Finding the Mobius Transformation that maps open unit disk onto itself

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.

Solution 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 $$f(z) = e^{i \theta}\frac{z-a}{1-\overline{a}z}, \qquad |a| < 1$$

Solution 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.