Lecture 6

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.

\(\varphi(z)\) is on the Riemann sphere.

**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.

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

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