Show that $f(z) $ is constant

Question

Suppose that $f(z)$ is entire with $a$ and $b$ positive constants. If $$f(z+a)=f(z+b i)=f(z)$$ for all $z$, show that $f(z)$ is constant .

Which theorem could I use ? Please give some hint thankyou

Answer 1

Let $w\in\mathbb{C}$ be arbitrary so $\tilde{w}=\tilde{u}+i\tilde{v}$ for real numbers $\tilde{u},\tilde{v}$. There exists unique $u\in[0,a)$,$v\in[0,b)$ and $m,n\in\mathbb{N}$ such that \begin{align*} \tilde{u} &= ma + u\\ \tilde{v} &= nb + v \end{align*} Then using the condition on $f$ we can see that \begin{align*} f(\tilde{w}) = f(\tilde{u}+i\tilde{v}) = f(ma + u+ i[nb + v]) = f(u +i[nb + v] ) = f(u+iv) = f(w) \end{align*} Here $w=u+iv$. Hence for any $\tilde{w}\in\mathbb{C}$, we may find a point $w\in B$ where $B=\{u+iv: u\in[0,a],v\in[0,b]\}$ such that $f(\tilde{w})=f(w)$. Thus $f$ attains all of its values on the compact set $B$. Since $f$ is entire, then $f$ is continuous and thus bounded on $B$ and hence $f$ is entire and bounded on $\mathbb{C}$. Thus by Liouville's Theorem we have that $f$ is a constant function.

Answer 2

For all $z\in\mathbb{C}$, we can find $w$ with $|\text{Re}(w)|\leq a$ and $|\text{Im}(w)|\leq b$ such that $f(z)=f(w)$. Now $f$ is bounded on $[-a,a]\times[-b,b]$.