 Exterior quarter-disc hodograph for a downward right-angle turn

A unit-speed free layer turning downward around a reentrant wall corner has complex velocity $q=u-iv$ in the first-quadrant exterior of the unit disc. Upstream $q=1$, downstream $q=i$, and the corner has infinite speed because its fluid angle is $3\pi /2$. Squaring $q$, applying $(s-1)/(s+1)$, and squaring again maps this domain to the upper half-plane with upstream, corner and downstream at $0,1,\infty$. The potential strip map $e^{\pi w}$ has the same normalization, proving the formula. The sign of the turn and the fluid corner angle determine whether the relevant hodograph is an exterior or interior quarter disc.