Conformal mapping of $z$-plane to lens-shaped region

Question

I have $$f(z)=\frac{1-(1-z)^{1/4}}{1+(1-z)^{1/4}}$$ I want to prove that $f(z)$ is a conformal mapping of the $z$-plane minus the real interval $[1, +\infty)$ onto the lens-shaped region bounded by two circular arcs that intersect at right angles at $w=\pm 1$.

I think I should find the distance between $\pm i$ and the points in the region, but I got a messy answer so it doesn't seem to work.

Answer 1

1. $1-z$ takes $\mathbb{C}\setminus[1,\infty)$ onto $\mathbb{C}\setminus(-\infty,0]$.
2. $(\,\cdot\,)^{1/4}$ takes this last domain to the region in the right hand plane between the straight lines $x\pm y=0$.
3. Finally, $\dfrac{1-(\,\cdot\,)}{1+(\,\cdot\,)}$ is a Möbius transformation. It takes $0\to1$ and $\infty\to-1$. The lines $x\pm y=0$ are transformed in circles that pass through the points $\pm1$, and cut at these points with a right angle. Moreover, they cut the real axis at a $45º$ angle.