If $z_1 - z_2 \in \Lambda$, then $f(z_1) - f(z_2) \in \Lambda'$. Show that $f(z) = az + b$ Let $f$ be an entire function. Let $\Lambda, \Lambda' \subset \mathbb{C}$ be lattices with periods $1,\tau$ and $1, \tau'$ respectively. For any $z_1, z_2 \in \mathbb{C}$ with $z_1 - z_2 \in \Lambda$, we have $f(z_1) - f(z_2) \in \Lambda'$. Prove that $f$ is an affine function, so $f(z) = az + b$.
This is an exercise I want to solve, but I'm unsure how. I thought about using a function $g(z,w) := f(z+w) - f(z)$. If $f$ is really an affine function, then $g$ should be constant if we fix $w$. If we fix $z$, then $g$ should behave like a linear function. Also, if $w \in \Lambda$, then also $g(z,w) \in \Lambda'$. These were my thoughts so far, but I don't really see how to continue from here, or how to prove the first two claims. Can somebody give me a hint, what I should consider?
Edit: If we fix $w$ with $w \in \Lambda$, then $g(z,w) \in \Lambda'$, but since $\Lambda'$ is a discrete set, if we change $z$ a little bit, then we still must stay on the same point in $\Lambda'$, we cannot "jump" to another point because $g$ is continous. So we see that $g_w(z)$ must be constant.
 A: Hints: Take $g_\omega(z)=f(z + \omega)-f(z)$ for any $\omega\in\Lambda$. Is $g_\omega$ continuous? What does $g_\omega$'s image look like (discrete)? Then $g_\omega$ is ... Now differentiating $g_\omega$ gives $g_\omega'=f'(z+\omega)-f'(z)$. What's the value of this derivative? So is $f'$ periodic? holomorphic? Can we apply Liouville's theorem here?
Well I will just put the details here.

 As you said, for arbitrary $\omega\in\Lambda$, we have $g_\omega$ constant. Therefore, $g'_\omega=0=f'(z+\omega)-f'(z)$ for any $z$ and $\omega\in \Lambda$. Thus, $f'$ is holomorphic (by $f$'s holomorphy$\implies $analyticity) and $\Lambda$-periodic. Therefore, it is bounded and by Liouville's theorem, this map is constant. Therefore, its antiderivative $f$ is of the form $z\mapsto mz+b$ for some $m,b\in\mathbb{C}$.

The exact same argument can be used to prove that all holomorphic maps between two elliptic curves $\varphi: \mathbb{C}/\Lambda\to\mathbb{C}/\Lambda'$ takes the form $[z]\mapsto [mz+b]$. The only thing you need to do is to lift $\varphi$ to some holomorphic map $\tilde{\varphi}:\mathbb{C}\to\mathbb{C}$ of the torus' universal covering space where the lift is denoted by $f$ in your case.
