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
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
Again, the condition $ad-bc\neq 0$ ensures that we do not have a constant function. The transformation $T(z)=1/z$ is 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:
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:
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
Equation (\ref{cross}) defines (implicitly) a linear fractional transformation that maps distinct 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
- If $z=z_1,$ then the right-hand side of (\ref{cross-02}) is zero. So $w=w_1.$
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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
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
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
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 transformation defined in (\ref{upper}). Here we start in the disk $|w| \leq 1.$ Drag the slider to apply the transformation. Then drag around the point $z_0$ and observe what happens. You can also 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 transformation 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.
Classification of Möbius Transformations
A fundamental way to understand linear fractional (Möbius) transformations is through their fixed points. A point $z_0$ is called a fixed point of a transformation $T$ if \[ T(z_0) = z_0. \] For a transformation \[ T(z) = \frac{az+b}{cz+d}, \qquad (ad-bc \neq 0), \] the fixed points are obtained by solving \[ \frac{az+b}{cz+d} = z, \] which leads to the quadratic equation \[ cz^2 + (d-a)z - b = 0. \]
Thus, a Möbius transformation has either one or two fixed points in the extended complex plane. The number and nature of the fixed points (distinct or repeated) are determined by the discriminant of this quadratic. This observation leads to a natural classification.
1. Transformations with two distinct fixed points
If the quadratic equation has two distinct solutions, then the transformation has two distinct fixed points. In this case, the transformation is conjugate to a simple dilation, rotation, or a combination of both, of the form \[ S(z) = kz, \qquad k \in \mathbb{C}\setminus\{0,1\}. \]
- If $k > 0$ (real positive) and $k \neq 1$, the transformation is called a hyperbolic transformation. In this case $|k| \neq 1$ and $k$ is real.
- If $|k| = 1$ but $k \neq 1$ (i.e., $k = e^{i\theta}$ with $\theta \not\equiv 0 \mod 2\pi$), the transformation is called an elliptic transformation.
- If $k$ is neither real positive nor of unit modulus (i.e., $k \notin \mathbb{R}^+$ and $|k| \neq 1$), the transformation is called a loxodromic transformation. Such transformations combine rotation and dilation.
Geometrically, hyperbolic transformations act as scalings (stretching along geodesics), elliptic transformations act as rotations, and loxodromic transformations are a composition of both. All these preserve the set of circles and lines after a suitable change of coordinates.
2. Transformations with one (double) fixed point
If the quadratic equation has a repeated root, then the transformation has exactly one fixed point in the extended complex plane. In this case, the transformation is conjugate to a translation \[ S(z) = z + c, \qquad c \neq 0. \]
Such transformations are called parabolic transformations.
3. Classification using the trace
Another useful classification arises by associating the transformation with the matrix \[ \begin{pmatrix} a & b \\ c & d \end{pmatrix}, \qquad ad - bc \neq 0. \]
Since multiplying the matrix by a nonzero scalar does not change the Möbius transformation, we can normalize the matrix so that its determinant equals $1$ (i.e., work with $SL(2,\mathbb{C})$, the special linear group). After this normalization, the classification depends on the value of the trace $\tau = a + d$:
- Elliptic: $\tau \in \mathbb{R}$ and $|\tau| \lt 2.$
- Parabolic: $\tau = \pm 2$ (the fixed point is unique).
- Hyperbolic: $\tau \in \mathbb{R}$ and $|\tau| > 2$.
- Loxodromic: $\tau \notin \mathbb{R}$ (i.e., $\operatorname{Im}(\tau) \neq 0$).
This classification is particularly important in geometry and dynamical systems, where it describes the qualitative behavior of iterates of the transformation. Hyperbolic and loxodromic transformations exhibit chaotic behavior on the Riemann sphere, while elliptic transformations are periodic, and parabolic transformations are neither periodic nor chaotic but have a single neutral fixed point.
The following animation provides a visual illustration of the classification of Möbius transformations. Select the graphic view with the mouse and then press the keys:
1. Elliptic, 2. Hyperbolic, 3. Loxodromic, 4. Parabolic, 5. Show/Hide Sphere.
Exercise 3: Find and classify the fixed points of the transformation \[ T(z) = \frac{2z+1}{z+1}. \] Determine whether it is elliptic, parabolic, or hyperbolic.
Solution
- Fixed points: $\dfrac{1\pm\sqrt{5}}{2}$ (two distinct real points).
- Matrix form: $\begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$, determinant $= 1$ (already normalized). Trace $\tau = 3 \in \mathbb{R}$, $|\tau| > 2$.
- Classification: Hyperbolic.