Applications of inverse of Klein's $j$-invariant

Question

The Klein $j$-invariant $$j(\tau) = q^{-1} + 744 + 196884q + \cdots$$ is a weight $0$ modular function holomorphic for $\tau$ in the upper-half plane $\mathbb{H}$. I understand that $j$ is important, given that two elliptic curves are isomorphic over $\mathbb{C}$ if and only if they have the same $j$-invariant.

We know that $j$ is a bijection from the fundamental domain $\operatorname{SL}_2(\mathbb{Z}) \backslash \mathbb{H}$ to $\mathbb{C}$.

I am wondering, what are some interesting applications of the inverse $j^{-1}$? What are some situations in which computing $j^{-1}$ of some complex number arises?

Answer 1

If $g_2$ and $g_3$ are the Weierstrass elliptic invariants, then the corresponding half-periods ratio is given by $$ \tau = j^{-1}\left(\frac{1728 g_2^3}{g_2^3 - 27 g_3^2}\right) $$ except in some particular cases.

Answer 2

Here is an inverse of Klein’s Invariant j. First put it in terms of Modular Lambda: $$y=\text j(z)=256 \frac{(\lambda^2(z)-\lambda(z)+1)^3}{\lambda^2(z)(1-\lambda(z))^2}$$

Now put the Modular Lambda function in terms of the Inverse Nome:

$$256 \frac{(\lambda^2(z)-\lambda(z)+1)^3}{\lambda^2(z)(1-\lambda(z))^2} = 256 \frac{(\text q^{-1}(e^{i\pi z})^2-\text q^{-1}(e^{i\pi z}) +1)^3}{\text q^{-1}(e^{i\pi z})^2(1-\text q^{-1}(e^{i\pi z}))^2} $$

Now invert $256\frac{(y^2-y+1)^3}{y^2(1-y)^2}$ with radicals like so:

$$\lambda(z)=\text q^{-1}(e^{i\pi z})=\frac12\pm\frac12\sqrt{\frac1{192}\sqrt[3]{y^3-2304y^2+12288\sqrt3\sqrt{1728y^2-y^3}+884736y}-\frac{1536y-y^2}{192\sqrt[3]{y^3-2304y^2+12288\sqrt3\sqrt{1728y^2-y^3}+884736y}}+\frac y{192}-3}$$

Therefore, two branches with the Elliptic Integral $\text K(k)$ are:

$$\text j^{-1}(z)=i\frac{\text K\left(\frac12\mp\frac12\sqrt{\frac1{192}\sqrt[3]{z^3-2304z^2+12288\sqrt3\sqrt{1728z^2-z^3}+884736z}-\frac{1536z-z^2}{192\sqrt[3]{z^3-2304z^2+12288\sqrt3\sqrt{1728z^2-z^3}+884736z}}+\frac z{192}-3}\right)}{\text K\left(\frac12\pm\frac12\sqrt{\frac1{192}\sqrt[3]{z^3-2304z^2+12288\sqrt3\sqrt{1728z^2-z^3}+884736z}-\frac{1536z-z^2}{192\sqrt[3]{z^3-2304z^2+12288\sqrt3\sqrt{1728z^2-z^3}+884736z}}+\frac z{192}-3}\right)}$$

where $-,+$ or $+,-$ is taken. Hopefully there are no typos.