Möbius Transformations

Consider a transformation $f: \mathbb{C} \to \mathbb{C}$ with

\[f(z) = {az + b \over cz + d}\]

for some complex numbers $a, b, c$ and $d$ with $ad - bc \not= 0$.

As

\[f(z) - f(w) = {(ad - bc)(z - w) \over (cz + d)(cw + d)}\]

By having $ad - bc \not= 0$, we ensure $f$ is not constant and is injective, i.e. $f(z) = f(w)$ implies $z = w$.

However, if $c \not= 0$, $f(-d/c)$ is not defined. It makes composition hard because each transformation will have its own undefined point. In order to make $f$ complete, we need to add the “point at infinity” to $\mathbb{C}$ to form the extended complex plane.

Definition. The set $\mathbb{C}_{\infty} = \mathbb{C} \cup \Set{\infty}$ is called the extended complex plane.

Definition. A Möbius transformation is a map of $\mathbb{C}_\infty \to \mathbb{C}_\infty$ defined by

\[f(z) = {az + b \over cz + d}\]

for $a, b, c, \in \mathbb{C}$ and $ad - bc \not= 0$.

If $c \not= 0$,

\[f(-{d \over c}) = \infty \quad \text{and} \quad f(\infty) = {a \over c}\]

If $c = 0$,

\[f(\infty) = \infty\]

Composition

Proposition. The composition of two Möbius maps is again a Möbius map.

Proof.

Let

\[f(z) = {az + b \over cz + d} \quad \text{and} \quad g(z) = {\alpha z + \beta \over \gamma z + \delta}\]

Then

\[\begin{align*} g(f(z)) &= {\alpha (az + b) + \beta (cz + d) \over \gamma (az + b) + \delta (cz + d)} \\ &= { (\alpha a + \beta c)z + (\alpha b + \beta d) \over (\gamma a + \delta c)z + (\gamma b + \delta d)} \end{align*}\]

where we note that $(\alpha a + \beta c)(\gamma a + \delta c) - (\alpha b + \beta d)(\gamma b + \delta d) = (ad - bc)(\alpha \gamma - \beta \delta) \not= 0$.

We also need to check a few special cases regarding $\infty$ which are omitted.

In other words, all Möbius maps are closed under composition.

Inverse

Proposition. The Möbius map

\[f^{-1}(z) = {dz - b \over -cz + a}\]

is the inverse of $f(z)$.

Proof.

From the result of composition, we have

\[f^{-1}(f(z)) = { (ad - bc)z + (bd - bd) \over (-ac + ac)z + (-bc + ad) } = z\]

Similarly, $f^{-1}(f(z)) = 1$.

Also, $f^{-1}(f(-d/c)) = f^{-1}(\infty) = -d/c$ and $f^{-1}(f(\infty)) = f^{-1}(a/c) = \infty$. Hence, $f^{-1}(z)$ is the inverse of $f(z)$.

Proposition. Möbius maps are bijections.

$f(z)$ is injective as shown above under the condition $ad - bc \not= 0$.

Consider

\[g(z) = {dz - b \over -cz + a}\]

we have $f(g(z)) = g(f(z)) = z$.

In fact, the Möbius maps form a group under composition and is called Möbius Group.

Existence and Uniqueness

Clearly we have

\[f(z) = {az + b \over cz + d} = {\lambda az + \lambda b \over \lambda cz + \lambda d}\]

so the representation is not unique (similar to fractions). However,

Proposition. Suppose $a, b, c, d, a’, b’, c’, d’ \in \mathbb{C}$ with $(ad - bc)(a’d’ - b’c’) \not= 0$. If there exists at least three values of $z \in \mathbb{C}$ such that $cz + d \not= 0$, $c’z + d’ \not= 0$ and

\[{az + b \over cz + d} = {a'z + b' \over c'z + d'}\]

then

\[\begin{pmatrix} a' & b' \\ c' & d' \\ \end{pmatrix} = \lambda \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix}\]

for some non-zero complex number $\lambda$, i.e. the representation is determined up to a complex scalar multiple.

Proof.

The existence of three values of $z \in \mathbb{C}$ satisfying the above conditions implies that the quadratic equation

\[(az + b)(c'z + d') = (a'z + b')(cz + d)\]

has three distinct solutions, which means we can equate the coefficients and have $ac’ = a’c$, $bc’ + ad’ = a’d + b’c$ and $bd’ = b’d$. It is equivalent to the existence of a complex number $\mu$ such that

\[\begin{pmatrix} d & -b \\ -c & a \\ \end{pmatrix} \begin{pmatrix} a' & b' \\ c' & d' \\ \end{pmatrix} = \begin{pmatrix} \mu & 0 \\ 0 & \mu \\ \end{pmatrix}\]

and hence

\[\begin{pmatrix} a' & b' \\ c' & d' \\ \end{pmatrix} = {\mu \over ad - bc} \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix}\]

It suggests that the general Möbius map should have three degrees of freedom.

Lemma. A Möbius map is an identity map if it has three fixed points.

Proof.

Consider the quadratic equation $az + b = z(cz + d)$ having three distinct solutions, it implies $c = b = 0$ and $a = d$, i.e. the identity map.

Theorem. Let $\Set{z_1, z_2, z_3}$ and $\Set{w_1, w_2, w_3}$ be triples of distinct points in $\mathbb{C}_\infty$. Then there is a unique Möbius map $f$ such that $f(z_i) = w_i$.

Proof.

Suppose $z_i \not= \infty$, consider

\[g(z) = \left( {z_3 - z_2 \over z_3 - z_1} \right) {z - z_1 \over z - z_2}\]

we have $g(z_1) = 0$, $g(z_2) = \infty$ and $g(z_3) = 1$. If any of the $z_i = \infty$, by choosing $z_4$ distinct from $z_i$, we can construct $g$ base on $z’_i = 1 / (z_i - z_4)$.

Similarily, we can construct $h(z)$ such that $h(w_1) = 0$, $h(w_2) = \infty$ and $h(w_3) = 1$ and $f = h^{-1}g$ is the map satisfying $f(z_i) = w_i$.

Suppose $f$ and $f’$ are Möbius map such that $f(z_i) = w_i = f’(z_i)$. As the map $f’^{-1}f$ fixes each $z_i$, it is the identity map and hence $f = f’$.

This can be used to construct the Möbius map provided three distinct points. If we have

\[\begin{align*} f(z_1) &= 0 \\ f(z_2) &= \infty \\ f(z_3) &= 1 \\ \end{align*}\]

Then

\[f(z) = {z_3 - z_2 \over z_3 - z_1} { z - z_1 \over z - z_2}\]

Decomposition

Theorem. Every Möbius transformation can be expressed as the composition of at most four maps, each of which is of one of the forms

\[z \mapsto az, \quad z \mapsto z + b, \quad z \mapsto 1/z\]

Proof.

If $c = 0$ and $d \not= 0$, then $f = f_2f_1$ where

\[\begin{align*} f_1(z) &= (a/d)z \\ f_2(z) &= z + b/d \end{align*}\]

If $c \not= 0$, then $f = f_4f_3f_2f_1$ where

\[\begin{align*} f_1(z) &= z + (d/c) \\ f_2(z) &= 1/z \\ f_3(z) &= -((ad - bc)/c^2)z \\ f_4(z) &= z + a/c \end{align*}\]

These simple maps are rotations, dilations, translations and complex inversion.

Reference