Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
Maybe you should ask for a different way to think about it rather than a better way. Personally I think this is a really good way to look at it!! –  Eric Naslund Feb 25 '11 at 3:29
    
Fair enough, and thanks! I guess I was just wondering if this is a standard approach, because I couldn't think of a different way to solve it. –  user1736 Feb 25 '11 at 3:59
    
One potential problem with (3) is that it only applies to inversion in another concentric circle. What if you invert in a different circle? Then the radii don't behave this way. On the other hand, maybe these are the only inversions you have to care about: maybe inversion in a different circle will never take the concentric circles to concentric circles. (I don't actually know but I suspect this to be the case.) –  aaron Feb 25 '11 at 4:22
    
Sorry but could you clarify what you mean by inversion into a different circle (maybe an example of a specific map would help)? –  user1736 Feb 25 '11 at 4:25
    
An inversion operation is always with respect to some circle (or line). Inversion in a circle of radius $R$ fixes the circle, takes the center to infinity, and inverts distances to the center (so a point $r$ from the center ends up $R^2/r$ from the center). But not all distances get inverted. So if you have two concentric circles and you invert about another concentric circle, you will end up with two concentric circles, and your (3) would work. Whereas if you invert about a random circle, (3) doesn't really make sense . . . but the resulting circles probably won't be concentric anyway. –  aaron Feb 27 '11 at 16:25

2 Answers 2

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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.