Möbius transformations and concentric circles Given a Möbius transformation that maps one pair of concentric circles to another pair of concentric circles, why is the ratio of the radii preserved through the map? 
I thought about how Möbius transformations are compositions of rotations, scaling, inversion, and translation, and that intuitively, these types of maps shouldn't change the ratio of radii between two circles. 
Would it be correct to just say that if $\frac{r_1}{r_2}$ is the ratio of radii between the two circles, then 
1) The radii are invariant under translation, $z \mapsto z+a$, so $\frac{r_1}{r_2}$ stays the same
2) Under scaling by a factor $z \mapsto az$, $\frac{ar_1}{ar_2} = \frac{r_1}{r_2}$ 
3) Under inversion, $z \mapsto \frac{1}{z}$, $\frac{1/r_1}{1/r_2} = \frac{r_2}{r_1}$
Or is there a different/better way to think about this problem? 
 A: You may assume that all four circles are centered at the origin. The $x$-axis $l_1$ and the $y$-axis $l_2$ can be considered as circles that intersect the first pair of circles at $90^\circ$. They will be mapped by $f$ onto two circles that intersect the second pair of circles at $90^\circ$, and it is easy to see that this is only possible if the circles $f(l_i)$ are again lines through the origin. As $l_1$ and $l_2$, as well as $f(l_1)$ and $f(l_2)$, intersect at $0$ and $\infty$ it follows that $f$ either keeps $0$ and $\infty$ fixed or interchanges these two points. This in turn implies $f(z)=c z$ or $f(z)=c/z$ for a suitable $c\ne 0$. In both cases the ratio between the larger and the smaller radius of the two circles stays the same.
A: If you think of the two concentric circles as bounding an annulus $A_1$, and then of their images as bounding an annulus $A_2$, then the Möbius transformation is inducing a conformal transformation between $A_1$ and $A_2$, and it is a general fact the ratio of the inner and outer radii of an annulus is a conformal invariant.
This places the property you are asking about in a broader context, although I haven't answered the question as to why this general fact is true.  For the moment, let me defer to wikipedia for one proof of this fact.
A: Your (3) is correct only if both circles are centered at the origin. Below is a brief proof.
Let $T$ be the Möbius transformation in question. Let's say the pair of concentric circles $C_1$ and $C_2$ are both centered at $a$ of radii $r_1$ and $r_2$, and their images $T(C_1)$ and $T(C_2)$ are both centered at $b$ of radii of $r'_1$ and $r'_2$. Without loss of generality, we can let $a=b=0$ through translations.
The key observation is that 0 and $\infty$ are symmetric to each other relative to both $C_1$ and $C_2$. Therefore $T(0)$ and $T(\infty)$ must be symmetric relative to both $T(C_1)$ and $T(C_2)$. So we have $|T(0)| |T(\infty)| = {r'}_1^2 = {r'}_2^2$. Since $r'_1\neq r'_2$, this implies that either $T(0)=0$ and $T(\infty)=\infty$, or $T(0)=\infty$ and $T(\infty)=0$.
In the first case, $T$ is the combination of a rotation and a dilation, and $r_1/r_2=r'_1/r'_2$. In the second case, $T$ is the combination of a rotation and a dilation and an inversion, and $r_1/r_2=r'_2/r'_1$.
