Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / 7 / iv

On the specified ellipse, the semiaxes are $\sqrt{2}$ and $1$, so its elliptic coordinate is $u_0=\mathrm{arcosh}\sqrt{2}=\log (\sqrt{2}+1)$. Thus in the exterior,is a harmonic function with a homogeneous Dirichlet boundary condition on that ellipse. Choose the square root with $\sqrt{z^2-1}\sim z$ at infinity. Although a global complex logarithm requires a cut, its real part $\log |z+\sqrt{z^2-1}|$ is well defined throughout the exterior and gives the same $\varphi$.

Expanding at large $|z|$ yields $z+\sqrt{z^2-1}=2z[1-1/(4z^2)+O(|z{|}^{-4})]$. ThereforeIn particular the potential grows logarithmically in the distance, in every direction at infinity. It is the exterior solution; no harmonic continuation through the focal segment in the ellipse's interior is claimed.