Fixed point of a Möbius transformation

A fixed point of a Möbius transformation satisfies a quadratic equation on the Riemann sphere.

For $f(z)=(az+b)/(cz+d)$ with distinct finite fixed points of a Möbius transformation $\alpha ,\beta$, the fixed-point equation gives $f(z)-\alpha =(a-c\alpha )(z-\alpha )/(cz+d)$. Dividing the two identities gives $k=(a-c\alpha )/(a-c\beta )$, which is nonzero because $(a-c\alpha )(a-c\beta )=ad-bc\ne 0$. The identities extend over the Riemann sphere. This conjugates the transformation to a dilation.

For $f(z)=(az+b)/(z+1)$, the finite fixed points are the roots ofThus a repeated fixed point or two prescribed fixed points determine $a$ and $b$ by the usual sum and product of roots.

Every nonidentity finite-order Möbius transformation has exactly two fixed points on the Riemann sphere. A transformation with one repeated fixed point is conjugate to a nontrivial translation and therefore has infinite order.