This question already has an answer here:

Let $f$ be an entire function such that there exists $z_0, z_1$ $\mathbb{R}$-linearly independent so that $\forall z \in \mathbb{C}\ f(z+z_0)=f(z)=f(z+z_1)$. Prove that $f$ is constant.

I know that $f(z_0) = f(z_1) = f(0)$, so I guess I should prove that f is equal to $f(0)$ in a set with an accumulation point, but I don't know what to do next. Also, $f(z) = f(a z_0+b z_1)\ \forall\ a,b\ \in \mathbb{Z}$

Any hints?


marked as duplicate by Martin R, Crostul, Community May 12 '16 at 21:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ Hint: Liouville's theorem. $\endgroup$ – Wojowu May 12 '16 at 20:42
  • $\begingroup$ Thanks, it's duplicate. I'll close it! $\endgroup$ – Guillermo Mosse May 12 '16 at 21:32

Prove that

$$ \sup_{z \in \mathbb{C}} |f(z)| = \sup_{t_1,t_2 \in [0,1]} |f(t_1z_1 + t_2z_2)| $$

and deduce that $f$ is bounded. Finish with Liouville's theorem.


Not the answer you're looking for? Browse other questions tagged or ask your own question.