Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2023 / Paper 1 / 11 / d / Solution

Set $w=e^z\ne 0$. The hyperbolic cosine identity gives $w+w^{-1}=2i$, hence the quadratic equationLet $L=\ln (1+\sqrt{2})>0$. The first root has modulus $1+\sqrt{2}$ and complex argument $\pi /2$; the second has modulus $\sqrt{2}-1=e^{-L}$ and complex argument $-\pi /2$. Taking all complex logarithm values givesThe two families are distinct because their real parts are respectively $L$ and $-L$. This exhausts the solutions because both roots for $w$ and all possible complex arguments have been included.