Conformal transformations

A linear fractional transformation maps lines and circles to lines and circles.

We start by showing that inversion maps lines and circles to lines and circles. Given \(z\) and \(w=\frac{1}{z}\) we define \(x,y,u\) and \(v\) by

\(\hspace{2cm}z=x+iy\hspace{1cm}\text{ and }\hspace{1cm}w=\frac{1}{z}=\frac{x-iy}{x^{2}+y^{2}}=u+iv.\)

So

\(\hspace{2cm}u=\frac{x}{x^{2}+y^{2}}\hspace{1cm}\text{ and }\hspace{1cm}v=-\frac{y}{x^{2}+y^{2}}\)

Now, every circle or line can be described by the equation

\(\hspace{5cm}Ax+By+C\left( x^{2}+y^{2}\right) =D\)

(If \(C=0\) it describes a line, otherwise a circle). We convert this to an equation in \(u,v\) as follows

\(\hspace{5cm}Ax+By+C\left( x^{2}+y^{2}\right) =D\)

\(\hspace{4cm}⇔\frac{Ax}{x^{2}+y^{2}}+\frac{By}{x^{2}+y^{2}}+C=\frac{D}{x^{2}+y^{2}} \)

\(\hspace{4cm}⇔Au-Bv+C=D\left( u^{2}+v^{2}\right).\)

In the last step we used the fact that

\(\hspace{4cm}u^{2}+v^{2}=\left\vert w\right\vert ^{2}=\frac{1}{\left\vert z\right\vert^{2}}=\frac{1}{\left( x^{2}+y^{2}\right) }.\)

Then we have shown that a line or circle in \(x,y\) is transformed to a line or a circle in \(u,v\). This shows that inversion maps lines and circles to lines and circles.

Now, to prove that an arbitrary fractional linear transformation maps lines and circles to lines and circles, we factor it into a sequence of simple transformations.

Fist suppose that \(c=0\). So,

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

Since this is just translation, scaling and rotation, it is clear it maps circles to circles and lines to lines.

Now suppose that \(c≠0.\) Then

\(\hspace{3cm}T\left( z\right) =\frac{az+b}{cz+d}=\frac{\frac{a}{c}\left( cz+d\right) +b-\frac{ad}{c}}{cz+d}=\frac{a}{c}+\frac{b-\frac{ad}{c}}{cz+d}.\)

So, \(w=T(z)\) can be computed as a composition of transforms

\(\hspace{1,5cm}z\mapsto w_{1}=cz+d\mapsto w_{2}=\frac{1}{w_{1}}\mapsto w=\frac{a}{c}+\left(b-\frac{ad}{c}\right) w_{2}.\)

We know that each of the transforms in this sequence maps lines and circles to lines and circles. Therefore the entire sequence does also.

Mapping \(z_{j }\) to \(w_{j}\)

It turns out that for two sets of three points \(z_{1},z_{2},z_{3}\) and \(w_{1},w_{2},w_{3}\) there is a fractional linear transformation that takes \(z_{j}\) to \(w_{j}\). We can construct this map as follows. Let

\(\hspace{5cm}T_{1}\left( z\right) =\frac{\left( z-z_{1}\right) \left( z_{2}-z_{3}\right)}{\left( z-z_{3}\right) \left( z_{2}-z_{1}\right) }.\)

Notice that

\(\hspace{3cm}T_{1}\left( z_{1}\right) =0,\hspace{1,5cm}T_{1}\left( z_{2}\right) =1,\hspace{1,5cm}T_{1}\left( z_{3}\right) =\infty .\)

Likewise let

\(\hspace{5cm}T_{2}\left( w\right) =\frac{\left( w-w_{1}\right) \left( w_{2}-w_{3}\right)}{\left( w-w_{3}\right) \left( w_{2}-w_{1}\right) }\)

Notice that

\(\hspace{3cm}T_{2}\left( w_{1}\right) =0,\hspace{1,5cm}T_{2}\left( w_{2}\right) =1,\hspace{1,5cm}T_{2}\left( w_{3}\right) =\infty .\)

Now \(T\left( z\right) =T_{2}^{-1}\circ T_{1}\left( z\right)\) is the required map.