Prove analytic function on $\mathbb{C}\setminus [-1,1]$

Question

The question I have is:

Let $\Omega$ denote the open set obtained by removing the interval $[-1,1]$ from $\mathbb{C}$. Prove that there is an analytic function $F(z)$ on $\Omega$ such that $F(z)^2=\frac{z+1}{z-1}$. Hint: What is the image of $\Omega$ under the map $\frac{z+1}{z-1}$?

I know we have to use LFTs, but I honestly don't know where to start. I guess I'm just unable to picture what the image of the domain is under the given map. Any help would be appreciated.

Answer 1

$F$ is a Mobius transformation which maps $\mathbb{C} \backslash [-1,1]$ conformally onto $\mathbb{C} \backslash [-\infty,0]$. The analytic square root function defined by $re^{i\theta} \to \sqrt{r}e^{i\theta/2}$ is well defined there. So just compose these functions.