# Characterization of linear fractional transformations that maps the unit disc into itself

I am reading the paper "ADJOINTS OF COMPOSITION OPERATORS ON HILBERT SPACES OF ANALYTIC FUNCTIONS" by MARIA J. MARTIN AND DRAGAN VUKOTIC. In Section 1.1 they say the linear fractional transformation $\phi(z)=\frac{az+b}{cz+d}$ maps the unit disc into itself if and only if $$|b\overline{d}-a\overline{c}|+|ad-bc|\leq |d|^2-|c|^2 .$$

I can't neither prove it nor see the reference. Could anyone please suggest something.

• This question deals with a similar topic. Their function is explicitly in the form of a Blashke Factor though. – Mark Sep 3 '16 at 19:35

You must have $|d|^2-|c|^2>0$ or else there is a pole inside the disk.
To get the formula (I remember doing this long time ago, it slowly comes back) there is a little trick, to precompose by a suitable conformal map preserving the disk but making our transformation linear. So define $$M(z) = \frac{z-\alpha}{1-\bar{\alpha} z}$$ using the value: $\alpha = \bar{c}/\bar{d}$ (modulus <1). Then $M\in {\rm Aut} ({\Bbb D})$ and we get (you should carry out this calculation on a piece of paper):
$$f\circ M(z) = \frac{ b\bar{d}-a \bar{c}}{d\bar{d}-c\bar{c}} + z \frac{d a-cb}{d\bar{d}-c\bar{c}}=q+rz$$ The image of ${\Bbb D}$ is now clearly a disk with center $q$ and radius r. The condition then reads $|q|+r\leq 1$ which translates into the inequality that you stated above.
the easy thing to remember is that $\phi(z)=\frac{az+b}{cz+d}$ maps the upper half plane to itself when $a,b,c,d$ are real and $ad-bc > 0.$
$$\left( \begin{array}{rr} i & 1 \\ 1 & i \end{array} \right)$$ takes the (open) upper half plane to the disc. Then $$\left( \begin{array}{rr} 1 & i \\ i & 1 \end{array} \right)$$ takes disc to half plane.