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.
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.