Past exam of the applied complex variables course of the University of Oxford / 2011 / 1 / b / ii / Solution

Starting from the hodograph plane, take the complex powerIt takes the velocity sector bijectively to the upper half of the unit disk. The boundary points representing the upstream end, the corner and the downstream end become $-1$, zero and one. Next use the Möbius transformationThe diameter $-1<s<1$ maps to the negative real axis, while for $s=e^{i\theta }$ on the upper semicircle,which is on the negative imaginary axis. The interior maps to the third quadrant. Squaring therefore gives a conformal bijection from that quadrant to the upper half-plane:The upstream end, corner and downstream end map to zero, one and infinity, respectively. The two walls map to the positive real axis, and the free boundary maps to the negative real axis.

Independently, the exponential function $\eta =e^{\pi w}$ takes the potential strip to the same upper half-plane with the same boundary correspondence. The strip height fixes the coefficient $\pi$, and the normalization $w(B)=0$ fixes the remaining positive multiplicative constant in $\eta$ to one. ConsequentlyOne can make the branches explicit. In the strip, $p=e^{\pi w/2}$ lies in the first quadrant, whereas $T=-p$. Thuswith the power taking arguments in $(0,\pi )$ to $(0,3\pi /4)$. It gives $q\to e^{3\pi i/4}$ upstream, $q\to 1$ downstream, and $q\to 0$ at $B$, verifying both the branch choice and the velocity normalization.