= Solution
Choose paths from $0$ to $z$ in the cut plane and deform them continuously with the endpoint, never crossing the specified cut. With $0\leq\arg z<2\pi$, use the sheet reached from the positive real axis; when an endpoint is moved into the left half-plane, the path passes around the upper endpoint $i$ of the cut. The cut makes all such admissible paths homotopic with fixed endpoints, so the integral is single valued and its endpoint derivative is $(1+z^2)^{-1/2}$; hence it is <holomorphic function>[analytic].
For $z=-\sinh u$, continue the endpoint counterclockwise from the positive real axis through the upper half-plane. In
$$
\operatorname{Arcsinh}z
=\log\!\left(z+\sqrt{1+z^2}\right),
$$
the continued square root equals $-\cosh u$ at $z=-\sinh u$. The logarithm is reached with argument $\pi$, and therefore
$$
\boxed{\operatorname{Arcsinh}(-\sinh u)=u+i\pi}.
$$
The identities
$$
\sinh(w+2\pi i)=\sinh w,\qquad
\sinh((2k+1)\pi i-w)=\sinh w
$$
show directly why continuation without an argument restriction is multivalued. The complete set of values of the <multivalued inverse hyperbolic sine> at $\sinh u$ is
$$
\boxed{u+2\pi ik\quad\hbox{and}\quad
-u+(2k+1)\pi i,\qquad k\in\mathbb Z}.
$$
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