Right-half-plane slit map ($z=(1-Z^2{)}^{-1/2}$)

With the square root positive for $Z$ on the positive imaginary axis, the displayed conformal map takes the upper half-plane bijectively to $\{\mathrm{Re}z>0\}\setminus [1,\infty )$. 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.