Can someone explain me this proof (complex analysis)?

Question

I need to prove that gives a function $f\in H(\Bbb C)$ such that $f(z)=f(z+i)=f(z+1) \forall z\in\Bbb C$, then $f$ is costant. The proof is the following:

Be $\gamma = [0, 1] + [1, 1 + i] + [1 + i, i] + [i, 0]$ and $z \in\mathbb C$ then $z$ can be written as $z=a+s+ti$ with $s,t \in\mathbb Z, a \in Int\gamma$. For assumption we have: $f (z) = f (a + s + ti) = f (a + s + (t − 1)i) = \dots = f (a + s) = f (a)$. This results in the maximum principle $|f(z)| = |f(a)| \le ||f||\gamma = c$ Accordingly, $f$ is limited and therefore constant according to Liouville's theorem.

Answer 1

The proof uses these steps:

• For every $z\in\Bbb C$, there is some $a\in\operatorname{Int}\gamma$ such that $f(z)=f(a)$.
• Since $\operatorname{Int}\gamma$ is compact, the restriction of $f$ to it is bounded.
• It follows from the previous steps that $f$ is bounded.
• Since it is bounded, it is constant, by Liouville's theorem.