MATH3401

The Möbius transformation \(T(z) = w = \dfrac{az+b}{cz+d}\; (*)\), with \(ad-bc \neq 0\), can be written as \[ A z w + B z + C w + D = 0 \] with \(A=c, B=-a, C=d\) and \(D=b\); called the implicit form.

Case I: \(c=0\). \(T\) is a bijection (1-1 and onto): \(\C\ra\C\).

Case II: \(c\neq 0\). \((*)\) \(\Rightarrow z=\dfrac{-dw+b}{cw-a}\). Thus \[ T^{-1}(w)= \dfrac{-dw+b}{cw-a}. \]

Then in Case II, \(T\) is 1-1 and onto \(\C \setminus \left\{-d/c\right\} \ra \C \setminus \left\{a/c\right\}\).

Question: Can we extend \(T\) to a function \(\C \ra \C\) in Case II? In particular, so extension is 1-1 and onto.

Ans: Yes! Take \(T\left(-d/c\right)=a/c\). Not great: discontinuous.

Extend \(\C\) to the extended complex plane, denoted by \(\conj{\C}\),
"by adding a point at infinity", which we call \(\infty\).

Define \(T\left(-d/c\right)=\infty\) and \(T\left(\infty\right)=a/c\).

This extends \(T\) to a map \(\conj{\C} \ra \conj{\C}\), which is 1-1 & onto.

Remark: \(\conj{\C}\) is a topological space and the given extension is continuous.

Remark 1: Given 3 distinct points in \(\conj{\C}\), \(z_1,z_2,z_3\) and 3 distinct points in \(\conj{\C}\), \(w_1,w_2,w_3\), \(\exists !\) (there exists a unique) Möbius transformation \(T\) such that \(T(z_j)=w_j\).

Here \(T=w(z)\) is given by \[ \frac{(w-w_1)(w_2-w_3)}{(w-w_3)(w_2-w_1)} = \frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}\;(**) \]

Note 1: In practice, may be easier to solve directly for \(a,b,c,d\) rather than use \((**)\).

Note 2: How does this work with \(\infty\)?

\(T(\infty)=\dfrac{a}{c}\) and \(T\left(-\dfrac{d}{c}\right)=\infty\), rigorously later.

\(T(\infty)=w \iff \ds \lim_{|z|\ra \infty}T(z) = w\).

\(T(\xi)=\infty \iff \ds \lim_{z\ra \xi }\dfrac{1}{T(z)} = 0\).