My question regards how we expess $z=sin^{-1}(w)$ as a natural logarithm in the complex plane.
What I know is that we start off with $sin(z)=w$, we know that $sin(z) = \frac{e^{iz}-e^{-iz}}{2i}$, and through this we can create the equation $ {e^{2iz}-2iwe^{iz}-1=0}$.
What I have found is that when we solve for z using the quadratic formula, we get the result $z = -i\ln(iw\pm\sqrt{1-w^2})$. However, from what I've seen on a multitude of sites, apparently $z=sin^{-1}(w)$ is solely expressed as $z = -i\ln(iw+\sqrt{1-w^2})$.
Why is the negative branch of the square root not considered? I cannot find the reasoning behind it.
If you check for instance Matlab's Documentation
The inverse sine is defined as:
$$ \sin^{−1}(z)=−i \log\left[i z+(1−z^2)^{1/2}\right] $$
Also keep in mind that if this is supossed to be a function $f:\mathbb{C}\rightarrow \mathbb{C}$, it can only return one complex number, so we can't consider two branches at the same time.