Solution
= Solution
The exponential definition of <hyperbolic sine> gives $e^z-e^{-z}=0$, or $e^{2z}=1$. Taking every branch of the <complex logarithm> gives
$$
\boxed{\sinh z=0\iff z=i\pi k,\qquad k\in\mathbb Z.}
$$
For <hyperbolic cosine>, multiply $\cosh z+1=0$ by $2e^z$ to obtain $(e^z+1)^2=0$. Thus $e^z=-1$, giving
$$
\boxed{\cosh z+1=0\iff z=i(2k+1)\pi,\qquad k\in\mathbb Z.}
$$