Can we characterize the Möbius transformations that maps the unit circle into itself?

Question

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?

Answer 1

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.

Answer 2

$|z|=1$ 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 mobius has value outside the unit circle.

I believe any characterization of such mobius 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 analized. 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 comparision 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.

Answer 3

$z\mapsto \frac{i-iz}{z+1}$ 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

Answer 4

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 $$ f(z) = \frac{ e^{i \theta} z + \gamma}{\bar{\gamma} z + e^{-i \theta}}. $$

Answer 5

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 $$|b\overline{d}-a\overline{c}|+|ad-bc|\leq |d|^2-|c|^2 $$