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The Möbius transformations are the maps of the form $$ f(z)= \frac{az+b}{cz+d}.$$ Can we characterize the Möbius transformations that map the unit circle into itself?

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What have you tried so far? An obvious way is to try to solve $|f(e^{i\varphi})|=1$ –  fgp Oct 8 '12 at 16:09
    
I mean $|z|<1 \Rightarrow |f(z)|<1$ –  Daniel Oct 8 '12 at 16:15
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$\{z \in \mathbb{C}:\ |z|<1\}$ is the unit disk, not the unit circle! –  Mercy Oct 8 '12 at 16:30
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@Daniel'kay. Please fix the question to say "unit disc", then! Anyway, the same question applies. Have you tried to solve $|f(re^{i\phi})| < 1$ for $r < 1$? –  fgp Oct 8 '12 at 16:37
    
I think the set of analytic functions mapping the unit disc to itself reduce to Mobius transformations. –  PAD Oct 8 '12 at 19:26

6 Answers 6

up vote 6 down vote accepted

Consider the function $$f(z)=\frac{e^{i \theta}(z-a)}{1- \bar {a}z}$$ where $a$ is in interior of the disk.

Now we have two parts to prove:

  1. It maps Unit circle to unit circle and $a$ to $0$.Easy.

  2. Every Möbius transformation that preserves the unit disk must be of the above form.

This can be proved quite easily by noting that every Möbius transformation is uniquely determined by its action on $3$ points. Take the points be $1,0, \infty$. Try.

If you are stuck, comment.

thanks.

EDIT: I misread the question to be sending unit disc instead of circle.

Please ignore.

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Yes to the three points. Input is on the disk though, so no to $\infty$, though I realize the form is a transformed one...(+1 vote from me) –  adam W Oct 8 '12 at 19:47
    
I don't understand. Möbius transforms are $\mathbb C_\infty \to \mathbb C_\infty$. Hence we may use any $3$ numbers. –  TheJoker Oct 8 '12 at 19:57
    
He asks about the disk into the disk, so he is only interested in $|z|<1$ and that is why I am questioning the use of $\infty$. So if your function works (looks good to me), does it describe the entire set of possible functions? –  adam W Oct 8 '12 at 20:57
    
These functions are sometimes called Blaschke factors. –  Alice Dec 3 '13 at 21:22

Let me try to answer the OP's original question; i.e., to find all the Möbius transformations that map the unit circle to itself.

First, we have the result in one of the other answers that the Möbius maps sending the unit disc to itself consist of precisely the functions $$f(z)=\frac{e^{i \theta}(z-a)}{1- \bar {a}z}$$ where $a$ is in interior of the disk.

Now, recall that Möbius transformations are actually holomorphic automorphisms of the Riemann sphere. So by continuity, any of such transformations as above map the unit circle to itself. It remains to be seen what other Möbius transformations preserve it. But then, they send circles to circles, and if it sends some point in the interior of the unit disc to another point in the unit disc, by continuity, the interior unit disc is sent to itself, and the transformation belongs to the above class. On the other hand, if some point in the interior of the disc is sent outside, then if $f$ is the transformation, $1/f$ belongs to the previous class.

So the answer is that the Möbius transformations sending the unit circle to itself are precisely the Möbius transformations sending the unit disc to itself, and their multiplicative inverses.

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$|z|=1$ an arbitrary point on the unit circle (I am assuming that you meant to talk about complex points on the complex plane). So your reworded question is, does $a,b,c,d$ exist such that $$|z|=1 \Rightarrow \left|\frac{az+b}{cz+d}\right|=1$$ This is, using laws of multiplication within the norm, same as: $$|z|=1 \Rightarrow \left|az+b\right|= \left|cz+d\right|$$ And it looks to me that of all possibilities, none include shifts of any sort since they would shift the unit circle away from the origin. Thus one possibility is a rotation of the points of the unit circle:

$$f(z)=\frac{az}{d}$$ Where $\left| \frac{a}{d}\right|=1$. For the unit disk, $|z| \le 1$. It becomes a matter of the two linear functions. It would require $$\left|az+b\right| \le \left|cz+d\right| \quad $$ For every $|z| \le 1$. For the complex plane, that means the scaling and shifting of the disks, such that the one remains entirely within the other (well almost, as I can imagine a possibility of the particular values for a certain $z$ not satisfying the inequality).

If it is any more of a help, I imagine it as a scaling and shifting, but the shift must be one that does not "outrun" the scaling and shifting of the other. Since if the "race from the zero point" is ever being won by the numerator, the Möbius has value outside the unit circle.

I believe any characterization of such Möbius functions would involve a separate rotation for both the numerator and denominator, as that does not alter the magnitude. It would align the two (numerator and denominator) to be aligned if you will, so that the magnitude characterization of the function may be analyzed. So it would be considering: $$f'(z) = \frac{r_n(az+b)}{r_d(cz+d)}$$ where $r_n$ and $r_d$ are any magnitude one values that make analysis more convenient. The $f'(z)$ would then exhibit the exact same behavior in the magnitude, and each $r$ separately represents a rotation. Then, if they are chosen correctly (the fastest in the "race" vs the "slowest" or similar), the function may be considered only along a single path, and it would become the comparison of two lines, if one always has magnitude larger than the other in the range [-1,1], then their (magnitude) ratio is always less than one.

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Seems like you're guessing in a couple points. –  Graphth Oct 8 '12 at 16:47
    
This isn't right. For example, there is a function of this form that leaves $1$ and $-1$ fixed and maps $i$ to $\frac35+i\frac45$. –  Michael Hardy Oct 8 '12 at 17:18
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And $z\mapsto \frac1z$ obviously leaves the unit circle fixed. –  MJD Oct 8 '12 at 17:39
    
Sorry, I may have meant to say one possibility, wasn't too sure it was the only possibility. Good correction, thank you. –  adam W Oct 8 '12 at 18:01
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@mjd The question is actually about the unit disc... if you look at the comments to the question. –  Graphth Oct 8 '12 at 18:15

Most answers seem to be characterizing Möbius transformations which map the unit disk onto itself, which is relatively well-known. If you are asking which map the disk into itself, the article here gives a simple proof that $|z|<1 \Rightarrow |f(z)|<1$ if and only if $$|b\overline{d}-a\overline{c}|+|ad-bc|\leq |d|^2-|c|^2 $$

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Tried your link, but Google blocked the book from me. Though I am glad someone mentioned the difference between into and onto. –  adam W Oct 8 '12 at 23:25

These transformations make a group, isomorphic to $PSL_2(\mathbb R),$ which takes the upper half plane to itself. The general form, with complex numbers $\alpha, \beta$ and $|\alpha| > |\beta|,$ is $$ f(z) = \frac{\alpha z + \beta}{\bar{\beta} z + \bar{\alpha}}. $$ This is the result of taking real numbers $a,b,c,d$ with $ad-bc > 0$ and calculating $$ \left( \begin{array}{rr} 1 & -i \\ -i & 1 \end{array} \right) \cdot \left( \begin{array}{rr} a & b \\ c & d \end{array} \right) \cdot \left( \begin{array}{rr} 1 & i \\ i & 1 \end{array} \right) = \left( \begin{array}{rr} (a+d) +(b-c)i & (b+c) +(a-d)i \\ (b+c) + (d-a)i & (a+d) + (c-b)i \end{array} \right). $$

We need the modulus of $\alpha$ to be the larger so that $|f(0)| < 1.$ For your own comfort, check that $f(1), f(-1), f(i), f(-i)$ all have modulus $1.$

To get down to three real variables underlying the thing, we may divide through by the positive real number $|\alpha|,$ thereby demanding $\alpha = e^{i \theta}$ have modulus $1,$ then $|\beta| < 1,$ using a new variable $\gamma$ with $|\gamma| < 1$ we have $$ f(z) = \frac{ e^{i \theta} z + \gamma}{\bar{\gamma} z + e^{-i \theta}}. $$

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$z\mapsto \frac{i-iz}{z+1}$ maps the unit circle to the real axis and its interior to the upper half plane. The maps $f(z)=\frac{az+b}{cz+d}$ fixing the upper half plane are possibly easier to describe, and you can combine them with the "disc to halfplane" and "halfplane to disk$ maps:

  • It must map $0$ to $\infty$ or a real number, hence $d=0$ or $\frac bd\in\mathbb R$.
  • It must map $\infty$ to $\infty$ or a real number, hence $c=0$ or $\frac ac\in\mathbb R$.
  • The $z$ with $f(z)=0$ must be $\infty$ or real, hence $a=0$ or $\frac ba\in\mathbb R$.

Also note that you may set one nonzero number wlog. to be $1$. This will hopefully helpp you

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