Prove that any entire conformal map satisfying $f(z + 2) = f(z) + 2$ and $f(z + 2iM) = f(z) + 2iM'$ for some real $M, M'$ is linear.
I wanted to check if my approach is correct: By subtracting $f(0)$ from both sides on each of the equations given, we may assume that $f(0) = 0$. Then by Cauchy integral formula we have that, if $a_i$ is the coefficient of $z^i$ of $f$ in its Taylor series expansion, $|a_i| \leq \frac{1}{r} \displaystyle\sup_{|z| = r} |f(z)|$. Suppose the supremum is achieved at $z = \alpha$. Then we have that $f(\alpha) = 2 + f(\alpha - 2) = 4 + f(\alpha - 4) = \cdots = 2m + f(\alpha - 2m) = 2m + 2iM' + f(\alpha - 2m - 2iM) = 2m + 4iM' + f(\alpha - 2m - 4iM) = \cdots = 2m + 2nM' + f(\alpha - 2m - 2nM)$ where $n,m$ are chosen so that $\alpha - 2m - 2nM$ is contained in the ball of radius $\max(2,2M)$. As $r$ grows, $n$ and $M$ grows at the same rate as $r$ so that if $i \geq 2$ we have that $a_i = 0$. Thus $f$ is linear.
I am not sure if my proof is rigorous enough. Could you check / make this proof more rigorous? Is there any other way to solve this question? Thank you