 Lemniscatic integral

Choose the holomorphic square root equal to one at zero on the unit disc. The resulting elliptic integral maps that disc conformally onto the square with vertices $\pm B,\pm iB$, where $B={\int }_0^{1}dt/\sqrt{1-t^4}$. For the inverse Cayley transform between the half-plane and disk $h(z)=i(1+z)/(1-z)$, the square of the derivative of $F_1\circ h$, with $F_1^{\prime }(w)=1/\sqrt{w(1-w^2)}$, is $2i/(1-z^4)$. Thus the two square maps differ by an affine change of coordinate.