Domain of the function $f(z) = \sqrt{z^2 -1}$ What will be the domain of the function $f(z) = \sqrt{z^2 -1}$?
My answers are: $(-\infty, -1] \cup [1, \infty)$ OR $\mathbb{R} - \lbrace1>x\rbrace$ OR $\mathbb {R}$, such that $z \nless 1$.

The first part of your answer (before the "or") is correct:
The domain of your function, in $\mathbb R$ is indeed $(-\infty, -1]\cup [1, \infty).$  That is, the function is defined for all real numbers $z$ such that $z \leq -1$ or $z \geq 1$.
Did you have any particular reason you included: this as your answer, along with "or...."? Did you have doubts about the above, that you were questioning whether the domain is not $(-\infty, -1] \cup [1, \infty)$?
Why is the domain $\;\;(-\infty, -1] \cup [1, \infty) \subset \mathbb R\;$? 
Note that the numbers  strictly contained in $(-1, 1)$, when squared, are less than $1$, making $\color{blue}{\bf z^2 - 1 < 0}$, in which case we would be trying to take the square root of a negative number - which has no definition in the real numbers. So we exclude those numbers, the $z \in (-1, 1)$ from the domain, giving us what remains. And so we have that our function, and its domain, is given by:
$$f(z) = \sqrt{z - 1},\quad z \in (-\infty, -1] \cup [1, \infty) \subset \mathbb R$$ 
