Lecture 5
Example 2: \(z\mapsto \dfrac{1}{z}\) on \(\C_*\).
Define \(\xi (z) = \dfrac{z}{|z|^2}\) on \(\C_*\); \(\;\;\eta(\xi)=\conj{\xi}\) on \(\C\).
\(\eta \circ \xi (z) = \eta \left( \xi \left( z \right) \right) = \conj{\left( \dfrac{z}{|z|^2} \right)} = \dfrac{\conj z }{ |z|^2 } \)
\(\;\;\,\quad\quad = \dfrac{\conj z}{z \conj z} = \dfrac{1}{z}\). \(\qquad z\in \C_*\)
\(\xi\) is called inversion (w.r.t the unit circle),
\(\eta\) is reflection in the real axis.
For \(w=\dfrac{1}{z} = \dfrac{\conj z }{|z|^2}\), we have
\(x+i\,y \mapsto u+i\,v\) and \( w = \dfrac{x-i\,y}{x^2+y^2} \), so \[ u = \dfrac{x}{x^2+y^2} \; (1)\quad \text{ & } \quad v = \dfrac{-y}{x^2+y^2} \; (2) \]
We can use this to show that \(1/z\) maps circles and
lines in the
\(z\)-plane to circles and lines in the \(w\)-plane.
Note: Circles and lines in the \(z\)-plane can be represented as \[ A(x^2+y^2)+Bx+Cy +D = 0 \;\;(*) \]
with \(A,B, C, D\in \R\), and \(B^2+C^2>4AD\).
If \(A = 0\) then we have a line. If \(A \neq 0\) then we have a circle.
\(*\; f: \Omega \ra \C\) is 1-1 or injective says \[f(z) = f(\xi)\Rightarrow z=\xi.\]
\(*\; f: \Omega \ra \Lambda\subseteq \C\) is onto or surjective says given \(\eta \in \Lambda\), there exists (at least one) \(z\in \Omega: f(z)=\eta\).
Definition: Let \(a,b,c,d\in \C\) with \(ad-bc\neq 0\). Then \[ w = T(z) = \frac{az+b}{cz+d} \] is a Möbius (or linear fractional) transformation.
Natural domain of definition:
\(*\; c=0\): dom\((w)=\C\) (note \(c= 0 \Rightarrow d\neq 0\));
\(*\; c\neq 0\): dom\((w)=\C\setminus \{-d/c\}\).
Goal: Understand \(T\) geometrically
Case I: \(c=0\).
Claim: \(T\) maps \(\C \ra \C\) 1-1 & onto.
Proof. (i) 1-1: Suppose \(T(z)= T(\xi)\;\; (**).\;\) We want to show: \(z=\xi\). From \((**)\) we have \[ \frac{a}{d}z+\frac{b}{d} = \frac{a}{d}\xi +\frac{b}{d} \implies z=\xi. \]
(ii) onto: given \(w\in \C\) we need \(z\in \C\) such that \(T(z)=w\).
📝 Check \(z=\frac{d}{a}\left(w-\frac{b}{d}\right)\) works.
Goal: Understand \(T\) geometrically
Case II: \(c\neq 0\).
\(w = \dfrac{az+b}{cz+d} =\) \( \dfrac{a\left(z+\dfrac{d}{c}\right)- \dfrac{ad}{c}+b}{c\left(z+\dfrac{d}{c}\right)}\) \(=\dfrac{a}{c}+\left(\dfrac{bc-ad}{c}\right)\cdot \dfrac{1}{cz+d}\)
So in this case, \(T\) is the composition of the mappings \[ Z= cz+d; W = \frac{1}{Z}; \;\text{ & } \; w = \frac{a}{c} + \frac{bc-ad}{c}W. \]
So in both cases I & II, \(T\) is a composition of maps previously studied.