Mobius Transformation viewed as a mapping on $\bar {\mathbb C}$

Question

The question I have on hand is as follows:

Suppose that a Mobius Transformation z $\to \frac{az + b}{cz + d}$ (viewed as a mapping on $\bar {\mathbb C}$) maps $\infty \to \infty$. What information does this yield about any of the coefficients a,b,c and d?

I have come up with the following answer:

With the understanding that T($\infty$) = $\frac{a}{c}$ and T($\frac{-d}{c}$) = $\infty$, I had the equation:

$\frac{a}{c}$ = $\frac{-d}{c}$

ac = -dc

To comply with the condition that ad - bc $\neq$ 0,

c = 0, a $\neq$ 0, d $\neq$ 0, b = K, for K $\in \mathbb R$

Any opinions on whether I came up with somewhat a proper deduction? Thanks

Answer 1

The formula $T(z)={az+b\over cz+d}$is not valid for $z=\infty$ or $T(z)=\infty$. For image and preimage of the point $\infty$ exception handling applies. Your arguments therefore do not catch the fish.

If $c\ne0$ then the denominator $cz+d$ vanishes at $z_0=-{d\over c}\in{\mathbb C}$, hence $T(z_0)=\infty$ by the "exception handling rules". Since we don't want that we have $c=0$ as a necessary condition for $T(\infty)=\infty$. If $c=0$ then the condition $ad-bc\ne0$ enforces $ad\ne0$. We therefore can write $$T(z)={a\over d}z+{b\over d}=a'z+b',\quad a'\ne0\ .$$ Conversely, any map $T(z)=a'z+b'$ with $a'\ne0$ is a Moebius transformation satisfying $T(\infty)=\infty$.

Answer 2

A Mobius transformation preserves the set of generalised circles, particularly the set containing circles and sets of the form $L \cup \{ \infty \}$ where $L$ is a line in the complex plane (note that none of the circles contain $\infty$). As such, the set of lines must be preserved by a Mobius transformation that maps $\infty$ to $\infty$, which means that the Mobius transformation must be an affine map. In particular, it must take the form $z \mapsto az + b$, with $a \neq 0$.