Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I would like to understand the map of $\mathbb{C}$ to $\mathbb{C}$ that results by iterating inversion in a unit circle. Let $f(z)$ for $z \in \mathbb{C}$ invert $z$ in a unit circle centered on $q_1$, and then in a unit circle centered on $q_2$, etc. Here is an example:
            Inversion
The black path shows the trajectory of one point $z$, which ends up inside the $q_5$ disk. The blue vectors show the complete map's effect on a number of (random) points. The end result resembles an inversion in the $q_5$ circle, but I imagine the plane is partitioned into regions that behave similarly due to their relationship to the several circles of inversion. Is this a correct way to view this map? And if so, is there a natural diagram to elucidate the partition?

I wonder if instead it may be better to view this map as an approximation to another map? For there is a continued-fraction representation of $f(z)$, as follows. The inversion of $z$ in a unit circle centered on $q$ can be expressed as $q + \frac{1}{z^* -q^*}$, where ${\cdot}^*$ represents the complex conjugate operation. Then iterating this (say, five times) leads to this expression: $$f(z) = q_5 + \frac{1} {q_4^*-q_5^* + \frac{1} {q_3-q_4+ \frac{1} {q_2^*-q_3^*+ \frac{1} {q_1-q_2+\frac{1}{z^*-q_1^*} } } } } $$

I realize I have not posed a sharp question. I am just seeking some approach that helps understand the composition of circle inversions, a bit far from my expertise. Thanks for pointers or ideas!

share|improve this question
3  
As far as I can see, the composition of $n$ inversions is a Möbius transformation, or a Möbius transformation followed by conjugation, depending on the parity of $n$. It thus should not involve any partitions. –  Emil Jeřábek Dec 14 '11 at 15:42
    
@Emil: Interesting! And yet different $z$s follow rather disparate paths depending on whether they land in or out of specific disks. Which is not to say I doubt you--rather, I appreciate this insight all the more! –  Joseph O'Rourke Dec 14 '11 at 16:31
1  
This is just a bit of free association, but your question reminds me a bit of the notion of a Schottky group en.wikipedia.org/wiki/Schottky_group . I learned about them from the fantastic book "Indra's Pearls" by Mumford, Series, and Wright. –  j.c. Dec 14 '11 at 18:23

2 Answers 2

up vote 4 down vote accepted

The maps that you get by composing multiple inversions are exactly the Möbius transformations and their reflections.

In terms of complex number coordinates for the planes, the compositions of even numbers of inversions are the fractional linear transformations $$z\mapsto \frac{az+b}{cz+d}$$ (where $a$, $b$, $c$, and $d$ are complex numbers with $ad-bc\ne 0$, and multiplying all four of them by the same scalar leaves the transformation unchanged). The compositions of odd numbers of inversions are their complex conjugates.

share|improve this answer

I am not sure what you are looking for, however this wikipedia page seems a propos.

http://en.wikipedia.org/wiki/Generalized_continued_fraction

share|improve this answer
    
Thanks, Igor, yes, I looked at that wealth of information earlier without gaining significant insight. Which is not to say it might not help! –  Joseph O'Rourke Dec 14 '11 at 16:07

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.