= Conformal map to a right half-plane with a horizontal slit
{title2=$z=(1-Z^2)^{-1/2}$}
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 $\pm1$ 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.
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