Value of the J Invariant at $\frac{1+\sqrt{-163}}{2}$

Question

For a while I've wanted to be able to show why $e^{\pi\sqrt{163}}\approx 744+640320^3$, but I have no idea how to show that $j(\frac{1+\sqrt{-163}}{2})=-640320^3$. I considered using the fact that $\mid \eta(\frac{1+\sqrt{-163})}{2})\mid^{4}=\frac{1}{326\pi}\prod_{n=1}^{162}\Gamma({\frac{n}{163}} )^{(\frac{n}{163})}$ (exponent is the legendre symbol) but I'd also need $\eta(\sqrt{-163})$ and also I have no idea how the simplification would work out. I've seen values of J being related to roots of polynomials and was wondering if that would be the easiest way to compute it. If anyone could explain methods that could be used to compute it I'd appreciate it.

Answer 1

$j(\frac{1+\sqrt{-163}}{2})=-640320^3$

$j(\frac{1+\sqrt{-163}}{2})=-2^{15}*J_{163}^{3}$

$j^{1/3}=-2^{5}*J=-2^{5}\big(\frac{G^{8}}{8}-\frac{1}{2G^{16}}\big)$.

$J=\frac{G^{8}}{8}-\frac{1}{2G^{16}}$

is the $J$ of Ramanujan

$\frac{4-G^{24}}{8G^{16}}=-20010$

E pertanto $G_{163}$ is that of Paramanand Singh.