1. ## linear fractional transformations

A fixed point of a transformation w=f(z) is a point Zo such that f(Zo)=Zo, Show that every linear transformation, with the exception of the identity transformation w=z, has at most two fixed points in the extended plane.

2. Originally Posted by mndi1105
A fixed point of a transformation w=f(z) is a point Zo such that f(Zo)=Zo, Show that every linear transformation, with the exception of the identity transformation w=z, has at most two fixed points in the extended plane.
If $\displaystyle f$ is a linear transformation then $\displaystyle z\mapsto az+b$. Say that $\displaystyle f(z) = z$. If $\displaystyle z\not = \infty$ then $\displaystyle az+b = z \implies z(a-1) = b$.
• If $\displaystyle a=1$ and $\displaystyle b\not = 0$ then the equation is not solvable and so $\displaystyle f$ has no fixed points in $\displaystyle \mathbb{C}$. It does however have $\displaystyle \infty$ as a fixed point on the Riemann sphere.
• If $\displaystyle a=1$ and $\displaystyle b=0$ then $\displaystyle f$ is the identity.
• If $\displaystyle a\not = 1$ then $\displaystyle \tfrac{b}{a-1}$ is a fixed point in $\displaystyle \mathbb{C}$ the otherfixed point is $\displaystyle \infty$ because $\displaystyle f(\infty) = \infty$ if $\displaystyle a\not = 0$ because otherwise $\displaystyle f(\infty) = 0$ is not a fixed point.

3. Originally Posted by mndi1105
A fixed point of a transformation w=f(z) is a point Zo such that f(Zo)=Zo, Show that every linear transformation, with the exception of the identity transformation w=z, has at most two fixed points in the extended plane.
The question asks about linear transformations, but the thread is headed "linear fractional transformations", in other words those of the form $\displaystyle z\to\frac{az+b}{cz+d}$. In that case, a fixed point must satisfy $\displaystyle z=\frac{az+b}{cz+d}$. This is clearly a quadratic equation for z, except when b=c=0 and a=d, when it becomes identically true.