11.7: Fractional Linear Transformations

Definition: Fractional Linear Transformations

A fractional linear transformation is a function of the form

$$T(z) = \dfrac{az + b}{cz + d} \nonumber$$

where $a$, $b$, $c$, and $d$ are complex constants and with $ad - bc \ne 0$.

These are also called Möbius transforms or bilinear transforms. We will abbreviate fractional linear transformation as FLT.

Simple Point

If $ad - bc = 0$ then $T(z)$ is a constant function.

Proof

The full proof requires that we deal with all the cases where some of the coefficients are 0. We’ll give the proof assuming $c \ne 0$ and leave the case $c = 0$ to you. Assuming $c \ne 0$, the condition $ad - bc = 0$ implies

$$\dfrac{a}{c} (c, d) = (a, b). \nonumber$$

So,

$$T(z) = \dfrac{(a/c) (cz + d)}{cz + d} = \dfrac{a}{c}. \nonumber$$

That is, $T(z)$ is constant.

Extension to $\infty$. It will be convenient to consider linear transformations to be defined on the extended complex plane $C \cup \{ \infty \}$ by defining

$$\begin{array} {rcl} {T(\infty)} & = & {\begin{cases} a/c & \text{ if } c \ne 0 \\ \infty & \text{ if } c = 0 \end{cases}} \\ {T(-d/c)} & = & {\infty \ \ \ \ \ \ \text{ if } c \ne 0.} \end{array} \nonumber$$

Example $\PageIndex{1}$: Scale and Rotate

Let $T(z) = az$. If $a = r$ is real this scales the plane. If $a = e^{i \theta}$ it rotates the plane. If $a = re^{i \theta}$ it does both at once.

Figure $\PageIndex{1}$: Multiplication by $a = re^{i \theta}$ scales by $r$ and rotates by $\theta$. (CC BY-NC; Ümit Kaya)

Note that $T$ is the fractional linear transformation with coefficients

$$\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a & 0 \\ 0 & 1 \end{bmatrix}. \nonumber$$

(We’ll see below the benefit of presenting the coefficients in matrix form!)

Example $\PageIndex{2}$: Scale and Rotate and Translate

Let $T(z) = az + b$. Adding the $b$ term introduces a translation to the previous example.

Figure $\PageIndex{2}$: The map $w = az + b$ scales, rotates and shifts the square. (CC BY-NC; Ümit Kaya)

Note that $T$ is the fractional linear transformation with coefficients

$$\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a & b \\ 0 & 1 \end{bmatrix}. \nonumber$$

Example $\PageIndex{3}$: Inversion

Let $T(z) = 1/z$. This is called an inversion. It turns the unit circle inside out. Note that $T(0) = \infty$ and $T(\infty) = 0$. In the figure below the circle that is outside the unit circle in the $z$ plane is inside the unit circle in the $w$ plane and vice-versa. Note that the arrows on the curves are reversed.

Figure $\PageIndex{3}$: The map $w = 1/z$ inverts the plane. (CC BY-NC; Ümit Kaya)

Note that $T$ is the fractional linear transformation with coefficients

$$\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}. \nonumber$$

Example $\PageIndex{4}$

Let

$$T(z) = \dfrac{z - i}{z + i}. \nonumber$$

We claim that this maps the $x$-axis to the unit circle and the upper half-plane to the unit disk.

Solution

First take $x$ real, then

$$|T(x)| = \dfrac{|x - i|}{|x + i|} = \dfrac{\sqrt{x^2 + 1}}{\sqrt{x^2 + 1}} = 1. \nonumber$$

So, $T$ maps the $x$-axis to the unit circle.

Next take $z = x + iy$ with $y > 0$, i.e. $z$ in the upper half-plane. Clearly

$$|y + 1| > |y - 1|, \nonumber$$

so

$$|z + i| = |x + i(y + 1)| > |x + i(y - 1)| = |z - i|, \nonumber$$

implying that

$$|T(z)| = \dfrac{|z - i|}{|z + i|} < 1. \nonumber$$

So, $T$ maps the upper half-plane to the unit disk.

We will use this map frequently, so for the record we note that

$T(i) = 0$, $T(\infty) = 1$, $T(-1) = i$, $T(0) = -1$, $T(1) = -i$.

These computations show that the real axis is mapped counterclockwise around the unit circle starting at 1 and coming back to 1.

Figure $\PageIndex{4}$: The map $w = \dfrac{z - i}{z + i}$ maps the upper-half plane to the unit disk. (CC BY-NC; Ümit Kaya)

Lines and circles

Theorem $\PageIndex{1}$

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

Before proving this, note that it does not say lines are mapped to lines and circles to circles. For example, in Example 11.7.4 the real axis is mapped the unit circle. You can also check that inversion $w = 1/z$ maps the line $z = 1 + iy$ to the circle $|z - 1/2| = 1/2$.

Proof

We start by showing that inversion maps lines and circles to lines and circles. Given $z$ and $w = 1/z$ we define $x, y, u$ and $v$ by

$$z = x + iy \ \ \text{ and } \ \ w = \dfrac{1}{z} = \dfrac{x - iy}{x^2 + y^2} = u + iv \nonumber$$

So,

$$u = \dfrac{x}{x^2 + y^2} \ \ \text{ and } \ \ v = -\dfrac{y}{x^2 + y^2}. \nonumber$$

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

$$Ax + By + C(x^2 + y^2) = D \nonumber$$

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

$$\begin{array} {cll} {} & \ & {Ax + By + C(x^2 + y^2) = D} \\ {\Leftrightarrow} & \ & {\dfrac{Ax}{x^2 + y^2} + \dfrac{By}{x^2 + y^2} + C = \dfrac{D}{x^2 + y^2}} \\ {\Leftrightarrow} & \ & {Au - Bv + C = D(u^2 + v^2).} \end{array} \nonumber$$

In the last step we used the fact that

$$u^2 + v^2 = |w|^2 = 1/|z|^2 = 1/(x^2 + y^2). \nonumber$$

We have shown that a line or circle in $x, y$ is transformed to a line or circle in $u, v$. This shows that inversion maps lines and circles to lines and circles.

We note that for the inversion $w = 1/z$.

1. Any line not through the origin is mapped to a circle through the origin.
2. Any line through the origin is mapped to a line through the origin.
3. Any circle not through the origin is mapped to a circle not through the origin.
4. Any circle through the origin is mapped to a line not through the origin.

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

First suppose that $c = 0$. So,

$$T(z) = (az + b)/d. \nonumber$$

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

Now suppose that $c \ne 0$. Then,

$$T(z) = \dfrac{az + b}{cz + d} = \dfrac{\dfrac{a}{c} (cz + d) + b - \dfrac{ad}{c}}{cz + d} = \dfrac{a}{c} + \dfrac{b - ad/c}{cz + d} \nonumber$$

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

$$z \ \ \mapsto \ \ w_1 = cz + d \ \ \mapsto \ \ w_2 = 1/w_1 \ \ \mapsto \ \ w = \dfrac{a}{c} + (b - ad/c) w_2 \nonumber$$

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

$$T_1 (z) = \dfrac{(z - z_1)(z_2 - z_3)}{(z - z_3)(z_2 - z_1)}. \nonumber$$

Notice that

$T_1(z_1) = 0$, $T_1 (z_1) = 1$, $T_1 (z_3) = \infty$.

Likewise let

$$T_2(w) = \dfrac{(w - w_1) (w_2 - w_3)}{(w - w_3)(w_2 - w_1)}. \nonumber$$

Notice that

$T_2(w_1) = 0$, $T_2(w_2) = 1$, $T_2 (w_3) = \infty$.

Now $T(z) = T_{2}^{-1} \circ T_1 (z)$ is the required map.

Correspondence with Matrices

We can identify the transformation

$$T(z) = \dfrac{az + b}{cz + d} \nonumber$$

with the matrix

$$\begin{bmatrix} a & b \\ c & d \end{bmatrix}. \nonumber$$

This identification is useful because of the following algebraic facts.

1. If $r \ne 0$ then $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ and $r \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ correspond to the same FLT.
   $Proof$. This follows from the obvious equality
   $$\dfrac{az + b}{cz + d} = \dfrac{raz + rb}{rcz + rd}. \nonumber$$
2. If $T(z)$ corresponds to $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ and $S(z)$ corresponds to $B = \begin{bmatrix} e & f \\ g & h \end{bmatrix}$ then composition $T \circ S(z)$ corresponds to matrix multiplication $AB$.
   $Proof$. The proof is just a bit of algebra.
   $$\begin{array} {rcl} {T \circ S(z)} & = & {T(\dfrac{ez + f}{gz + h}) = \dfrac{a((ez + f)/(gz + h)) + b}{c((ez + f)/(gz + h)) + d} = \dfrac{(ae + bg)z + af + bh}{(ce + dg) z + cf + dh}} \\ {AB} & = & {\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} e & f \\ g & h \end{bmatrix} = \begin{bmatrix} ae + bg & af + bh \\ ce + dg & cf + dh \end{bmatrix}} \end{array} \nonumber$$
   The claimed correspondence is clear from the last entries in the two lines above.
3. If $T(z)$ corresponds to $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ then $T$ has an inverse and $T^{-1} (w)$ corresponds to $A^{-1}$ and also to $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$, i.e. to $A^{-1}$ without the factor of $1/\text{det}(A)$.
   $Proof$. Since $A A^{-1} = I$ it is clear from the previous fact that $T^{-1}$ corresponds to $A^{-1}$. Since
   $$A^{-1} = \dfrac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \nonumber$$

Fact 1 implies $A^{-1}$ and $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ both correspond to the same FLT, i.e. to $T^{-1}$.

Example $\PageIndex{5}$

1. The matrix $\begin{bmatrix} a & b \\ 0 & 1 \end{bmatrix}$ corresponds to $T(z) = az + b$.
2. The matrix $\begin{bmatrix} e^{i \alpha} & 0 \\ 0 & e^{-i \alpha} \end{bmatrix}$ corresponds to rotation by $2\alpha$.
3. The matrix $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ corresponds to the inversion $w = 1/z$.