Past exam of the mathematics course of the University of Cambridge / 2026 / ib / Paper 1 / 11 / 11.2C / iii / Solution

For $\zeta =R{e}^{i\theta }$,
$x=\frac{c}{2}(R+R^{-1})\cos \theta$ and $y=\frac{c}{2}(R-R^{-1})\sin \theta$. Thus $a=c(R+R^{-1})/2$, $b=c(R-R^{-1})/2$. Nested level curves and the behaviour at infinity show that exteriors correspond.

Solved by gpt-5.6-sol high.