I'm well aware that the Möbius transformations that take the unit disk to itself, $f:D \to D$, are given by $$ f(z) = \frac{e^{i \theta}(z-\alpha)}{1-\bar{\alpha}z}, $$ where $\theta\in [0,2\pi)$ and $|\alpha|<1$. Moreover, it's easy to see that regardless of the value for $\theta$, we have $f(\alpha) = 0$.

I'm also aware that Möbius transformations, in general, are completely determined by where they map three points; and that if you want to find the particular transformation that sends $z_1 \mapsto w_1$, $z_2\mapsto w_2$, and $z_3\mapsto w_3$, it is given by $$ f(z) = \frac{az+b}{cz+d}, $$ where $$ a = \det \begin{pmatrix} z_1 w_1 & w_1 & 1\\ z_2 w_2 & w_2 & 1\\ z_3 w_3 & w_3 & 1 \end{pmatrix}, \quad b = \det \begin{pmatrix} z_1 w_1 & z_1 & w_1\\ z_2 w_2 & z_2 & w_2\\ z_3 w_3 & z_3 & w_3 \end{pmatrix} $$ $$ c = \det \begin{pmatrix} z_1 & w_1 & 1\\ z_2 & w_2 & 1\\ z_3 & w_3 & 1 \end{pmatrix}, \quad d = \det \begin{pmatrix} z_1 w_1 & z_1 & 1\\ z_2 w_2 & z_2 & 1\\ z_3 w_3 & z_3 & 1 \end{pmatrix}. $$

So naturally, shouldn't there be some way of explicitly constructing a Möbius transformation that preserves the unit disk and takes $z_i \mapsto w_i$, for $i=1,2,3$, with $|z_i|,|w_i|<1$ given?

The problem that I'm running into is that the formula given above in terms of determinants is not preserving the unit disk. I'm using Mathematica to plot the image of a number of coordinate curves, and it's clear that the disk is not being preserved.

What am I missing? How should I choose my values of $\alpha$ and $\theta$ to make the correct transformation? In this case there are two degrees of freedom, yet in the other case there seem to be four. How do I reconcile this? Trying to equate the two formulas, and setting up a system of equations like $$ a = e^{i\theta}, \quad b = -\alpha e^{i\theta}, \quad c= -\bar{\alpha}, \quad d=1 $$ doesn't seem very helpful either.

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    $\begingroup$ How do you conclude that there must be a Mobius transformation that maps $D$ to $D$ and maps $z_i$ to $w_i$ for $i=1,2,3$. I think this is false. The unique Mobius transformation that maps $z_i$ to $w_i$ for $i=1,2,3$ may not map $D$ to $D$. $\endgroup$ Nov 4, 2018 at 0:10

1 Answer 1


What you're asking for is impossible.

First of all, we know that some Möbius transformations do not take the unit disc to itself, for example $f(z)=\frac{z+1}{0z+1} = z+1$.

On the other hand, if we pick any three distinct points $z_1,z_2,z_3$, and then compute $w_i=f(z_i)$ for $i=1,2,3$, it follows that $f$ is the unique Möbius transformation such that $w_i=f(z_i)$ for $i=1,2,3$.

So, using $f(z)=z+1$, let $z_1=1$, $z_2=2$, $z_3=3$. Then compute $w_1=2$, $w_2=3$, $w_3=4$. With those values of the $z_i$'s and $w_i$'s, what you want is doomed to failure.


Regarding your additional question asked in the comments regarding the Appolonian gasket, what that link is saying is that in this image, once you have chosen the outer circle $C$ and three points $x,y,z$ on the circle $C$, the next three circles inside $C$, tangent to each other, and tangent to $C$ at $x,y,z$, are determined by the choices of $C,x,y,z$. Also, all subsequent circles in the construction of the gasket are also determined. Then comes this sentence:

Since there is a Möbius transformation which maps any three given points in the plane to any other three points, and since Möbius transformations preserve circles, then there is a Möbius transformation which maps any two Apollonian gaskets to one another.

What this means is that if instead you had instead chosen three other points $x',y',z'$ on the circle $C$, then the Möbius transformation that takes $x,y,z$ to $x',y',z'$ must also take $C$ to $C$. And, it must take the first three circles of the $x,y,z$ gasket to the first three circles of the $x',y',z'$ gasket. And this continues, by induction, as you go down to deeper and deeper levels of circles.

  • $\begingroup$ I started going down this path because I was reading about Apollonian gaskets, and saw the comment about how you can use Möbius transformations to take any one gasket to any other, since these transformations preserve circles and each gasket is uniquely determined by the 3 points of tangency. So how would I go about explicitly finding this kind of a transition between two gaskets? $\endgroup$
    – Patch
    Nov 4, 2018 at 0:26
  • $\begingroup$ There might be a question there, and you might think about how to formulate it. But it's a very different question from the one that you asked in this post. $\endgroup$
    – Lee Mosher
    Nov 4, 2018 at 1:18
  • $\begingroup$ But if you can find a conformal mapping from one Apollonian gasket to another, isn't that going to make a conformal mapping from the unit disk to itself? Am I wrong? $\endgroup$
    – Patch
    Nov 4, 2018 at 1:39
  • $\begingroup$ Since I don't know what comment you are talking about regarding Apollonian gaskets, I don't have any response to your question regarding them. However I do think that you should think about some actual examples of actual Möbius transformations if you really want an answer to the question that you posted. $\endgroup$
    – Lee Mosher
    Nov 4, 2018 at 2:03
  • $\begingroup$ Here is what I was referring to: en.wikipedia.org/wiki/… $\endgroup$
    – Patch
    Nov 4, 2018 at 2:08

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