is there a Mobius transformation with no fixed points?
I have the equation $$\frac{az+b}{cz+d}=z\implies cz^2+dz-az-b=0$$ given the fixed points when this is true. So if we set $c\ne 0$ we get two roots and thus two fixed points, but in the case of $c=0$ we get one fixed point. I can't see where there is a case with no fixed points?
Best Answer
If $c\neq 0$ there are one or two fixed points by the quadratic formula. If $c=0$ and $d\neq a$, then $z=\dfrac{b}{d-a}$ is a fixed point. If $c=0$ and $d=a$, and $b=0$ and every point is a fixed point, because the transformation is the identity. If $c=0$, $d=a$, and $b\neq 0$, then it will fix the point at infinity.