Explicit disk map of a slit half disk ($u^2=\frac{4(z^2+1)}{z^2+16},F(z)=\frac{u+u^{-1}+i}{u+u^{-1}-i}$)

For the upper half disk $|z|<2$ slit along $z=iy$, $0<y\le 1$, squaring maps onto $|w|<4$ minus $[-1,4)$. After scaling the radius-four disc, the disk automorphism $v=4(w+1)/(w+16)$ sends this slit to $[0,1)$. Its square root with argument in $(0,\pi )$ maps onto the upper half unit disc. There $u+u^{-1}$ maps bijectively onto the lower half-plane: its imaginary part is $\mathrm{Im}u(1-|u{|}^{-2})<0$, and the two roots of its inverse equation are reciprocal, with exactly one inside the unit circle. The final Möbius map gives the displayed conformal disk bijection.

In the explicit slit map, right-side approach to $i/2$ gives the positive root, and left-side approach gives the negative root. The corresponding lower-half-plane boundary values are $\pm 25/(2\sqrt{21})$, so the two unit-circle image values are distinct. Every other conformal bijection onto the disk differs by a disk automorphism, which extends bijectively to the unit circle. Thus a sequence approaching that slit point has exactly one or both of two boundary cluster values, according to which sides occur infinitely often.