Show that $f(z) $ is constant 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
 A: 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.
A: 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 $[-a,a]\times[-b,b]$.
