Finding the Mobius Transformation that maps open unit disk onto itself

Question

Hi I am trying to find all the Mobius transformations that map unit open disk onto itself i.e., if $|z|<1$ then $|f(z)|<1$ where $f(z)=\frac{az+b}{cz+d}$. I did so far \begin{align*} &\Big|\frac{az+b}{cz+d}\Big|<1\\ \Rightarrow & |az+b|<|cz+d|\\ \Rightarrow & |az+b|^2<|cz+d|^2\\ \Rightarrow & |az+b||\bar{a}\bar{z}+\bar{b}|<|cz+d||\bar{c}\bar{z}+\bar{d}|\\ \Rightarrow & |a|^2|z|^2+|b|^2+2\text{Re }(az\bar{b})<|c|^2|z|^2+|d|^2+2\text{Re }(cz\bar{d}) \end{align*}

After that I am stuck. Can anyone help me. I would be obliged...Thanks in advance.

Answer 1

There is a derivation of the bijective Möbius transformations $\mathbb{D} \to \mathbb{D}$, the non bijective ones are easily found from this

• Without loss of generality, you can look at $$g(z) = C\frac{z-A}{z-B}, \qquad A \in (0,1),\ |B| > 1,\ C > 0$$

  $f(z) = \frac{az+b}{cz+d} = \frac{a}{c}\frac{z+b/a}{z+d/c}$ and $g(z) = e^{-i\text{arg} C} f(-ze^{i\text{arg}(b/a)})$ has the form $C\frac{z-A}{z-B}, A > 0, C > 0$. Since $g(z)$ is bijective $\mathbb{D} \to \mathbb{D}$, the point where $g(\rho) = 0$ must be inside $\mathbb{D}$, so $A \in \mathbb{D} \implies A \in (0,1)$. Finally $|B| > 1$ since $g(z)$ is holomorphic on $\mathbb{D}$ and $B \ne A$.

• By the maximum modulus principle (not hard to prove) you have that $g$ maps $\partial \mathbb{D} = \{|z|=1\}$ to itself, so that $$1 = |g(\pm i)| = |C| \frac{|\pm i-A|}{|i-B|} = |C| \frac{\sqrt{1+A^2}}{|\pm i-B|} $$ i.e. $|B-i| = |B+i| \implies B \in \mathbb{R}$,

  and $1 = |g(1)| = C \frac{1-A}{|B-1|}= |g(-1)| = C \frac{1+A}{|B+1|}$ means $|B-1| < |B+1|$ so that $B > 0 \implies B > 1$.

• Finally since $A \in (0,1),B > 1, C > 0$, you have $g(1) < 0, g(-1) > 0$ and $|g(1)| = |g(-1)| =1 $ means $$g(1) =C \frac{1-A}{1-B} =-1,\qquad g(-1) = C\frac{-1-A}{-1-B} = 1$$ i.e. $\frac{1-A}{B-1} = \frac{1+A}{B+1} \implies (1-A)(B+1) = (1+A)(B-1) \implies AB = 1$ and $f(-1) = 1 \implies C = \frac{1+1/A}{1+A} = 1/A$.

• Going back to $f(z)$, you get the desired general form $$f(z) = e^{i \theta}\frac{z-a}{1-\overline{a}z}, \qquad |a| < 1$$

Answer 2

Hints: If you say onto then you actually want $|z|=1$ to be mapped to $|f(z)|=1$ (as well as being non-trivial and mapping interior to interior). Taking $z=e^{i\theta}$ and looking for equality you need $ |a|^2 + |b|^2 = |c|^2 + |d|^2 $ and for all real $\theta$: $$ { \rm Re} \left((a\bar{b}-c\bar{d}) e^{i\theta} \right)=0 $$ The latter implies $a\bar{b}-c\bar{d}=0$ from which you may continue.

Answer 3

This is a classical issue about what is called "Blaschke factors".

Their general form is

$$f(z)=e^{i \theta}\frac{z-a}{1-\overline{a}z} \ \ |a|<1 \ \ 0 \leq \theta < 2 \pi.$$

Here are two references:

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

(Conformal map of the unit disk onto itself which is not 1 to 1)

See the very nice pictures on Blaschke factors after page 111 in Visual "Complex Functions: An Introduction with Phase Portraits" by Elias Wegert (Google book)

Answer 4

Mobius transformations take circles to circles. So the transformation $T$ such that $T(D) =D$ takes the unit circle $\tau$ to another circle $\tau'$. Since $T$ maintains orientation, $T$ maps $D$ to either inside or outside of $\tau'$. Since $T(D) = D$, it follows that $T$ must map $D$ to inside of $\tau'$. It follows that $\tau' = \tau$.

Following @H.H.Rugh 's idea, by equating $|\frac{az+b}{cz+d}|=1$ we obtain $$ |a|^2 + |b|^2 = |c|^2 + |d|^2 \tag{1} \label{eq1} $$ $$ a\bar{b} = c\bar{d} \tag{2} \label{eq2} $$

assuming $c = 0$ we have $$ab = 0 \land bc = 0 \tag{3} \label{eq3}$$ $$|a|^2 = |d|^2 \tag{4} \label{eq4}$$

Since $ad -bc \neq 0 \land bc = 0$, we have $ad \neq 0$

Therefore $a \neq 0 \land d \neq 0$. We get

$$T(z) = \frac{az}{d}$$

Since $|a| = |d|$, we get $$T(z) = \lambda z$$ wher $|\lambda| = 1$

Suppose $c \neq 0$. Then we have $$\bar{d} = \frac{a\bar{b}}{c} \tag{5} \label{eq5}$$ $$ \implies |d|^2 = \frac{|a|^2|b|^2}{|c|^2} \tag{6} \label{eq6}$$

Substitute $\eqref{eq6}$ in $\eqref{eq1}$ and obtain after multiplying both sides by $|c|^2$ and gathering common terms $$(|a|^2 - |c|^2)(|b|^2 - |c|^2) = 0$$

Suppose $|a| = |c|$. Substituting $\eqref{eq5}$, we get

$$T(z) = \frac{az+b}{cz+\frac{\bar{a}b}{\bar{c}}}= \frac{a\bar{c}z + b\bar{c}}{|c|^2z+\bar{a}b}$$

Let $c = e^{i\theta}a, \theta \in \mathbb{R}$. Then

$$T(z) = \frac{|a|^2 e^{-i\theta}z + \bar{a}b e^{-i\theta}}{|a|^2z+\bar{a}b} = e^{-i\theta}\frac{|a|^2 z + \bar{a}b}{|a|^2z+\bar{a}b} = e^{-i\theta}$$

In this case T is not a mobius transformation. Therefore we must necessarily have $|b| = |c|$. Let $c = be^{i\theta}, \theta \in \mathbb{R}$. Then

$$ T(z) = \frac{a\bar{b}e^{-i\theta}z + |b|^2e^{-i\theta}}{|b|^2z + \bar{a}b} $$ $$ = e^{-i\theta}\frac{a\bar{b}z + |b|^2}{|b|^2z + \bar{a}b} $$ $$ =e^{-i\theta}\frac{\bar{b}}{b}\frac{az + b}{\bar{b}z + \bar{a}} $$ $$ = \lambda \frac{az + b}{\bar{b}z + \bar{a}} \tag7 \label{eq7} $$ where $|\lambda| = 1$. Suppose $z_1 = T(1), z_2 = T(i), z_3 = T(-1)$. By orientation principle, the direction $z_1 \rightarrow z_2 \rightarrow z_3$ should also be clockwise. This means $Im (\frac{z_3 - z_2}{z_2 - z_1}) > 0$.

Evaluating each $z_i$ for $i \in \{1,2,3\}$ $$ z_1 = \lambda\frac{a+b}{\bar{a}+\bar{b}} \tag8 \label{eq8} $$

$$ z_2 = \lambda\frac{ai+b}{\bar{a}+\bar{b}i} \tag9 \label{eq9} $$

$$ z_3 = \lambda\frac{a-b}{\bar{b}-\bar{a}} \tag{10}\label{eq10} $$

Using $\eqref{eq8}$, $\eqref{eq9}$ and $\eqref{eq10}$, we get

$$ \frac{z_3 - z_2}{z_2 - z_1} = \frac{i+1}{i-1}\frac{\bar{a}+\bar{b}}{\bar{b}-\bar{a}} $$ $$=\sqrt{2}\frac{[i(a\bar{b} -\bar{a}b) + i (|a|^2 - |b|^2)]} {|b - a|^2} \tag{11} \label{eq11}$$

So $Im (\frac{z_3 - z_2}{z_2 - z_1}) > 0$ implies $|a| > |b|$. It means $a \neq 0$. Let $ b = -\alpha a$, where $|\alpha| < 1$. Substituting in $\eqref{eq7}$ we get

$$ T(z) = \lambda\frac{az-\alpha a} {-\bar{\alpha}\bar{a}z + \bar{a}} $$ $$ =\lambda\frac{a}{-\bar{a}}\frac{z-\alpha} {\bar{\alpha}z - 1} $$ Therefore $$ T(z) = e^{i\theta}\frac{z-\alpha}{\bar{\alpha}z - 1}, |\alpha| < 1, \theta \in \mathbb{R} $$