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Let $\psi$ be an elliptic function with periodic lattice $\mathbb{Z}[\omega]= \mathbb{Z}\omega \oplus \mathbb{Z}$, a pole of order $2$ at $0$, and simple zeros at $\pm\dfrac{\omega-1}{3}$. Here $\omega= \exp(2 \pi i/3)$.

Assume that $\psi(1/3)=1$.

Why is it true that $\psi\left( \frac{\omega}{3} \right) \psi\left( \frac{2-\omega}{3}\right)= 1? $

Meow
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    What do you understand by a period lattice $\mathbb{Z}[\omega]$ ? That the periods are $1$ and a non real complex number $\omega$? – Alonso Delfín Sep 18 '15 at 12:13
  • @AlonsoDelfín Yes, though giving an explicit basis of the lattice is of course unnecessary. – Meow Sep 18 '15 at 18:13
  • Sorry but I am a little confused by your notation. Here $\omega$ is a fixed non real complex number? or $\omega$ is a point of the set $ \Omega = { \omega : \psi(z+\omega)=\psi(z)}$ and you want to see why your assertion is true for any $\omega \in \Omega$? Also in any case does $1 \in \Omega$ ? – Alonso Delfín Sep 18 '15 at 18:49
  • @AlonsoDelfín Apologies for the confusion, I'll edit. Is it clear now? – Meow Sep 18 '15 at 21:59

1 Answers1

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You can check that $\psi(\omega z)/\psi(z)$ is $\Bbb Z[\omega]$-periodic and has no zero nor pole, so it is a constant.

Near $z = 0$, $\psi(z) \sim a/z^2$ for some nonzero complex number $a$, and so $\psi(\omega z) \sim a/\omega^2 z^2 = \omega a/z^2$, hence $\psi(\omega z)= \omega \psi(z)$, forall $z \in \Bbb C$.

Applying this repeatedly to $z=1/3$ gives you $\psi(\omega/3) = \omega$ and $\psi((2-\omega)/3) = \omega^2$.

mercio
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