How to Express the domain of the mult-valued function $\sqrt{1-z^{2}}$?

Question

In the analysis of the function $w(z)=\sqrt{1-z^{2}}$ I've set the branch cuts from $-\infty$ to $-1$ and from $1$ to $\infty$. I would like to know how can I express the domain of this function when the branch cut is set this way?

Answer 1

As a subset of $\mathbb C$ the domain is just what you described, i.e., $\mathbb C\setminus(-\infty, -1]\setminus[+1,+\infty).$

If you want a domain where the multivalued square root becomes single-valued, use two copies of $\mathbb C$ with the open half lines doubled and the points $\left\{-1,+1\right\}$ taken away, and identify the half line that is the boundary of the upper half plane from one copy with the half line that is the boundary of the lower half plane from the other copy.

As a set,

$$D=\mathbb C_1\setminus\left\{-1,+1\right\}_1\cup\mathbb C_2\setminus\left\{-1,+1\right\}_2$$

where the indices 1 and 2 are used to distinguish disjoint copies.

The function $f$ can be defined single-valuedly on $D$ and the values it takes on the two indexed versions of a complex number are each other's negatives.