Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2012 / Paper 1 / 12 / d / Solution

Using the exponential definition of hyperbolic cosine, multiply $e^z+e^{-z}=-2$ by the nonzero quantity $e^z$. This yields $(e^z+1{)}^2=0$, hence $e^z=-1$.

If $z=x+iy$, its complex modulus requires $e^x=1$, so $x=0$; its phase then requires $y=(2k+1)\pi$. Thus all complex solutions areThere are no real solutions, since real $\cosh x\ge 1$.