# Fixed points of self-conformal mappping

Given a conformal self map f (analytic function from unit disc to itslef that is one to one and onto), such that it is not identity. I need to show that either f has two fixed point on the boundary or one fixed point inside that unit disc.

Thank you beforehand for your help!

• Are you familiar with the Schwarz lemma? – Siméon Nov 27 '13 at 12:24
• Or, perhaps, the maximum modulus principle? – Jonathan Y. Nov 27 '13 at 12:26
• @Siméon ,@Jonathan Y. yes I am familiar with it. I see its use only in finding general form of conformal self map of the unit disc...other than that I cant think of any relation between the Schwarz lemma and this problem. – Nasibabuba Nov 27 '13 at 19:25

There are two standard ways of writing automorphisms of the unit disk, first

$$Tz = \frac{az+b}{\overline{b}z+\overline{a}};\quad \lvert a\rvert^2 - \lvert b\rvert^2 = 1,$$

and second

$$Tz = e^{i\varphi} \frac{z-w}{1-\overline{w}z};\quad w \in \mathbb{D}, \varphi \in \mathbb{R}.$$

Pick whichever form you prefer, and solve the quadratic equation $Tz = z$.

• Oh now I get it! Thank you! – Nasibabuba Nov 27 '13 at 19:44
• Yep, it isn't really tedious. Note however, that you have to treat the case that $0$ is a fixed point ($T$ is a rotation) separately, to avoid $\frac00$. In your case, if $a=0$ in the first equation. – Daniel Fischer Nov 27 '13 at 19:48
• sorry i deleted previous comment,just that I solved it and got too excited..thank you! – Nasibabuba Nov 27 '13 at 19:50

If I solve this quadratic equation I will reduce problem to solving $$z^2 \bar{a}+(e^{i\phi}-1)z-e^{i/phi}a=0$$ $$z^2 +\frac{(e^{i\phi}-1)}{\bar{a}}z-e^{i/phi}\frac{a}{\bar{a}}=0$$ The roots of this equation satisfy: $$|z_1z_2|=1$$ $$|z_1||z_2|=1$$ If$|z_1|<1$ then $|z_2|>1$ thus there will only one fixed point inside the unit disc or: $|z_1|=|z_2|=1$ both roots are on the boundary.