Fixed Points of a Möbius Transformation

Question

in "Complex Analysis" by Freitag & Busam there is an exercise which asks to proof that each Möbius transformation has exactly one or two fixed points. However, if $M(z)=\frac{az+b}{cz+d}$ we can choose $c=0$, $b\neq 0$ and $a=d\neq 0$. Then $M(z)=z+b/a\neq z$, so in that case there is no fixed point. In the solution section it is just stated that the condition for a fixed point is $z(cz+d)=az+b$ which is quadratic in $z$ and therefore there are one or two solutions. Am I missing something?

Answer 1

For the Möbius transformation: $$M(z)=\frac{az+b}{cz+d}$$ the coefficients $a,b,c$ and $d$ are complex numbers satisfying $ad-bc\neq 0$ since it is a rational map of degree $1$.

• If $c\neq0$ then we can use the quadratic formula to find the fixed point/s, that is $ cz^2+(d-a)z-b=0 $

• If $c=0$, $b\neq 0$ and $a=d\neq0$ , then the fixed point will be the point at infinity

• If $c=0$, $b=0$ and $a=d\neq0$, then every point is a fixed point since we have the identity map

• If $c=0$, $b=0$ and $a\neq d$, then the fixed point is $0$

• If $c=0$, and $d\neq a$, then the fixed point is $z=\frac{b}{d-a}$