# 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.

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.

• Sorry I mean open unit disk. Then how should I proceed. – mint Sep 25 '16 at 17:14
• If you want the map to be onto, then it is the same. The map must extend continuously to the boundary and map the boundary onto the boundary. You may then proceed as above. Another (possibly helpful) hint: The last condition also implies : $|a| \; |b| = |c| \; |d|$ which together with sum of squares limit the possibilities quite a lot. – H. H. Rugh Sep 25 '16 at 17:21

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

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:

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)