Conformal transformations

We can identify the transformation

\(\hspace{5cm}T\left( z\right) =\frac{az+b}{cz+d}\)

with the matrix .

this identification is useful because of the following algebraic facts.

1. \(\hspace{0,5cm}\) If \(r≠0\) then and \(r\) correspond to the same FLT.

Proof : This follows from the obvious equality

\(\hspace{4cm}\frac{az+b}{cz+d}=\frac{raz+rb}{rcz+rd}.\)

2. \(\hspace{0,5cm}\) If \(T(z)\) corresponds to and \(S(z)\) corresponds to then composition \(T∘S(z)\) corresponds to matrix multiplication \(AB.\)

Proof : The proof is just a bit of algebra.

\(\hspace{1cm}T\circ S\left( z\right) =T\left( \frac{ez+f}{gz+h}\right) =\frac{a\left(\frac{\left( ez+f\right) }{\left( gz+h\right) }\right) +b}{c\left( \frac{\left( ez+f\right) }{\left( gz+h\right) }\right) +d}=\frac{\left(ae+bg\right) z+af+bh}{\left( ce+dg\right) z+cf+dh}\)

\(\hspace{2cm}\)

The claimed correspondence is clear from the last entries in the two lines above.

3. \(\hspace{0,5cm}\) If \(T(z)\) corresponds to then \(T\) has an inverse and \(T^{-1}(w)\) corresponds to \(A^{-1}\) and also to i.e . to \(A^{-1}\) without the factor of \(\frac{1}{det(A)}\).

Proof : Since \(AA^{-1}=I\) it is clear from the previous fact that \(T^{-1}\) corresponds to \(A^{-1}\). Since

\(\hspace{5cm}A^{-1}=\frac{1}{ad-bc}\)

Fact \(1\) implies \(A^{-1}\) and both correspond to the same FLT, i.e. to \(T^{-1}\).

1. \(\hspace{0,5cm}\)The matrix corresponds to \(T(z)=az+b.\).

2. \(\hspace{0,5cm}\) The matrix corresponds to rotation by \(2α.\)

3. \(\hspace{0,5cm}\)The matrix corresponds to the inversion \(w=\frac{1}{z}.\)