= Solution
Simplify the real component to $u=(1-x^2-y^2)/D$. Thus $u=0$ is the unit <circle> with $(1,0)$ excluded from the mapping's domain. Its upper semicircle has $v>0$ and its lower semicircle has $v<0$; the point $(-1,0)$ maps to $u=v=0$. Inside the <disk>, $u>0$, so the principal arctangent is unambiguous. Setting $\alpha=V$ gives
$$
\boxed{\Psi(x,y)=\frac{2V}{\pi}\arctan\frac{2y}{1-x^2-y^2},\qquad x^2+y^2<1}.
$$
The <Laplace equation> follows from (b). Approaching an open upper <boundary> arc makes the ratio tend to $+\infty$, giving $+V$; the lower arc gives $-V$. This solution is $V(2F-1)$ in terms of the <harmonic measure of a semicircle in a disk> $F$.
The <boundary> data jump at $(\pm1,0)$, so no path-independent continuous <boundary> value can be imposed at those two junctions. The displayed bounded <harmonic function> is the solution with the specified limits on the open arcs; those endpoint discontinuities are part of the idealized <capacitor> geometry.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2011/paper-2-conformal-potential.png]
{title=Harmonic ray contours in the right half-plane and the potential in a disk with opposite semicircle boundary values}
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