Is complex $sin ^{-1} z$ multivalued for real numbers bigger than one?

Question

I am trying to understand the complex sin function. Let $a>1$ be a real number. How many solutions does

$\sin z= a$

has inside a circle of radius $R$ centered at the origin? Could it have infinite solutions?

Answer 1

Given a solution to the equation we may construct an infinite set of solutions by adding multiples of $2\pi$ to the real part of the solution since: $$ \sin(z + 2k\pi) = \dfrac{e^{iz + i2k\pi} - e^{-iz - i2k\pi}}{2i} = \dfrac{e^{iz} - e^{-iz}}{2i} = \sin(z). $$ In fact there are two such sets: Let $z = \theta - i\log r$ and $\omega = r e^{i\theta}$ then $$ \sin(z) = a \implies \omega^2 - 2ia \omega - 1 = 0, $$ yielding two solutions for $\omega$ $$ \omega_1 = r_1 e^{i(\theta_1 + 2k\pi)}, \quad \omega_2 = r_2e^{i(\theta_2 + 2k\pi)}, $$ and a parameterised solution space for $z$ $$ z\in\{\theta_i + 2k\pi + i\log r_i:k\in\mathbb{N}, i=1,2\}. $$ To answer your question directly: if $|z|<R$ then the solution space is finite.

Answer 2

If it took the same value $a$ at an infinite subset $S$ of $|z|\leq R$, then $S$ would have a limit point inside $|z|\leq R$ and $\sin z$ would agree with $z=a$ on a set with a limit point. Therefore, $\sin z$ would be constant, so the answer is "no."

Answer 3

Yes, the complex arcsine function is, indeed, multivalued - but so too is, strictly speaking, its real counterpart. That $\sin^{-1}(x)$, for $x \in \mathbb{R}$, ranges in $(-\pi, \pi]$ is an arbitrary convention. Any interval of width $2\pi$ will work and give an equally good "arcsine".

In the complex domain, we need to do a bit more arbitrary choosing. The points $z = \pm 1$ are branch points indeed, and so what we do is to, in addition to setting what values are taken at real inputs in $[-1, 1]$, to take branch cuts along $(-\infty, -1)$ and $(1, \infty)$. We then extend it to be continuous from above and below, respectively, on these intervals. These arbitrated properties are enough to single out a unique, maximal extension of the real arcsine as a function, and that becomes the meaning of the symbol $\sin^{-1}(z)$ when the input is a complex number, or also, a real number outside of $[-1, 1]$.