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I was going through the proof of the fact that an entire doubly periodic function is constant (using Liouville's theorem). My question is, do I have to assume the periods are independent over $\mathbb R$? For example, if I have an entire function $f$ with $f(z+\lambda_1)=f(z+\lambda_2)=f(z)$ for all $z\in\mathbb C$ with the $\lambda_1,\lambda_2$ being independent over, say $\mathbb Q$, but not necessarily over $\mathbb R$, does it still follow that $f$ is constant? I felt the proof I'm reading uses somewhere the fact that the period parallelogram is generated by two complex numbers forming a $\mathbb R$ basis of $\mathbb C$. What happens if I take them to be only independent over $\mathbb Q$? They may no longer form a basis for $\mathbb C$ but it seems that even then if $f$ is entire with $f(z+\lambda_1)=f(z+\lambda_2)=f(z)$ for $\lambda_1,\lambda_2$ independent over $\mathbb Q$, it has to be constant. Why is that true?

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If $\lambda_1,\lambda_2$ are $\Bbb Q$-linearly independent, but $\Bbb R$-linearly dependent, then $\lambda_2=c\lambda_1$ with some irrational real $c$. Then the set $\Bbb Z+c\Bbb Z$ is dense in $\Bbb R$, i.e., for arbitrary $\alpha\in \Bbb R$, we find sequences $n_k,m_k\in\Bbb Z$ such that $n_k+m_kc\to \alpha$. But $$f(0)=f(n_k\lambda_1+m_k\lambda_2)=f((n_k+m_kc)\lambda_1)\to f(\alpha\lambda_1)$$ as $k\to\infty$. As $\alpha\in\Bbb R$ was arbitrary, $f|_{\Bbb R}$ is constant. It follows that $f$ is constant.

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  • $\begingroup$ I do not understand. Why is $\mathbb Z+c\mathbb Z$ dense in $\mathbb R$? And it's not clear to me what's a sequence in $\mathbb Z$. $\endgroup$ – adrija Nov 25 '16 at 10:05

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