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Let $h(\zeta )=i(1+\zeta )/(1-\zeta )$, the inverse Cayley transform between the half-plane and disk. Direct calculation givessoOn the unit disc, choose the holomorphic square root of $1-{\zeta }^4$ equal to one at zero. Both sides define nonvanishing holomorphic derivatives, and their quotient has constant square. The branch chosen in part ii gives $(F_1\circ h{)}^{\prime }(0)=1+i$. ConsequentlyAn affine change of image coordinate preserves conformality, so the lemniscatic integral maps the disc bijectively onto a square.

The normalization specifies the square exactly. Put $B=F(1)={\int }_0^{1}dt/\sqrt{1-t^4}$. The substitution $x=t^2$ gives $A=2B$. The integral satisfies $F(-z)=-F(z)$ and $F(iz)=iF(z)$, so its four corner values are $B,iB,-B,-iB$. ThusIn particular the image is a square rotated through $\pi /4$ relative to the one in part ii, and its side length is $\sqrt{2}B$. These symmetries also give $F_1(i)=(1+i)B$ from the affine identity.

Use polar coordinates $x=r\cos \theta$, $y=r\sin \theta$. Away from the crossing, the lemniscate of Bernoulli has $r^2=\cos 2\theta$. An explicit parametrization of its four quarter-arcs iswhere the two signs are independent. Indeed $x^2+y^2=t^2$ and $x^2-y^2=t^4$. The four arcs meet at the origin and at the two lobe endpoints, covering the entire curve.

For the upper-right quarter, take $0\le \theta \le \pi /4$, $r=\sqrt{\cos 2\theta }$. Differentiating gives $dr/d\theta =-\sin 2\theta /r$. Thus the arc length element satisfiesSet $t=r$, which decreases from one to zero. Since $|dt/d\theta |=\sqrt{1-t^4}/t$, it follows that $ds=|dt|/\sqrt{1-t^4}$. All four quarter-arcs have the same length, soThe endpoint singularity is proportional to $(1-t{)}^{-1/2}$ and is integrable. This is the geometric origin of the lemniscatic integral.