Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2017 / Paper 1 / 11 / d / Solution

Set $w=e^z\ne 0$. The hyperbolic cosine equation becomes $w+w^{-1}=-4$, hence $w^2+4w+1=0$. Its two polynomial roots are $w=-2\pm \sqrt{3}$, both negative. Taking every value of the complex logarithm yieldsSince $(2-\sqrt{3})(2+\sqrt{3})=1$, their real logarithms have opposite signs. All solutions are thereforeThe positive real constant is also $\mathrm{arcosh}2$, the inverse hyperbolic cosine of $2$. Substitution gives $\cosh z=-\cosh (\ln (2+\sqrt{3}))=-2$ for both signs and every integer $k$.