Say I wish to prove that every möbius transformation of the unit disk onto itself can be written in the form

$A(z) = e^{i\theta}\frac{z+a}{1+\bar{a}z}$, where $\theta$ is a real number and $a$ is a point in the unit disk which is defined to be $\mathbb{D} = \{z \in \mathbb{C} : |z|<1\}$.

Now I already know what the general element of a möbius transform that preserves the extended real line is any one of the four möbius transforms of the following kind (I'll only state one)

$m(z)= \frac{az+b}{cz+d}, a,b,c,d \in \mathbb{R}$ and $ad-bc = 1$.

Now let $p$ be a möbius transform that maps the extended real line to the unit circle. It can be shown that the choice of $p$ does not matter, so i'll just take $p(z) = \frac{z-i}{z+i}$.

Now then I know that any möbius transform that preserves the unit circle is of the form $p \circ m \circ p$, where $m$ is any one of the möbius transforms that preserves the extended real line, $\mathbb{\bar{R}}$. Of course to check if a point in the unit circle still remains in the unit circle, I'll have to check and if it does not I can apply combine the above with $K(z) = \frac{-1}{z}$.

However, here's the problem. With many choices of $m$ in fact for all 4 I would have to check and write down $p^{-1} (z)$ and do compositions of functions and other things. The algebra is messy, but what's even worse is it tells me nothing of whether the coefficient of $z$ is a complex number that is within the disk (as what is asked to be proved).

Anyone got any ideas, as brute force is the wrong way to approach the problem? Please do not give me a definite answer as I would most certainly like to finish the problem myself.


"Better to do the right problem wrong than the wrong problem right" - Richard Hamming

  • $\begingroup$ Isn't enough to determine (for your various possibilities) that zero is sent to a point inside the unit circle? By continuity you can rule out an "inversion" of the unit circle. Alternatively of course you could check that the point at infinity remains outside the unit circle after Möbius transformation is applied. $\endgroup$ – hardmath Apr 20 '11 at 14:26

Your idea is good. It ought to work out once you crunch through the formulas. Alternatively, you could use the fact that any Möbius transformation is determined by its action on $1,0,\infty$. With that in mind, it's easy to show that every Möbius transformation that preserves the unit disk must be of the above form. Then you need to show that everything of the above form preserves the unit disk. (As requested, this is just a hint.)

  • $\begingroup$ Say I try and show the converse of the theorem above, namely that if $A(z)$ is of the form above, it preserves the unit circle and sends a point $-a$ inside $\mathbb{S}^1$ to itself (i.e. it preserves the unit disk). This is because $A(-a) = 0$ and $|A(z)| = \frac{|1+a\bar{z}|}{|1+\bar{a}z|} = \frac{|1+a\bar{z}|^2}{|1+\bar{a}z|^2}$. We conclude that the only real numbers equal to their squares is either $0$ or $1$, and hence $|A(z)|=1$. The possibility of 0 is excluded as this would mean that $a\bar{z} = -1$ which means $|a| = 1$, contradicting $a$ being a point in the disk. $\endgroup$ – user38268 Apr 21 '11 at 0:42
  • $\begingroup$ Now the more difficult part is to show that every möbius transform that preserves the unit disk is of the form above. $\endgroup$ – user38268 Apr 21 '11 at 0:44

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