= Solution
{c}
The <complex potential of a two-dimensional potential flow> maps the fluid region to the strip $0<\operatorname{Im}w<1$. Its lower boundary is the wall, with $B$ at $w=0$; upstream is $\operatorname{Re}w\to-\infty$ and downstream is $\operatorname{Re}w\to+\infty$. The free streamline is the upper strip boundary.
Write $q=w'=u-iv$. The horizontal wall gives positive real velocities running from $q=1$ to the stagnation value $q=0$ at $B$. The vertical wall gives negative imaginary velocities from zero to $-i$. On the <free surface> $|q|=1$, with the angle turning continuously from zero to $-\pi/2$. Thus the <hodograph plane> domain is the fourth-quadrant quarter of the <unit disk>, bounded by those two radii and the unit-circle arc.
The following successive conformal maps straighten it. First $r=q^2$ maps to the lower half <unit disk>. Next $s=(r-1)/(r+1)$ maps to the third quadrant, since $\operatorname{Re}s=(|r|^2-1)/|r+1|^2<0$ and $\operatorname{Im}s=2\operatorname{Im}r/|r+1|^2<0$. Finally $s^2$ maps onto the <upper half-plane>. The three wall landmarks $q=1,0,-i$ go to $0,1,\infty$ respectively.
The strip map $e^{\pi w}$ also maps onto the <upper half-plane>, with these same landmarks $0,1,\infty$. The unique half-plane automorphism fixing all three is the identity. Therefore
$$
\boxed{e^{\pi w}=\left(\frac{q^2-1}{q^2+1}\right)^2,\qquad q=w'.}
$$
The <free surface> maps to the negative real axis in either construction, as required by its <streamfunction> value one. This completes the <right-angle free-streamline hodograph map> construction.
\Image[/past-exam-of-the-perturbation-methods-and-applied-complex-variables-course-of-the-university-of-oxford/2010-hodograph-map.png]
{title=The potential strip and quarter-disk velocity hodograph for right-angle free-streamline flow}
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