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