4.3 Mobius Transformations

Möbius (Linear Fractional) Transformations

A Möbius transformation is a rational function of the form:

S (z) = \frac{a z + b}{c z + d}, where a d - b c \neq 0

Its derivative:

S^{'} (z) = \frac{a d - b c}{(c z + d)^{2}}

Inverse Transformation

Solve for $z$ from $w = \frac{a z + b}{c z + d}$ :

Multiply both sides:

w (c z + d) = a z + b \Rightarrow w c z + w d = a z + b

Bring $z$ terms together:

w c z - a z = b - w d \Rightarrow z (w c - a) = b - w d

So the inverse is:

z = \frac{d w - b}{- c w + a} = S^{- 1} (w)

Special Cases

$z + a$ is a translation
$\frac{1}{z}$ is an inversion
$S (\infty) = \frac{a}{c}$
$S (- \frac{d}{c}) = \infty$

Notes

Möbius maps are conformal (angle-preserving) wherever defined.
They map circles and lines to circles and lines.

Möbius Decomposition

Let:

S (z) = \frac{a z + b}{c z + d}, a d - b c \neq 0

Case 1: $c = 0$

S (z) = \frac{a z + b}{d} = \frac{a}{d} z + \frac{b}{d}

→ translation + scaling

Case 2: $c \neq 0$

Rewrite:

\frac{a z + b}{c z + d} = \frac{a}{c} + \frac{b c - a d}{c^{2}} \cdot \frac{1}{z + \frac{d}{c}}

This is:

$z \mapsto z + \frac{d}{c}$ (translation)
$\mapsto \frac{1}{z + \frac{d}{c}}$ (inversion)
$\mapsto \frac{b c - a d}{c^{2}} \cdot \dots$ (dilation/rotation)
$\mapsto + \frac{a}{c}$ (translation)

Conclusion

Any Möbius map = composition of:

translation
inversion
dilation
rotation

Any Möbius transformation maps circles and lines to circles and lines.

The inversion has a magic property: it maps the set of {lines and circles} to itself. Clearly translations, dilation and rotation already do this, but inversion ( $z \mapsto 1 / z$ ) does this:

A line through the origin maps to a line through the origin.
A line not through the origin maps to a circle through the origin.
A circle through the origin maps to a line not through the origin.
A circle not through the origin maps to a circle not through the origin.

\underset{line, circle, etc.}{\underset{⏟}{C-Plane (C)}} \overset{Stereographic Projection}{\to} Riemann Sphere (C_{\infty}) \overset{Möbius Transformation}{\to}

\overset{Möbius Transformation}{\to} Riemann Sphere (C_{\infty}) \overset{Inverse Stereographic Projection}{\to} \underset{line, circle, etc.}{\underset{⏟}{C-Plane (C)}}

Proof

Circle mapped by inversion

Let a circle:

| z - a |^{2} = r^{2}

Apply inversion $z = \frac{1}{w}$ :

\begin{matrix} (*) & {| \frac{1}{w} - a |}^{2} = r^{2} \Rightarrow | 1 - a w |^{2} = r^{2} | w |^{2} \end{matrix}

Let $w = u + i v$ and expand:

\begin{aligned} | w |^{2} & = u^{2} + v^{2} \\ a w & = a (u + i v), \overset{―}{a w} = \overset{―}{a} (u - i v) \\ | 1 - a w |^{2} & = (1 - a u - i a v) (1 - \overset{―}{a} u + i \overset{―}{a} v) \\ = 1 - 2 ℜ (a) u - 2 ℑ (a) v + | a |^{2} (u^{2} + v^{2}) \end{aligned}

So from $(*)$ :

(| a |^{2} - r^{2}) (u^{2} + v^{2}) + A u + B v + 1 = 0

If $| a | = r$ → linear in $u, v$ → line
Else → quadratic → circle because reasons

Line mapped by inversion

Let a line: $A x + B y = C$ , i.e.,

A ℜ (z) + B ℑ (z) = C

Let $z = \frac{1}{w}$ with $w = u + i v$ :

\frac{1}{w} = \frac{u - i v}{u^{2} + v^{2}} \Rightarrow ℜ (\frac{1}{w}) = \frac{u}{u^{2} + v^{2}}, ℑ (\frac{1}{w}) = \frac{- v}{u^{2} + v^{2}}

So:

A \frac{u}{u^{2} + v^{2}} + B \frac{- v}{u^{2} + v^{2}} = C \Rightarrow A u - B v = C (u^{2} + v^{2})

If $C = 0$ → line
If $C \neq 0$ → circle

Conclusion

Inversion $z \mapsto \frac{1}{z}$ maps:

lines ↔ circles
lines through origin ↔ lines
circles through origin ↔ lines

So Möbius transformations (built from inversions + affine maps) preserve circles and lines.

The Cross Ratio & Canonical Möbius Map

Given any three distinct points $z_{2}, z_{3}, z_{4}$ in the extended complex plane $\hat{C} = C \cup \infty$ , there exists a unique Möbius transformation $S (z)$ that maps:

$z_{2} \mapsto 1$
$z_{3} \mapsto 0$
$z_{4} \mapsto \infty$

This map is used to define the cross ratio and is central in projective geometry and complex analysis.

$S (z_{2}) = (z_{2}, z_{2}, z_{3}, z_{4}) = 1$
$S (z_{3}) = (z_{3}, z_{2}, z_{3}, z_{4}) = 0$
$S (z_{4}) = (z_{4}, z_{2}, z_{3}, z_{4}) = \infty$

The Cross Ratio

Definition:
The cross ratio of four distinct points $z, z_{2}, z_{3}, z_{4}$ is:

(z, z_{2}, z_{3}, z_{4}) := S (z)

where $S$ is the unique Möbius transformation mapping:

z_{2} \mapsto 1, z_{3} \mapsto 0, z_{4} \mapsto \infty

$(z, z_{2}, z_{3}, z_{4})$ is not a list of points, it means cross-ratio of $z, z_{2}, z_{3}, z_{4}$

S (z) = \frac{(z - z_{3}) (z_{2} - z_{4})}{(z - z_{4}) (z_{2} - z_{3})}

Which is a möbius transform

None of $z_{2}, z_{3}, z_{4}$ are $\infty$

S (z) = \frac{(z - z_{3}) (z_{2} - z_{4})}{(z - z_{4}) (z_{2} - z_{3})}

This satisfies:

S (z_{2}) = 1, S (z_{3}) = 0, S (z_{4}) = \infty

If $z_{2} = \infty$ :

S (z) = \frac{z - z_{3}}{z - z_{4}}

If $z_{3} = \infty$ :

S (z) = \frac{z_{2} - z_{4}}{z - z_{4}}

If $z_{4} = \infty$ :

S (z) = \frac{z - z_{3}}{z_{2} - z_{3}}

Uniqueness of Such a Möbius Transformation

Let $T$ be another Möbius map with the same action:

T (z_{2}) = 1, T (z_{3}) = 0, T (z_{4}) = \infty

Then $S \circ T^{- 1}$ fixes:

1, 0, \infty

So it must be the identity transformation. Let’s verify that.

Assume

S T^{- 1} (z) = \frac{a z + b}{c z + d}

Then:

$S T^{- 1} (0) = 0$ :

\frac{a \cdot 0 + b}{c \cdot 0 + d} = \frac{b}{d} = 0 \Rightarrow b = 0

$S T^{- 1} (1) = 1$ :

\frac{a}{c + d} = 1 \Rightarrow a = c + d

$S T^{- 1} (\infty) = \infty$ :

\frac{a}{c} = \infty \Rightarrow c = 0

Now from $a = c + d$ and $c = 0$ , we get:

a = d

So the map becomes:

S T^{- 1} (z) = \frac{a z}{d} = z

Hence:

S = T

Summary

The transformation sending any 3 distinct points to $1, 0, \infty$ is a Möbius map.
This transformation is unique.

Theorem (Möbius Invariance)

For any Möbius transformation $T$ , we have:

(T z_{1}, T z_{2}, T z_{3}, T z_{4}) = (z_{1}, z_{2}, z_{3}, z_{4})

Proof

Let $S (z) = (z, z_{2}, z_{3}, z_{4})$ . Then define:

F := S \circ T^{- 1}

Then:

F (T z_{2}) = S (z_{2}) = 1, F (T z_{3}) = S (z_{3}) = 0, F (T z_{4}) = S (z_{4}) = \infty

So $F$ is the Möbius map sending $T z_{2} \mapsto 1$ , $T z_{3} \mapsto 0$ , $T z_{4} \mapsto \infty$
$\Rightarrow$ by uniqueness, $F = (z, T z_{2}, T z_{3}, T z_{4})$
Therefore:

(T z_{1}, T z_{2}, T z_{3}, T z_{4}) = F (T z_{1}) = S (z_{1}) = (z_{1}, z_{2}, z_{3}, z_{4})

Cross-Ratio Circle Test

The cross-ratio $(z_{1}, z_{2}, z_{3}, z_{4})$ is real $⟺$ the four points lie on a circle or a line.

Proof

\begin{aligned} (z_{1}, z_{2}, z_{3}, z_{4}) \in R & ⟺ (S z_{1}, S z_{2}, S z_{3}, S z_{4}) \in R \\ ⟺ (S z_{1}, 1, 0, \infty) \in R \\ ⟺ S z_{1} \in R \\ ⟺ S z_{1} \in line through 1, 0, \infty \\ ⟺ z_{1} lies on the circle or line through z_{2}, z_{3}, z_{4} \end{aligned}

Symmetry

Definition

Points $z$ and $z^{*}$ are symmetric w.r.t. the circle through $z_{1}, z_{2}, z_{3}$ iff:

(z^{*}, z_{1}, z_{2}, z_{3}) = \overset{―}{(z, z_{1}, z_{2}, z_{3})} or S z^{*} = \overset{―}{S z}

Note

If $(w, z_{2}, z_{3}, z_{4})$ are given, then:

(\overset{―}{w}, \overset{―}{z_{2}}, \overset{―}{z_{3}}, \overset{―}{z_{4}}) = \overset{―}{(w, z_{2}, z_{3}, z_{4})}

Theorem — Mobius Transformations Preserve Symmetry

If a Möbius transformation sends a circle $C_{1}$ to a circle $C_{2}$ , it sends symmetric pairs w.r.t. $C_{1}$ to symmetric pairs w.r.t. $C_{2}$ .

Möbius (Linear Fractional) Transformations

Inverse Transformation

Special Cases

Notes

Möbius Decomposition

Case 1: c=0

Case 2: c≠0

Conclusion

Any Möbius transformation maps circles and lines to circles and lines.

Proof

Circle mapped by inversion

Line mapped by inversion

Conclusion

The Cross Ratio & Canonical Möbius Map

The Cross Ratio

Uniqueness of Such a Möbius Transformation

Assume

Summary

Theorem (Möbius Invariance)

Proof

Cross-Ratio Circle Test

Proof

Symmetry

Definition

Note

Theorem — Mobius Transformations Preserve Symmetry

Case 1: $c = 0$

Case 2: $c \neq 0$