Suppose $a > 0$ and $b > 0$ are constants and a non-constant function $F(z)$ is such that $F(z+a) = F(z)$, and $F(z+bi) = F(z)$. Prove that $F(z)$ cannot be analytic in the rectangle $0\leq x\leq a,\ 0\leq y\leq b$ .
Use Liouville’s Theorem:
Suppose that for all $z$ in the entire complex plane,
(i) $f(z)$ is analytic and
(ii) $f(z)$ is bounded, i.e., $| f(z) | < M$ for some constant $M$.
Then $f(z)$ must be a constant.
I really don't know how to solve this excercise, Do I have to find an $M$?.
This problem is saying that $F(z)$ is a periodic function, right? so, it is true that has an upper bound.
Any help will be appreciated.