Mobius transformation from region between two intersecting circles to an annulus

Question

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?

Answer 1

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.

Answer 2

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.