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Let $w$ satisfy $\sin w=z$ and put $t=e^{iw}\ne 0$. The exponential definition of the complex sine givesThe quadratic formula therefore gives $t=iz\pm \sqrt{1-z^2}$. Taking the complex logarithm yields the multivalued complex inverse sine formulaThe 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 $k\in \mathbb{Z}$.

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.

Start with the germ $w=\mathrm{Arcsin}(z)$ selected in part (a) and continue it along paths in $\mathbb{C}\setminus \{-1,1\}$. At the two branch points,The monodromy reflection at a square-root branch point therefore shows that continuation once around $1$ sends $w$ to $\pi -w$, while continuation once around $-1$ sends it to $-\pi -w$. Composing these reflections produces every translation by an even multiple of $\pi$. Hence the branches of the complex inverse sine are

Every path can change sheets only by applying the two reflectionsTheir products satisfyThus an even number of reflections gives a translation by an integral multiple of $2\pi$, while an odd number gives a translated reflection. All possible values are thereforein agreement with the branches of the complex inverse sine.