Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 2 / 19 / c

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 $\alpha =V$ givesThe 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 $(\pm 1,0)$, 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.