How do I show that if the ratio of the periods of an elliptic function is irrational then the function reduces to a constant?

Question

I know that taking $\omega_1$ and $\omega_2$ periods of an elliptic function, the ratio $\frac{ω1}{ω2}$ has to be not real, in fact when $\frac{ω1}{ω2}$ is real the parallelogram collapse, when $\frac{ω1}{ω2}$ is rational the function reduces to a singly periodic function but how do I show that if $\frac{ω1}{ω2}$ is irrational then the function reduces to a constant?

Answer 1

For $f$ meromorphic and $w_1,w_2$ periodic and $w_1/w_2$ real irrational.

If $f$ has a pole on the real line at $a$ then it also has a pole a $a+ \Bbb{Z}w_1+\Bbb{Z} w_2$, which is dense in $\Bbb{R}$, whence $f$ isn't meromorphic, a contradiction.

Therefore $f$ has no pole on the real line, it is continuous, so that $f(a)=f(a+ \Bbb{Z}w_1+\Bbb{Z} w_2)$ and the density of $a+ \Bbb{Z}w_1+\Bbb{Z} w_2$ in $\Bbb{R}$ implies that $f(\Bbb{R})=f(a)$, ie. $f$ is constant on the real line whence on the whole of $\Bbb{C}$ by analytic (meromorphic) continuation.