Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 1 / 11 / b / Solution

Using the exponential definition of hyperbolic tangent, set $q=e^{2z}$:Here $q+1=1-i\ne 0$, so the resulting points are not poles. Taking all values of the complex logarithm gives $2z=i(-\pi /2+2\pi k)$. ThusEquivalently $\tanh (iu)=i\tan u$, so these are exactly the imaginary-axis solutions of $\tan u=-1$.