= Solution
Set $w=e^z\ne0$. The <hyperbolic cosine> equation becomes $w+w^{-1}=-4$, hence $w^2+4w+1=0$. Its two <polynomial roots> are $w=-2\pm\sqrt3$, both negative. Taking every value of the <complex logarithm> yields
$$
z=\ln|w|+i(2k+1)\pi,\qquad k\in\mathbb Z.
$$
Since $(2-\sqrt3)(2+\sqrt3)=1$, their real logarithms have opposite signs. All solutions are therefore
$$
\boxed{z=\pm\ln(2+\sqrt3)+i(2k+1)\pi,\qquad k\in\mathbb Z.}
$$
The positive real constant is also $\operatorname{arcosh}2$, the <inverse hyperbolic cosine> of $2$. Substitution gives $\cosh z=-\cosh(\ln(2+\sqrt3))=-2$ for both signs and every integer $k$.
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