Given the Möbius-Transformation $T: \mathbb{D} \rightarrow \mathbb{D}, T(z)=\eta\frac{a-z}{1-\overline az} $ for some fixed $a \in \mathbb{D}, \eta \in \partial \mathbb{D} $ I am trying to show that if $T$ has two fixed points in $ \mathbb{D}$, then $T$ must be the identity $id_ \mathbb{D}$ .
I have been manipulating fixed point equations for hours but I did not arrive at any insightful result.
So far I found that if $v,w$ are two distinct fixed points, then $\eta= \overline a(v+w)-1$. So $T(v+w)=a-(v+w)$. Then $T$ can only be the identity if $a=0$ and then also $\eta=-1$. So would the strategy perhaps be to show that $a, \eta$ can only have these values given the two distinct fixed points?
I know that here on SE there is a general solution for analytic functions, but it uses theorems like the Schwarz lemma which our lecture has not covered yet.
Any hint would be very much appreciated.
Best Answer
The equation $T(z)=z$ gives $$ \bar{a} z^2 +z(\eta +1) -\eta a=0 $$
First case $\bar{a}=0$, in this case $z(\eta +1) = 0$. If $\eta = -1$ we are done because then $T$ is the identity. If $\eta \ne -1$, then this solution has just one solution ($z=0$) and there is not two fixed points.
Second case $\bar{a} \ne 0$ in this case the equation become $$ z^2 +z \frac{(\eta +1)}{\bar{a}} -\eta \frac{a}{\bar{a}}=0 $$ If we note $z_1$ and $z_2$ the two roots of this equation we have that $z_1 z_2 =\eta \frac{a}{\bar{a}}$, but then computing the module we find that $|z_1 z_2|=|\eta \frac{a}{\bar{a}}|= |\eta| \frac{|a|}{|\bar{a}|}=1$. So there is no way that $|z_1|<1$ and $|z_2|<1$. Which mean there are no two solutions inside the unit disk.