= Right-half-plane slit map
{title2=$z=(1-Z^2)^{-1/2}$}
With the square root positive for $Z$ on the positive imaginary axis, the displayed <conformal map> takes the upper half-plane bijectively to $\{\operatorname{Re}z>0\}\setminus[1,\infty)$. Squaring, subtracting from one, taking the positive-real-part square root, and reciprocating identifies each intermediate slit domain. The boundary prevertices $-1,0,1$ correspond to the lower infinite end, the slit tip at one, and the upper infinite end. The inverse is $\sqrt{1-z^{-2}}$ with positive imaginary part.
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