A conformal map from the lens $|z-1|<2\cap|z+1|<2$ onto $\HH$

The two boundary circles meet at \[ z=\pm i\sqrt3. \] Set \[ T(z)=\frac{z-i\sqrt3}{z+i\sqrt3}. \] Then \(T\) sends the two intersection points to \(0\) and \(\infty\), so each of the two boundary circles is sent to a line through the origin, hence to a ray on the relevant boundary arc.

To identify the rays, note that \(-1\) lies on the circle \(|z-1|=2\) and \(1\) lies on \(|z+1|=2\). Direct computation gives \[ T(-1)=e^{2\pi i/3}, \qquad T(1)=e^{4\pi i/3}. \] Also \(0\) lies in the lens and \(T(0)=-1=e^{i\pi}\). Thus \(T\) maps the lens biholomorphically onto the sector \[ \frac{2\pi}{3}<\arg w<\frac{4\pi}{3}. \] After rotating by \(e^{-2\pi i/3}\) we obtain the sector \(0<\arg w<2\pi/3\). On this sector choose the branch of \(w^{3/2}\) with \(0<\arg(w^{3/2})<\pi\). Therefore \[ \boxed{ \Phi(z)= \left( e^{-2\pi i/3}\frac{z-i\sqrt3}{z+i\sqrt3} \right)^{3/2}} \] is a conformal bijection from the lens onto the upper half-plane.