This is a bit of a repost from an old question, but it doesn't seem like it was fully answered before and this is a bit of an abstraction from that post. I'm trying to show the following:
Let $f(z)$ be an entire function such that $f(z)=f(z+1)$ for all $z$. If there exists $c\in\mathbb{R}$ with $c$ not a multiple of $2\pi $ such that $|f(z)|\le e^{c|z|}$ for all $z$, then $f(z)=0$.
The previous post was here: Entire "periodic" function
I've tried using Zarrax's method in that post:
The periodicity of $f(z)$ allows us to write $f(z)=g(e^{2\pi i z})$, since $e^{2\pi i (x+iy)}=e^{2\pi i (x+1+iy)}$. Now we have: $\left|g(e^{2\pi i z})\right|\le e^{c|z|}$
Now make a change of variables, re-writing $z=(2\pi i)^{-1}\log z$: $\left|g(z)\right|\le e^{c\left|\frac{\log z}{2\pi i}\right|}$
Now assume $|z|\ge 1$: $\left|g(z)\right|\le e^{\frac{c}{2\pi}(\log |z|+2\pi)}=|z|^\frac{c}{2\pi}e^c$
So now I would say that this implies $g(z)$ has to be a polynomial, but I don't really see how I can derive a contradiction from that.