 Solution

For the velocity potential strip, $e^{\pi w/a}$ maps to the upper half-plane. The real Möbius transformationpreserves that half-plane and sends its marked endpoints to $-1,\infty ,0$ in the order $A/E,D,B/C$.

For the hodograph plane, first put $r=(q-1)/(q+1)$. Its image is the second quadrant, bounded by the negative real ray and positive imaginary ray. Squaring takes that quadrant to the lower half-plane, and negating restores the upper half-plane. Thushas exactly the same three marked values. Uniqueness of a conformal map to a half-plane with three assigned boundary values makes these maps equal. Cancel the minus signs to obtainThe sign in the intermediate map is important for identifying the upper, rather than lower, half-plane.