Möbius transformation maps inverse points to inverse points?

Question

Say $f$ is a Möbius transformation that maps the unit disk to the unit disk. The solution in this page, and in many other pages Möbius transformations on $D$ such that $f(D)=D$ says

"Let $z_0$ be the point s.t. $f(z_0)=0$. Then $f(1/\overline{z_0})=\infty$."

In a comment on the post, it says "one property of Möbius transformation is that inverse point maps to inverse point." How can this property be true? It is definitely wrong for a general Möbius transformation s.t. $z/(z-1)$. So I think it is only true in this problem. But why is it?

Answer 1

Inversion in this context means “inversion at a circle”. In the (extended) complex plane, inversion (or reflection) can be defined as follows:

Let $C$ be a circle (or line) in $\Bbb {\hat C}$, $z \in \Bbb {\hat C}$, and $T$ a Möbius transformation which maps $C$ onto $\Bbb R \cup \{ \infty \}$. Then $$ w = T^{-1}(\overline {T(z)}) $$ is the inverse point (or reflection point) of $z$ with respect to $C$.

It can be shown that this definition is independent of the choice of $T$ (because any Möbius transformation $S$ mapping the real axis onto itself satisfies $S(\bar z) = \overline {S(z)}$).

For lines, this definition coincides with the usual definition of reflection at a line. For circles, it coincides with “circle inversion”.

It also follows from the definition that this relation is preserved under Möbius transformations:

Let $C$ be a circle or line, and $T$ a Möbius transformation. If $w$ is the inverse point of $z$ with respect to $C$ then $T(w)$ is the inverse point of $T(z)$ with respect to $T(C)$.

In the case $T(C) = C$ we conclude that

Let $C$ be a circle or line, and $T$ a Möbius transformation which maps $C$ onto itself. If $z$, $w$ are inverse points with respect to $C$ then $T(z)$, $T(w)$ are also inverse points with respect to $C$.

and that is the precise meaning of “.... one property of Möbius transformation is that inverse point maps to inverse point.”

In the case of the unit circle we can choose $T(z) = i\frac{1+z}{1-z}$ as the mapping onto the real line, and a straight-forward calculation show that

$1/\overline z$ is the inverse point of $z$ with respect to the unit circle.

In particular, $z=\infty$ is the inverse point of $z=0$. If $f$ is a Möbiustransformation preserving the unit disk and $f(z_0) = 0$ then it follows from the above invariance that $f(1/\overline{z_0}) = \infty$.