Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 7A (Further Complex Methods)

Putand take the branch cut $[-1,1]$. On $\mathbb{C}\setminus [-1,1]$, choose the analytic branch of a square root whose upper boundary value isIts lower boundary value is $s(x-i0)=-\sqrt{1-x^2}$. This fixes the integrand to be $+1$ at $0^{+}$ as required.

For $0\le \arg z\le \pi$, start at $0^{+}$ and use any path to $z$ lying above the cut; path deformations within that region do not alter the integral by the Cauchy integral theorem. For $\pi <\arg z<2\pi$, start in the same way, continue counterclockwise around the branch point $-1$ from the upper bank to the lower bank, and then continue to $z$ through the lower half-plane. This second prescription is the continuation obtained as $argz$ increases through $\pi$.

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

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) satisfiesThe change in the primitive is consequentlyEquivalently, this is the composition of the two reflections described by the monodromy of the complex inverse sine. Thus, for this orientation,The reverse orientation gives $\mathrm{Arcsin}(z)+2\pi$. 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: