Linear Fractional Transformations

If $a,b,c,$ and $d$ are complex constants with $ad-bc\neq 0,$ then the transformation \begin{eqnarray}\label{mobius} T(z) = \frac{az+b}{cz+d} \end{eqnarray} is called linear fractional transformation, or Möbius transformation. Observe that we can also write (\ref{mobius}) in the form

\begin{eqnarray}\label{bilinear} Azw + Bz+Cw+D = 0 \qquad (AD-BC\neq 0); \end{eqnarray}

with $w = T(z).$ Since this alternative form is linear in $z$ and linear in $w,$ another name for linear fractional transformation is bilinear transformation.

If $c=0,$ then the condition $ad-bc\neq 0$ with equation (\ref{mobius}) becomes $ad\neq 0.$ So the transformation $T$ reduces to a nonconstant linear function $az+b.$ When $c\neq 0,$ then we can write equation (\ref{mobius}) as

\begin{eqnarray}\label{mobius-composition} T(z) = \frac{bc-ad}{c}\cdot \frac{1}{cz+d}+ \frac{a}{c} \qquad (ad-bc\neq 0) \end{eqnarray}

Again, the condition $ad-bc\neq 0$ ensures that we do not have a constant function. The transformation $T(z)=1/z$ us obviously a special case of (\ref{mobius}) when $c\neq 0.$ (We already covered this function in Chapter 3)

Equation (\ref{mobius-composition}) reveals that a linear transformation is a composition of the mappings:

\[ f(z)= \frac{bc-ad}{c}z+\frac{a}{c},\quad g(z) = \frac{1}{z}, \quad h(z) = cz+d. \]

That is, $T(z) = f\circ g \circ h(z) = f\left( g\left( h(z)\right)\right).$ Thus it follows that, regardless of whether $c$ is zero or nonzero, any linear fractional transformation maps circles and lines into circles and lines. You can easily verify this for the functions $f$ and $h.$ The function $g$ was already explored in the section: The Transformation $1/z$.

The domain of a linear fractional transformation $T$ given in (\ref{mobius}) is the set of all complex $z$ such that $z\neq d/c.$ Moreover, since \[ T'(z) = \frac{ad-bc}{\left(cz+d\right)^2}, \] with $ad-bc \neq 0,$ then linear fractional transformations are conformal on their domains.

When $c\neq 0,$ that is, when $T$ is not a linear function, it is often useful to view $T$ as a mapping of the extended complex plane:

\begin{eqnarray}\label{mobius-extended} T(z) = \left\{ \begin{array}{ll} \dfrac{az+b}{cz+d}, & z\neq -\dfrac{d}{c}, z\neq \infty\\ \infty, & z=-\dfrac{d}{c}\\ \dfrac{a}{c}, & z=\infty. \end{array} \right. \end{eqnarray}

In this case, $T$ is a one-to-one mapping of the extended complex plane. So there exists $T^{-1}$ such that $T^{-1}\left(T(z)\right) = z =T\left(T^{-1}(z)\right).$ In this case $T^{-1}$ is also a linear fractional transformation and is defined as \[ T^{-1}(w) = \frac{-dw + b}{cw -a} \qquad (ad-bc\neq 0). \]

In applications we often need to find a conformal mapping from a domain $D$ that is bounded by circles onto a domain $D'$ that is bounded by lines. Linear fractional transformations are, in particular, well-suited for such applications.

Cross Ratios

Consider the following equation

\begin{eqnarray}\label{cross} \frac{\left(w- w_1\right)\left(w_2- w_3\right)}{\left(w- w_3\right)\left(w_2- w_1\right)} = \frac{\left(z- z_1\right)\left(z_2- z_3\right)}{\left(z- z_3\right)\left(z_2- z_1\right)}. \end{eqnarray}

The two sides of this equation are knwon as cross ratios.

Equation (\ref{cross}) defines (implicitly) a linear fractional transformation that maps disctinct three points $z_1,$ $z_2,$ and $z_3$ in the finite $z$ plane onto three distinct points $w_1,$ $w_2,$ and $w_3,$ respectively, in the finite $w$ plane. We can easily verify this by writing equation (\ref{cross}) as

\begin{eqnarray}\label{cross-02} \left(z- z_3\right)\left(w- w_1\right)\left(z_2- z_1\right)\left(w_2- w_3\right) = \left(z- z_1\right)\left(w- w_3\right)\left(z_2- z_3\right)\left(w_2- w_1\right) \end{eqnarray}

If $z=z_1,$ then the right-hand side of (\ref{cross-02}) is zero. So $w=w_1.$
If $z=z_2,$ then we have the linear equation
\[ \left(w- w_1\right)\left(w_2- w_3\right)=\left(w- w_3\right)\left(w_2- w_1\right) \]
whose unique solution is $w=w_2.$
If $z=z_3,$ then the left-hand side of (\ref{cross-02}) is zero; and consequently, $w=w_3.$

It can be proved that equation (\ref{cross}) defines linear fractional transformation by expanding the products of (\ref{cross-02}) and writing the result in the form \[ Azw+Bz+Cw+D=0 \qquad (AD-BC\neq 0). \] As an exercise, prove that equation (\ref{cross}) actually defines the only linear mapping the points $z_1,$ $z_2,$ and $z_3$ onto $w_1,$ $w_2,$ and $w_3,$ respectively.

Example 1: To find a linear fractional transformation that maps $z_1=1,$ $z_2=i,$ and $z_3=-1$ onto $w_1=-1,$ $w_2=0,$ and $w_3=1,$ we use equation (\ref{cross}) to write

\begin{eqnarray*} \frac{\left(w+1\right)\left(0-1\right)}{\left(w- 1\right)\left(0+ 1\right)} = \frac{\left(z- 1\right)\left(i + 1\right)}{\left(z+1\right)\left(i- 1\right)}. \end{eqnarray*}

Solving for $w$ in terms of $z$ we obtain \[ w = T(z) = \frac{z-i}{iz-1}. \]

If equation (\ref{cross}) is modified properly, it can also be used when the point at infinity is one of the prescribed points in either the (extended) $z$ or $w$ plane. Suppose, for example, that $z_1=\infty.$ Since any linear fractional transformation is continuous on the extended plane, we need only replace $z_1$ on the right-hand side of equation (\ref{cross}) by $1/z_1,$ clear fractions, and let $z_1$ tend to zero. That is

\begin{eqnarray*} \lim_{z_1\to 0} \frac{\left(z- 1/z_1\right)\left(z_2 -z_3\right)}{\left(z-z_3\right)\left(z_2- 1/z_1\right)}\cdot \frac{z_1}{z_1} = \lim_{z_1\to 0} \frac{\left(z_1z- 1/z\right)\left(z_2 -z_3\right)}{\left(z-z_3\right)\left(z_1z_2- 1\right)} = \frac{z_2 -z_2}{z-z_3} . \end{eqnarray*}

Thus we obtain the desired modification of quation (\ref{cross})

\begin{eqnarray*} \frac{\left(w- w_1\right)\left(w_2- w_3\right)}{\left(w- w_3\right)\left(w_2- w_1\right)} = \frac{\left(z_2- z_3\right)}{\left(z- z_3\right)}. \end{eqnarray*}

This modification is obtained by simply deleting the factors involving $z_1$ in equation (\ref{cross}). It is easy to check that the same approach applies when any of the other prescribed points is $\infty.$

Example 2: To find a linear fractional transformation that maps $z_1=1,$ $z_2=0,$ and $z_3=-1$ onto $w_1=i,$ $w_2=\infty,$ and $w_3=1,$ we use the modification

\begin{eqnarray*} \frac{\left(w- w_1\right)}{\left(w- w_3\right)} = \frac{\left(z- z_1\right)\left(z_2- z_3\right)}{\left(z- z_3\right)\left(z_2- z_1\right)}. \end{eqnarray*}

of equation (\ref{cross}). Thus

\begin{eqnarray*} \frac{\left(w- i\right)}{\left(w- 1\right)} = \frac{\left(z- 1\right)\left(0+1\right)}{\left(z+1\right)\left(0- 1\right)}. \end{eqnarray*}

Solving for $w$ we obtain \[ w = T(z) = \frac{(i+1)z+(i-1)}{2z}. \]

Mappings of the upper half plane

The following transformation \begin{eqnarray}\label{upper} T(z) = e^{i\alpha} \frac{z- z_0}{z-\conj{z}}, \end{eqnarray} with $\alpha\in \R$ and $\Im(z_0)\gt 0,$ defines all linear fractional transformations that map the upper half plane $\Im(z)$ onto the open disk $|w| \lt 1 $ and the boundary $\Im(z)=0$ of the half plane onto the boundary $|w| = 1$ of the disk.

Use the applet below to explore the transfomation defined in (\ref{upper}). Here we start in the disk $|w| \leq 1.$ Drag the slider to apply the transfomation. Then drag around the point $z_0$ and observe what happens. You can aslo modify the value of $\alpha.$ What do you notice? What do you wonder?

Exercise 1: Based on your observations from the previous applet, what can you say about the transformation when $\Im(z_0) = 0$ or $\Im(z_0) \lt 0$? What happens when you modify the value of $\alpha$?

Exercise 2: Prove that the transfomation defined in (\ref{upper}) maps the half plane $\Im (z) \gt 0$ onto the disk $|w| \lt 1$ and the boundary of the half plane onto the boundary of the disk.