= Solution
Let the argument of $z$ increase by $2\pi$ along a counterclockwise path enclosing both branch points. On a large circle, the branch chosen in part (a) satisfies
$$
\sqrt{1-t^2}\sim-it,
\qquad
\frac1{\sqrt{1-t^2}}\sim\frac{i}{t}.
$$
The change in the primitive is consequently
$$
\lim_{R\to\infty}\oint_{|t|=R}\frac{dt}{\sqrt{1-t^2}}
=\oint\frac{i}{t}\,dt
=-2\pi.
$$
Equivalently, this is the composition of the two reflections described by the <monodromy of the complex inverse sine>. Thus, for this orientation,
$$
\boxed{\operatorname{arcsin}(e^{2\pi i}z)=\operatorname{Arcsin}(z)-2\pi.}
$$
The reverse orientation gives $\operatorname{Arcsin}(z)+2\pi$. Both continued values invert the same $z$, so $\sin(w-2\pi)=\sin w$ on an open set of inverse values. The <identity theorem> extends this equality to every $w\in\mathbb C$, proving the <periodicity of the complex sine>:
$$
\boxed{\sin(w+2\pi)=\sin w.}
$$
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