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I'm looking at "Mobius transformations" where $a,b,c,d\in\mathbb R$. I want to know for which $n$ there exists $a,b,c,d$ such that for $f(x) = \dfrac{ax+b}{cx+d}$,

$$f^n(x) = f(f(...(f(x)))) = x$$

and what relationships between $a,b,c,d$ are required. Or if it is for all $n$, if there is a pattern to these relationships.

For example,

$$f^1(x)=x \iff a-d=0, c= 0, b=0, a\neq 0$$

$$f^2(x)=x \iff a+d=0, a^2+bc\neq 0$$

I see that for $f^{2k}(x)$, we can get an iterative relationship from the above. With the same conditions as the $n=2$ case.

Did
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David P
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    The case $n=2$ is also solved by $b=c=0$, $a=\pm d\ne0$. // To solve the general case, consider the matrices $M(f)=\begin{pmatrix}a&b\c&d\end{pmatrix}$ then $M(fg)=M(f)M(g)$ for every Möbius transforms $f$ and $g$ hence $f^n$ is the identity if and only if $M(f)^n=t,I$ for some nonzero $t$. – Did Sep 10 '16 at 19:09

1 Answers1

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As said by @Did, you need to find coefficients a,b,c,d such that

$$M(f)^n=\begin{pmatrix}a&b\\c&d\end{pmatrix}^n=k\begin{pmatrix}1&0\\0&1\end{pmatrix}$$

Thinking to rotation matrices, there is an evident solution :

$$M=\begin{pmatrix}\cos(\frac{\pi}{n})&-\sin(\frac{\pi}{n})\\ \sin(\frac{\pi}{n})&\cos(\frac{\pi}{n})\end{pmatrix}$$

Otherwise said, a possible Möbius (or homographic) transformation is :

$$f_n(x)=\dfrac{\cos(\frac{\pi}{n})x-\sin(\frac{\pi}{n})}{\sin(\frac{\pi}{n})x+\cos(\frac{\pi}{n})}$$

Edit : This rotation is not unique in general. Let us take an example: if $n=12$, you can take any constant $K=1,2,\cdots 11$ in the following matrix

$$M=\begin{pmatrix}\cos(\frac{K\pi}{n})&-\sin(\frac{K\pi}{n})\\ \sin(\frac{K\pi}{n})&\cos(\frac{K\pi}{n})\end{pmatrix}$$

and have $M^n=\pm I_2.$

(following a very judicious remark of "studiosus") a very general type of non trivial matrices $M$ such that $M^n=\pm I_2$, at least among diagonalizable matrices is obtained by thinking to the conjugation operation, that doesn't change the eigenvalues that will still be $e^{iK\pi/n}$ and $e^{-iK\pi/n}$:

$$M=P\begin{pmatrix}\cos(\frac{K\pi}{n})&-\sin(\frac{K\pi}{n})\\ \sin(\frac{K\pi}{n})&\cos(\frac{K\pi}{n})\end{pmatrix}P^{-1}$$

for any invertible matrix $P.$

Jean Marie
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  • Does this characterize all such solutions? This is kind of a generalization of a challenge problem I was planning on posing to pre-calculus students. Evidently the case for n>2 seems to be out of reach for them – David P Sep 10 '16 at 20:41
  • I partially answer to your question by an edit to my answer. – Jean Marie Sep 10 '16 at 20:45
  • See as well the similar answer I gave some months ago to a similar question (http://math.stackexchange.com/q/1659401) and the very nice article I give as a reference at the end of this answer – Jean Marie Sep 10 '16 at 21:29
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    All are conjugate to the transformations of the form $f(z)= \exp(2\pi m/n)z$. – Moishe Kohan Sep 10 '16 at 21:46
  • @stodiosus You are right. I realize that the end of my answer (in the Edit part) lacks the "up to a conjugation". I correct it. – Jean Marie Sep 10 '16 at 21:58
  • Lastly: A matrix of finite order is always diagonalizable (just use Jordan normal form to see this). – Moishe Kohan Sep 12 '16 at 09:45
  • @studiosus 2 Thank you very much for this "last" argument I hadn't thought about. – Jean Marie Sep 14 '16 at 10:25