Q: A fractional linear transformation maps the annulus $r < \|z \| <1$ (where $r > 0$) onto the domain bounded by the two circles $\|z- \frac{1}{4} \|=\frac{1}{4}$ and $\|z \|=1$. Find $r$.
Attempt at solution: Consider a fractional linear transform $T(z)=\frac{az+b}{cz+d}$ Suppose that $c=0$. Then the transformation would be a scaling composed with a translation and hence would map concentric circles to concentric circles, which does not happen. Hence $c\neq 0$.
Thus, we can divide by it to get a map of the form $w=T(z)=\frac{Az+B}{z+D}$ Now, consider the $x$-axis in the $z$-plane. It intersects the two bounding circles of the annulus perpendicularly. Hence the image of the $x$-axis must intersect the images of the circles perpendicularly as well. We see in the $w$ plane the $x$-axis does that as well, so without loss of generality, we can take the $x$-axis in the $w$-plane to be the image of the $x$-axis in the $z$-plane.
This gives the relations $T(-1)=-1, T(-r)=0, T(r)=\frac{1}{2}, T(1)=1$. Using the first and fourth, we get $B-A=1-D, A+B=1+D$ and upon adding them together we get $2B=2$ hence $B=1$. Using the second relation, gives that $-Ar+1=0$, hence $A=1/r$.
The third relation gives $\frac{Ar+1}{r+D}=\frac{1+1}{r+D}=\frac{1}{2}$, hence $r+D=4$. Now using $D=4-r$ and $A=1/r$ in $A+B=1+D$ we have $\frac{1}{r}+1=1+(4-r)$ yielding the quadratic $r^{2}-4r+1=0$ with roots $2+\sqrt{3}, 2-\sqrt{3}$ and since $r< 1$ we must have $r=2-\sqrt{3}$.
Does this look okay?
Best Answer
I think you can consider the problem from another way: consider the inverse function which maps the unconcentric circles into concentric circles, which can be done by find the common symmetric points of two circles and map them to 0 and infinity.