= Solution
Use the exponential definition of the <hyperbolic tangent>:
$$
\frac{e^{2z}-1}{e^{2z}+1}=-i
\quad\Longrightarrow\quad
(1+i)e^{2z}=1-i
\quad\Longrightarrow\quad e^{2z}=-i.
$$
The <complex exponential> has modulus one here, so $\operatorname{Re}z=0$. Its arguments satisfy $2\operatorname{Im}z=-\pi/2+2k\pi$. Thus all solutions are
$$
\boxed{z=-\frac{i\pi}{4}+ik\pi,\qquad k\in\mathbb Z.}
$$
The denominator is then $1-i\ne0$, so no pole was admitted by cross-multiplication. Substitution also follows from $\tanh(i\theta)=i\tan\theta$ and the period $i\pi$.
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