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

Inverting $u+iv=\mathrm{arcosh}z$ gives $z=\cosh (u+iv)$, hence the elliptic coordinatesEliminating $v$ at fixed $u>0$ gives the ellipseIts focal distance is $\sqrt{{\cosh }^{2}u-{\sinh }^{2}u}=1$. At fixed $v$, eliminating $u$ instead gives the hyperbolaIts focal distance is also one, so both families have foci $(\pm 1,0)$. A fixed $v$ traces an appropriate half of the hyperbola; complementary coordinate values complete it. Degenerate values give the focal segment and portions of the coordinate axes. Away from these degeneracies the curves are orthogonal by part (i), or by ${\mathbf{r}}_u\cdot {\mathbf{r}}_v=0$ in the displayed parametrization. A branch of the inverse is used locally; its real part with $u\ge 0$ is single-valued outside the focal segment.