1

For my current project, I am looking into Möbius transformations from the unit disk onto itself. Such Möbius transformations can be found, for example, by specifying three points and their images on the unit circle. Assume that I have the four parameters $a$, $b$, $c$ and $d$ of a given Möbius transformation:

$$M(z) = \frac{az+b}{cz+d}$$

Sometimes, the Möbius transformation causes an inversion, i.e. a mapping from the inside of the unit disk to the outside of the unit disk and vice versa.

Now my question: Is there a way to recognize whether a Möbius transformation induces an inversion directly through the parameters $a$, $b$, $c$, and $d$?

J.Galt
  • 961
  • Have you tried writing $z$ in polar coordinates? – Kenneth Goodenough Nov 10 '20 at 22:35
  • 1
    Not yet, no. I'm quite new to the field of complex algebra. How would this help? – J.Galt Nov 10 '20 at 22:37
  • Related: https://math.stackexchange.com/q/4482480/96384, https://math.stackexchange.com/q/34071/96384, https://math.stackexchange.com/q/1132830/96384, https://math.stackexchange.com/q/1705159/96384, https://math.stackexchange.com/q/209308/96384. – Torsten Schoeneberg Mar 29 '24 at 04:54

1 Answers1

1

If the Möbius transformation maps $\frac12+\frac12i$ outside the unit disc, then it indicates that the transformation is an inversion. Now $$|M(\tfrac12+\tfrac12i)|^2 = \frac{(a+b)^2 + b^2}{(c+d)^2+d^2},$$ and when this is $>1$, you have that $\frac12+\frac12i$ is mapped outside the disc. Thus you could take your condition to be that $$(a+b)^2+b^2>(c+d)^2+d^2.$$

Edit: As suggested in runaway44's comment, we can obtain an easier condition by seeing where $0$ is mapped. Indeed, $$|M(0)| = b/d,$$ so we just need $|b| > |d|$.

Luke Collins
  • 8,748
  • 1
    Why not apply $M$ to $0$? – anon Nov 10 '20 at 23:03
  • 1
    @runway44 I'm a bit hazy on the details of the Mobius transform, I forget how zero is treated. That would certainly simplify the condition. – Luke Collins Nov 10 '20 at 23:06
  • 2
    $M(0)=b/d$ and then the condition is $|b|>|d|$. – anon Nov 10 '20 at 23:07
  • Oh yeah, that is actually a very simple solution, runway44. Do you want to post it? Then I can accept it as a solution. – J.Galt Nov 10 '20 at 23:11
  • Well these criteria are sufficient but certainly not necessary. By basic theory, there are tons of Möbius transformations which map $0$ to some $b/d$ in the unit disk, even to $0$, but (say) $0.5$ to $17$, or $-3i$, or $\infty$; or, $\dfrac{z}{z-1}$ maps $0.9$ to $-9$ whilst keeping $0$ fixed, etc. – Torsten Schoeneberg Mar 23 '24 at 21:35
  • Of course, if you already know that the transformation maps the unit circle to itself, then it maps the unit disk to itself iff $|b|<|d|$, while it exchanges the inside and the outside of the unit circle iff $|b|>|d|$. – Torsten Schoeneberg Mar 29 '24 at 05:24