Past exam of the mathematical methods I course of the University of Oxford / 2022 / 3 / Solution

Using the complex-angle identity,Since $\cosh y>0$, the real part can vanish only when $x=\pi /2+k\pi$. Then the imaginary part requires $\sinh y=(-1{)}^{k+1}$. Therefore all roots areSubstitution gives $\cos z=i$ for every integer $k$, and the positivity of $\cosh y$ shows that no other real parts were missed. The imaginary offsets are the two values $\pm \mathrm{arsinh}1$.