$\DeclareMathOperator\I I \DeclareMathOperator\E E \DeclareMathOperator\F F \def\L{\text L}$Some geometric problems require inverting elliptic E, especially $\E(x,2)$. One symbolic way to do so is using inverse beta regularized $\I^{-1}_x(a,b)$, $0\le x\le 1$, and the second lemniscate constant $\L_2$:
$$\E(x,2)=a\implies x=\frac12\sin^{-1}\left(\sqrt{\I^{-1}_\frac a{\L_2}\left(\frac12,\frac34\right)}\right)$$
However, many problems require solving $\E(x,2)=ia,a\in\Bbb R$ for which the above formula fails. It turns out using a negative parameter transform that:
$$\E(x,2)=\L_2+ia\implies x=\frac\pi2-\frac12\sin^{-1}\left(\sqrt{1-\I^{-1}_\frac a{\L_2}\left(\frac12,\frac34\right)}\right)$$
However, this is not the formula we want. It is possible to solve similar equations with elliptic F. Applying a complementary transform gives:
$$\E(x,-1)-\F(x,-1)=\L_2-\E\left(\sin^{-1}\left(\frac{\cos(x)}{\sqrt2}\right),2\right)$$
gives:
$$\E(x,-1)-\F(x,-1)=a\implies x=\\\sin^{-1}\left(\sqrt[4]{\I^{-1}_\frac a{\L_2}\left(\frac34,\frac12\right)}\right)\\ \E(x,-1)-\F(x,-1)=ia\implies x=\\i\sinh^{-1}\left(\sqrt[4]{\I^{-1}_\frac a{\L_2}\left(\frac34,\frac12\right)}\right)$$
If possible, what is the corresponding formula solving $\E(x,2)=i a$?