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While following the work of calculating $e^{\pi\sqrt{163}}$, I was stuck on this one step where you calculate what $j(\tau)$ is equal to.

Note: The j-function is defined as $$j(\tau)=\dfrac 1q+744+196884q+\cdots\tag1$$ Where $q=e^{2\pi i\tau}$. In this case, we have $q=-e^{-\pi\sqrt{163}}$ and $\tau=\frac{1+\sqrt{-163}}2$. So $(1)$ is supposed to become$$j\left(\dfrac {1+\sqrt{-163}}2\right)=-e^{\pi\sqrt{163}}+744+196884\cdot\left(-e^{-\pi\sqrt{163}}\right)+\cdots\tag2$$ However, $(2)$ was then evaluated into$$j\left(\frac{1+\sqrt{-163}}2\right)=-640320^3\tag3$$ Which is confusing to me.

Question:

  1. How does $-e^{\pi\sqrt{163}}+744+196884\left(-e^{-\pi\sqrt{163}}\right)+\cdots=-640320^3$
  2. How would you calculate $j\left(\frac {1+\sqrt{-67}}2\right)$ then?
  3. Is there a way to do all the calculations by hand?
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    The fact that $j(\tau)$ is an integer follows from the fact that $\mathbb Q(\sqrt{-163})$ has class number $1$. – 2017-01-26
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    @Mathmo123 Is there a special number $m$ such that $\mathbb{Q}(\sqrt{-m})=2$? How about three, four? – 2017-01-26

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Hint: It is an infinite summation. You already know that $q = -e^{-\pi\sqrt{\color{blue}{163}}}$. So try to edit your post and give the numerical value of the finite sums $F_n$, $$\begin{aligned} F_1 &= 1/q+744+196884q\\ F_2 &= 1/q+744+196884q+21493760q^2\\ F_3 &= 1/q+744+196884q+21493760q^2+864299970q^3 \end{aligned}$$ and so on. As $n$ increases, is the value of $F_n$ getting familiar? You should be able to answer your question by then.

For $d=67$, it's the same formula. Obviously just replace the $\color{blue}{163}$ in your $q$ with $67$.