When is the complex arccos real? Let $\theta=\arccos(x)$ for $x \in \mathbb{C}$. Is the following true?
$\theta \in \mathbb{R} \iff |x| \leq 1, \mbox{ and } x \in \mathbb{R} $
Or just
$\theta \in \mathbb{R} \implies |x| \leq 1, \mbox{ and } x \in \mathbb{R}$.
If the answer is yes, why is it true? If the answer is no, is there another condition that is equivalent to $\theta$ beeing real? And is there another condition that is equivalent(or implies) $|x| \leq 1$?
Note: I have just some basic knowledge in complex analysis.
 A: Cosines of real numbers are real as well as bounded absolutely by $1$. Thus properly,
$\theta\in\mathbb{R}\iff |x|\le1\color{blue}{\text{ and }x\in\mathbb{R}}.$
We can prove the equivalence by from the cosine as rendered in complex exponential. Thus $x=\cos\theta$ is rendered as
$x=\dfrac{\exp(i\theta)+\exp(-i\theta)}{2}$
Since the $\exp$ function is holomorphic and real for real arguments, $\exp(i\theta)$ and $\exp(-i\theta)$ must be a pair of complex conjugates, so their sum which comprises the numerator is real. These complex conjugates are also reciprocals because the arguments are negatives of each other, so they have modulus $1$. The Triangle Ineguality then forces the real numerator to have absolute value $\le2$. Thus $|x|\le 1$.
Moreover, both $x=1$ and $x=-1$ are actually realized for $\theta=0$ and $\theta=\pi$ respectively. By continuity/IVT  all real values between these bounds must also occur for some real $\theta$. Thus $|x|\le 1$ and $x\in\mathbb R$ is sufficient to generate a real $\theta$.
