= Solution
Start with the germ $w=\operatorname{Arcsin}(z)$ selected in part (a) and continue it along paths in $\mathbb C\setminus\{-1,1\}$. At the two branch points,
$$
\operatorname{Arcsin}(1)=\frac\pi2,
\qquad
\operatorname{Arcsin}(-1)=-\frac\pi2.
$$
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
$$
\boxed{\operatorname{arcsin}(z)
=n\pi+(-1)^n\operatorname{Arcsin}(z),
\qquad n\in\mathbb Z.}
$$
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