Mobius transformation from region between two intersecting circles to an annulus I'm trying to find a Mobius transformation from from the region between the circles $|z|=1$ and $|z+1| = \frac {4}{\sqrt(3)}$ to an annulus. I've tried to find three points in the original region that map to an annulus. Specifically, for an annulus $o<|z+p|<q$
$(-1 -\frac {4} {3^{(1/4)}}i , -1 +\frac {4} {3^{(1/4)}}i,-1 -\frac {4} {3^{(1/4)}}) $ should go to $(-p-qi, -p+qi, -p-q)  $
But the algebra that I get from that is so convoluted that I'm not able to obtain a solution. Is there a less algebraic way to go about this or a way to simplify the  possible mobius transformation to make the algebra simpler?
 A: A Mobius transformation
$$w=\frac{\zeta-\alpha }{1-\bar{\alpha }\zeta}\quad (|\alpha |<1) \tag{1}$$
maps $|\zeta|<1$ to $|w|<1$. We use this formula.
We map first the region between the circles $|z|=1$ and $|z+1|=\frac{4}{\sqrt{3}}$ (Fig.1 below) to the region between the circles $C$ and $|\zeta|=1$ (Fig.2 below) by $\zeta=f(z)=\frac{\sqrt{3}}{4}(z+1)$. This is only a translation and scaling. The intersection points of $C$ and the real axis are $0$ and $\frac{\sqrt{3}}{2}$.
Next we map $|\zeta|<1$ to $|w|<1$ by ($1$) with $\alpha \in \mathbb{R}$, that is, by
$$
w=g(\zeta)=\frac{\zeta-\alpha }{1-\alpha \zeta}\quad(\alpha \in \mathbb{R}).$$
The point $\zeta=\alpha $ is mapped to the point $w=0$. By the symmetry $\alpha $ should be real.
We determine the exact value of $\alpha $. We want to map the region in Fig.2 to an annulus in Fig.3, hence we must map the circle $C$ to a circle $|w|=r$. Thus $g$ must satisfy
$$
g\left(\frac{\sqrt{3}}{2}\right)=-g(0).
$$ 
By easy calculations we have $\alpha =\frac{\sqrt{3}}{3}$ and $g(\zeta)=\frac{\sqrt{3}\zeta-1}{\sqrt{3}-\zeta}$.
Thus we obtain $$
w=g(f(z))=\frac{3z-1}{\sqrt{3}(3-z)}$$
as a solution.

A: You mean two nested non-intersecting circles, not two intersecting circles.
If you invert around a point on one circle, you transform the region $A_1$ between your two
circles to the region $B_1$ between a straight line and a circle.  Do the same for an 
annulus $A_2$ and you get another region $B_2$ between a straight line and a circle.  Check what you need for $B_1$ and $B_2$ to be related by a translation and scaling.
