The range of $\arccos$ My question is whether or not the function $\arccos$ takes complex numbers to complex numbers? Specifically, if we identify $\mathbb{R}$ with the subset of the complex numbers which have zero imaginary part, then is it true that $z \in \mathbb{C}\setminus \mathbb{R}$ implies that $\arccos(z) \in \mathbb{C}\setminus\mathbb{R}$? 
I know that $\arccos$ can be written as 
$$ \arccos(z) = -i\log(z + \sqrt{z^2 - 1})$$
where one has to then fix a branch of $\log$ to work with, but I just haven't been able to show anything with this approach, which sort of makes me believe that it might not be true. Any insight (or potentially a counter-example) would be much appreciated!

It's much easier if you use the property $\cos (\arccos z) = z$ of $\arccos$. The contrapositive of
$$z\in\mathbb{C}\setminus \mathbb{R}\implies \arccos z \in \mathbb{C}\setminus\mathbb{R}$$
is
$$\arccos z\in \mathbb{R} \implies z \in \mathbb{R},$$
and we can immediately see that that is true since $\cos(\mathbb{R}) \subset\mathbb{R}$.
