# How to prove that $G_3>0$ in this case?

Let $\Lambda=\{a+be^{2\pi i/3}|a,b\in Z\}$, then $G_{3}(\Lambda)=\sum_{\omega\in\Lambda-\{0\}}\frac{1}{\omega^{6}}$ should be real and nonzero, but how can one prove that it's positive?

Moreover, in this case by considering the flexes of the corresponding cubic curve it should hold that: $P'(1/3;\Lambda)^{2}=420G_{3}(\Lambda)$,where $P'(z;\Lambda)=\sum_{\omega\in\Lambda}\frac{-2}{(z-\omega)^{3}}$. I expect that $P'(1/3;\Lambda)$ should be actually negative. Is this true and how to prove it if so?

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$\omega^6 = $$\omega^3$$^2$ – James Oct 5 '12 at 12:50
Hey James, I don't really understand... – Jun Su Oct 5 '12 at 14:03
If cubing is enough to ensure you have a real number, then squaring this will give you a positive number – James Oct 5 '12 at 14:06
But cubing doesn't – Jun Su Oct 5 '12 at 14:08