I wish to find all Möbius transformations $T(z)=(az+b)/(cz+d)$ that map the circle $C=\{z\in\Bbb C:|z|=R\}$ into itself.

My attempt: Is it sufficient to find all Möbius transformations $T$ such that $|T(R)|=1$, $|T(0)|\neq 1$ and $|T(\infty)|\neq 1$ ?

  • $\begingroup$ If you use an inversion-like map $z \mapsto \frac{1}{z-R}$, that maps the circle $|z| = R$ to the line $\operatorname{Re} z = -\frac{1}{2R}$, so that would reduce the problem to finding transformations that preserve that line. $\endgroup$ – Daniel Schepler Jul 5 '20 at 18:50
  • $\begingroup$ $T(z)={1 \over R} z$ satisfies the conditions in your attempt but does not map the circle to itself. $\endgroup$ – copper.hat Jul 5 '20 at 19:00
  • $\begingroup$ @MartinR would it be possible to set this up using the cross ratio? $\endgroup$ – User7238 Jul 5 '20 at 19:11
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    $\begingroup$ Does this answer your question? Sufficient conditions for a mobius transformation to map the unit circle to itself. $\endgroup$ – Martin R Jul 5 '20 at 20:06
  • $\begingroup$ The following is related but is answering a different question (with more constraints) math.stackexchange.com/a/209407/27978. $\endgroup$ – copper.hat Jul 5 '20 at 20:41

I use the property that bilinear transformation maps inverse points w. r. t a circle to inverse points w. r. t. the image circle

Suppose that $w = \frac{az+b}{cz+d }$ is a transformation mapping $|z|=R$ onto $|w|=R$

Now $w=0$ , $w=\infty$ are inverse points for $|w|=R$ and they are transforms of

$z=-\frac{b}{a}, z=-\frac{d}{c}$


$\implies -\frac{b}{a}, -\frac{d}{c}$

are inverse point for $|z|=R$

if we write $\alpha = -\frac{b}{a}$

inverse point of any point $\alpha$ W. R. T circle is $|z|=R$ is $\frac{R^2}{\bar{\alpha }}$

$ \implies \frac{-d}{c}= \frac{R^2}{\bar{\alpha }}$

So we rewrite $ w$

$w= \frac{a(z+b/a) }{c(z+d/c) }$

Using above relations

$w = \frac{a}{c} \frac{z-\alpha }{z- R^2/ \bar{\alpha }}$

$w= \frac{a\bar{\alpha }}{c}. \frac{z-\alpha }{\bar{\alpha }z-R^2}$

Let $K =\frac{a\bar{\alpha }}{c}.$

$$w = K \frac{z-\alpha }{\bar{\alpha }z-R^2}$$

Using fact $|w|=R, |z|=R$ you can easily verify $|K|=R^2$


If $T$ maps the $R$ circle into itself then $ z\mapsto {1 \over R} T(Rz)$ maps the unit circle into itself and a similar relationship holds in the opposite direction so we can assume that $R=1$.

We are looking for transformations with $ad \neq cb$ that the unit circle into itself. Note that we can assume $c=1$.

Suppose $T$ is such a transformation, then for $|z|=1$ we have $|az+b|^2=|z+d|^2$. In particular, $a \neq 0$ and $|a|^2 +|b|^2 + 2 \operatorname{re} (a\bar{b}z) = 1+|d|^2 + 2 \operatorname{re} (\bar{d}z)$ or $|a|^2 +|b|^2 + 2 \operatorname{re} ((a\bar{b}-\bar{d})z) -1 -|d|^2= 0$. Since this holds for all $|z|=1$ we must have $\bar{a}b=d$ and so $|a|^2 +|b|^2 -1 -|a|^2|b|^2 = 0$ which gives $(|a|^2-1)(|b|^2-1) = 0$.

Since $ad\neq cb$ we have $|a|^2b \neq b$ from which we get $|a|^2 \neq 1$ and so $|b| = 1$.

Hence I claim that $T$ has the form $T(z) = {az+e^{i \theta} \over z+\bar{a}e^{i \theta}}$ with $a \neq 0$ and $|a| \neq 1$.

It is straightforward to check that any such $T$ is a Möbius transformation and if $|z|=1$, then $T(z) = { 1\over e^{i \theta}z} { az + e^{i \theta} \over e^{-i\theta} + \bar{a} \bar{z}} = = { 1\over e^{i \theta}z} { az + e^{i \theta} \over \overline{ e^{i\theta} + {a} {z}} }$ and so $|T(z)| = 1$.

Hence $T$ maps the $R$ circle into itself iff $T$ has the form $T(z) = R{az+Re^{i \theta} \over z+\bar{a}Re^{i \theta}}$ with $R \neq 0$, $a \neq 0$ and $|a| \neq 1$.

  • $\begingroup$ @MartinR: If a Möbius transformations maps the unit circle into the unit circle then it maps it onto the unit circle. This is because the inverse mapping (which must exist since it is Möbius) maps the unit circle into the unit circle. $\endgroup$ – copper.hat Jul 5 '20 at 20:11
  • $\begingroup$ Yes, I deleted my comment as soon as I realized that I had misread the question (“unit circle” vs “unit disk”). $\endgroup$ – Martin R Jul 6 '20 at 6:48
  • $\begingroup$ @MartinR: While solving the problem, it took me a while to realise that Möbius means it is invertible. Things are always clear afterwards :-). $\endgroup$ – copper.hat Jul 6 '20 at 7:05

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