least as fast as $\sin z$ for large negative real values of $z$. So I'm confused... are there non-polynomial entire functions of order $0$ or not? And how can we see this.
There are many non-polynomial entire functions of order zero. You can show that the order $\rho$ can be computed as $$ \rho = \limsup_{n\to\infty} \frac{n\log n}{\log(1/|a_n|)} $$ where $a_n$ are the Maclaurin coefficients of $f$. To get a function of order $0$, you only have to make sure that $|a_n|$ tends sufficiently fast to $0$ as $n\to\infty$. For example, $$ f(z) = \sum_{n=0}^\infty e^{-n^2} z^n $$ has order $0$ (and is clearly not a polynomial).