= Solution
Set $w=e^z\ne0$. The <hyperbolic cosine> identity gives $w+w^{-1}=2i$, hence the <quadratic equation>
$$
w^2-2iw+1=0,\qquad w=i(1\pm\sqrt2).
$$
Let $L=\ln(1+\sqrt2)>0$. The first root has modulus $1+\sqrt2$ and <complex argument> $\pi/2$; the second has modulus $\sqrt2-1=e^{-L}$ and <complex argument> $-\pi/2$. Taking all <complex logarithm> values gives
$$
\boxed{z=L+i\left(\frac\pi2+2k\pi\right)\quad\text{or}\quad z=-L+i\left(-\frac\pi2+2k\pi\right),\qquad k\in\mathbb Z.}
$$
The two families are distinct because their <real parts> are respectively $L$ and $-L$. This exhausts the solutions because both roots for $w$ and all possible <complex arguments> have been included.
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