Doubly periodic entire function with periods that are linearly independent over $\mathbb{Q}$ is constant. Question
Let $f$ be entire and suppose there exists $\lambda_1,\lambda_2 \in \mathbb{C}$ linearly independent over $\mathbb{Q}$ such that $f(z+\lambda_1)=f(z)=f(z+\lambda_2)$. Then $f$ is constant.
Attempt
A similar question was asked here but I don't think the answer is satisfactory. I want to (hopefully) show that for any $\epsilon>0$ I can find integers $m_1,m_2$ such that $|m_1\lambda_1+m_2\lambda_2|<\epsilon$. Then I will be able prove the result as follows:
Let $z_0\in \mathbb{C}$. Then $g(z)=f(z)-z_0$ has a zero at $z_0$ and this zero must be isolated. Consequently we can find an $\epsilon>0$ such that the epsilon ball about $z_0$ contains only one zero ($z_0$) of $g(z)$. However, by the above, we can find $m_1,m_2$ so that $|m_1\lambda_1+m_2\lambda_2|<\epsilon$ and by assumption $g(z_0+m_1\lambda_1+m_2\lambda_2)=g(z_0)=0$ which is only possible if $f$ had been constant. 
 A: First assume that $\lambda_1,\lambda_2$ are linearly independent not only over $\mathbb Q$ but over $\mathbb R.$
The convex hull of $\{\,0,\,\lambda_1,\,\lambda_2,\,\lambda_1+\lambda_2\,\}$ is a parallelogram and thus a compact set. Entire functions are continuous. A continuous function on a compact set is bounded. By periodicity conjoined with the fact that $\lambda_1,\lambda_2$ are linearly independent over $\mathbb R$ and $\mathbb C$ is $2$-dimensional over $\mathbb R,$ we conclude that this entire function is bounded. Liouville's theorem says bounded entire functions are constant.
Now weaken the linear-independence hypothesis to being over $\mathbb Q.$ If $\lambda_1,\lambda_2$ are not linearly independent over $\mathbb R,$ then there is some irrational number $a\ne0$ such that $\lambda_1=a\lambda_2.$ Since $a$ can be approximated as closely as desired by rational numbers, one can show that $0$ can be approximated as closely as desired by integer linear combinations of $\lambda_1$ and $\lambda_2.$ This yields arbitrarily small periods of the periodic function, and for continuous functions, this means the function must be constant on the line $\{ c\lambda_1 : c\in\mathbb R\}.$ This implies its derivative is $0$ on that line. A holomorphic function whose derivative is $0$ on a set of points that has a limit point is constant.
