 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 $\tan u=-1$.