Given mobvius transformations of the form $$f(z) = \eta\frac{a-z}{1-\bar a z}$$ with $\vert \eta \vert = 1$ and $a\in\mathbb{D}$ ($\mathbb{D}= $ open unit disk). I have to proof that if this transformation have two fixed points $z_1,z_2\in\mathbb{D}$, $f(z)$ is already the identity $f(z)=z$.
We can consider $\eta\not=-1$ (that case is kinda trivial, no fixed points IN $\mathbb{D}$, but on the boarder). I calculated the fixed points for some examples and found a pattern: If one fixed point is $z$, the other one is $\frac{1}{\bar z}$. That would already proofs the claim. Because if $z$ is inside $\mathbb{D}$, than $\frac{1}{\bar z}$ is not. And since $f(z)$ has at most two fixed points we are done.
But I struggle very hard to proof that if one fixed point is $z$, $\frac{1}{z}$ is the other. I start with $$\eta\frac{a-z}{1-\bar a z}=z$$ and expand $\frac{z}{1}$ to $\frac{\vert z \vert^2}{\bar z}$ and then I do some algebra. I tried many different approaches to prove this, but since I am unable to do so I think that this is not true in general case.. Can anyone help me out? I dont necessarily need an entire solution, but some hints would be nice. Am I on the right path? Or is everything I did so far kinda useless?