Conformal map to a right half-plane with a horizontal slit ($z=(1-Z^2{)}^{-1/2}$)

The principal square root maps the upper $Z$ half-plane bijectively onto $\Re z>0$ with the ray $[1,\infty )$ removed. The prevertex zero is the slit tip, and $\pm 1$ represent the two ends at infinity. The inverse is the unique root $Z^2=1-z^{-2}$ with positive imaginary part. Pulling back ${\pi }^{-1}\arg ((Z-1)/(Z+1))$ gives the bounded harmonic function with value one on the slit banks and zero on the imaginary axis. The harmonic function $\Im [Z/(1-Z^2)]$ has zero finite boundary values but grows at infinity, explaining the role of boundedness.