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.


Suppose to the contrary, that $F$ is analytic in the rectangle $R$. Since $F$ is double-periodic, F is an entire function.

F is continouos on $R$ and $R$ is compact, hence $F$ is bounded on $R$ and therefore bounded on $ \mathbb C$.

Liouville says now: $F$ is constant.


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