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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 $d$?

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  • $\begingroup$ Have you tried writing $z$ in polar coordinates? $\endgroup$ – Kenneth Goodenough Nov 10 '20 at 22:35
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    $\begingroup$ Not yet, no. I'm quite new to the field of complex algebra. How would this help? $\endgroup$ – J.Galt Nov 10 '20 at 22:37
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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 $|b| > |d|$.

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    $\begingroup$ Why not apply $M$ to $0$? $\endgroup$ – runway44 Nov 10 '20 at 23:03
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    $\begingroup$ @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. $\endgroup$ – Luke Collins Nov 10 '20 at 23:06
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    $\begingroup$ $M(0)=b/d$ and then the condition is $|b|>|d|$. $\endgroup$ – runway44 Nov 10 '20 at 23:07
  • $\begingroup$ Oh yeah, that is actually a very simple solution, runway44. Do you want to post it? Then I can accept it as a solution. $\endgroup$ – J.Galt Nov 10 '20 at 23:11

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