Problem and Context:
$\def\K{\operatorname K}\def\F{\operatorname F} \def\sn{\operatorname{sn}} \def\B{\operatorname B} \def\I{\operatorname I} \def\E{\operatorname E}\def\am{\operatorname{am}}$
An inverse of elliptic E$(z,m)$, elliptic F$(z,m)$, and the incomplete beta B$_x(a,b)$ function uses the Jacobi amplitude am$(z,m)$ and inverse beta regularized I$^{-1}_x(a,b)$:
$$\E(z,\sqrt[-3]{-1})-\left(\frac12+\frac i{2\sqrt3}\right)\F(z,\sqrt[-3]{-1})=a\implies z=\am\left(\frac{\sqrt[12]{-1}}{2\cdot3^\frac34}\B_{\I^{-1}_{1-\sqrt[4]3\sqrt[3]4\sqrt[12]{-1}\B z}\left(\frac23,\frac 12\right)}\left(\frac13,\frac12\right)+\frac{\sqrt[12]{-1}(2\sqrt3+3i)}{2^\frac433^\frac34\pi}\Gamma^3\left(\frac13\right),\sqrt[-3]{-1}\right)\tag1$$
with the Baxter constant B shown here. An identity is:
$$\am\left(\frac{\sqrt[12]{-1}}{2\cdot3^\frac34}\B_y\left(\frac13,\frac12\right)+\frac{\sqrt[12]{-1}(2\sqrt3+3i)}{2^\frac433^\frac34\pi}\Gamma^3\left(\frac13\right),\sqrt[-3]{-1}\right)=\am\left(\sqrt[3]{-1}\F(w,\sqrt[-3]{-1})-\frac{\sqrt{11\sqrt3i-15}}{2^\frac{17}6\pi}\Gamma^3\left(\frac13\right),\sqrt[-3]{-1}\right),w=\sin^{-1}\left(\frac{\sqrt3\sqrt[6]{-1}}{\sqrt{\sqrt3\sqrt[3]{-iy}+\sqrt[3]{-1}+1}}\right)\tag2$$
and
$$\text{sn}\left(\frac{(-1)^\frac34}{2\cdot 3^\frac34}\B_\frac{\left(\frac3{x^2}-\sqrt[3]{-1}-1\right)^3i}{3\sqrt3}\left(\frac13,\frac12\right)+\frac{\sqrt[12]{-1}}{2^\frac43\sqrt[4]3\pi}\Gamma^3\left(\frac13\right),\sqrt[3]{-1}\right)=\sqrt[6]{-1}x\tag3$$
shown here as a concept proof with Jacobi sn$(z,m)$. $(1)$ almost displays $\am(x,m)$ periodicity with Elliptic K$(m)$:
$$\K\left(\sqrt[-3]{-1}\right)=\frac{\sqrt[4]3\Gamma^3\left(\frac13\right)}{2^\frac73\sqrt[12]{-1}\pi}, \K\left(\sqrt[3]{-1}\right)=\frac{\sqrt[4]3 \sqrt[12]{-1}\Gamma^3\left(\frac13\right)}{2^\frac73\pi} $$
Strategy:
Likely the sn$(u+v,m)$ formula, since $\sn\left(\frac{\sqrt[12]{-1}(2\sqrt3+3i)}{2^\frac433^\frac34\pi}\Gamma^3\left(\frac13\right),\sqrt[-3]{-1}\right)$ etc has a closed form, Jacobi transformations, and finally $\am(\F(x,m),m)=x$ will simplify $(1)$. However, I am not that experienced with elliptic functions, so help is wanted.
$(1)$ is likely expressible in terms of I$^{-1}_x\left(\frac23,\frac12\right)$ only without other special functions. What is a simplification for $(2)$, like in $(3)$?
Best Answer
$\def\sn{\operatorname{sn}}\def\cn{\operatorname{cn}}\def\am{\operatorname{am}}\def\dn{\operatorname{dn}}\def\m{\sqrt[-3]{-1}}\def\sgn{\operatorname{sgn}}$ Using inverse Jacobi sn$^{-1}(x,m)$: $$z=-\am(\sqrt[3]{-1}\text F(\sin^{-1}(u),\m)+c,\m)\iff-\sin(z)=\sn(\sqrt[3]{-1}\sn^{-1}(u,\m)+c,\m),c=-\frac{\sqrt{11\sqrt3i-15}}{2^\frac{17}6\pi}\Gamma^3\left(\frac13\right)$$ Use the sn$(u+v,m)$ formula with Jacobi cn$(x,m)$ and Jacobi dn$(x,m)$. A problem is that am$\left(\frac{\sqrt{11\sqrt3i-15}}{2^\frac{17}6\pi}\Gamma^3\left(\frac13\right),\m\right)=(a+b i)\infty;a,b>0$ with unknown $a,b$, so there is a limit. Each Jacobi function has the same $m$ omitted: $$\sn(\sqrt[3]{-1}\sn^{-1}(u,\m)+c,\m)=\\\lim_{c\to-\frac{\sqrt{11\sqrt3i-15}}{2^\frac{17}6\pi}\Gamma^3\left(\frac13\right)}\frac{\cn(\sqrt[3]{-1}\sn^{-1}(u))\dn(\sqrt[3]{-1}\sn^{-1}(u))\sn(c)+\cn(c)\dn(c)\sn(\sqrt[3]{-1}\sn^{-1}(u))}{1-\m(\sn(\sqrt[3]{-1}\sn^{-1}(u))\sn(c))^2}$$ with Jacobi transformations: $$\sn(\sqrt[3]{-1}\sn^{-1}(u,\m),\m)=\sqrt[3]{-1}\sgn(u)\sqrt{\frac{u^2}{(-1)^\frac23u^2+1}}$$ $$\cn(\sqrt[3]{-1}\sn^{-1}(u,\m),\m)=\frac1{\sqrt{(-1)^\frac23u^2+1}}$$ $$\dn(\sqrt[3]{-1}\sn^{-1}(u,\m),\m)=\sqrt{\frac{1-u^2}{(-1)^\frac23u^2+1}}$$
Let’s mimic the divergence of the Jacobi functions at $x=c$:
$$\lim_{c\to\infty}\frac{a(1+i)c+b ((1-i)c)^2}{1- p((1-i)c)^2}=-\frac bp,b=\sqrt p= \sqrt[3]{-1}\sgn(u)\sqrt{\frac{u^2}{(-1)^\frac23u^2+1}}\implies-\sin(z)=-\frac1{ \sqrt[3]{-1}\sgn(u)\sqrt{\frac{u^2}{(-1)^\frac23u^2+1}}}$$
Therefore simplify into:
$$\boxed{\operatorname E(z,\sqrt[-3]{-1})-\left(\frac12+\frac i{2\sqrt3}\right)\operatorname F(z,\sqrt[-3]{-1}) =a\implies z=\sin^{-1}\left(\frac{\sqrt[12]{-1}}{\sqrt[4]3}\sqrt{1-\sqrt[3]{\operatorname I^{-1}_{1-\sqrt[4]3\sqrt[3]4\sqrt[12]{-1}\text Ba}\left(\frac23,\frac12\right)}}\right),0\le \sqrt[12]{-1}a\le \frac1{\sqrt[4]3\sqrt[3]4\text B}}$$
shown here with the Baxter constant B