= Solution
Let $w$ satisfy $\sin w=z$ and put $t=e^{iw}\ne0$. The exponential definition of the <complex sine> gives
$$
\frac{t-t^{-1}}{2i}=z,\qquad t^2-2izt-1=0.
$$
The <quadratic formula> therefore gives $t=iz\pm\sqrt{1-z^2}$. Taking the <complex logarithm> yields the multivalued <complex inverse sine> formula
$$
\boxed{w=-i\log\left(iz\pm\sqrt{1-z^2}\right).}
$$
The two quadratic roots have product $-1$, so neither is zero and their logarithms exist as multivalued quantities. If $w_0$ is one local inverse value, the complete set of values is $w_0+2\pi k$ and $\pi-w_0+2\pi k$, with $k\in\mathbb Z$.
\b[Branch qualification:] the source's $\pm$ formula describes the <branches of the complex inverse sine>, not two interchangeable expressions for a single chosen inverse branch. At $z=0$, the plus sign with the usual local choices gives $0$, whereas the minus sign with $\log(-1)=i\pi$ gives $\pi$. Thus both signs cannot simultaneously equal the principal inverse at zero. A principal <complex inverse sine> requires compatible square-root and logarithm branches, normalized to $w_0=0$ at zero and continued on the chosen cut domain.
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