Möbius Transformations

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Möbius Transformations

Definition. A linear fractional transformation is a rational function of the form $T(z) = \frac{az + b}{cz + d}$ where the coefficients $a, b, c, d \in \mathbb{C}$ are complex numbers.

Definition. In addition, if the coefficients satisfy the condition $ad - bc \ne 0$, the function is specifically called a Möbius transformation.

To understand why this condition is absolutely vital, we can arrange the coefficients into a $2 \times 2$ matrix: $ M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$

The quantity ad - bc is exactly the determinant of this matrix $M=\left( \begin{matrix}a&b\\ c&d\end{matrix}\right)$.

• If $ad-bc \neq 0$, the matrix is invertible. That means the transformation is non-degenerate: iIt genuinely moves points around the complex plane in a meaningful, bijective way, preserving mathematical structure.
• If $ad-bc = 0$, the matrix is singular (non-invertible). This is a massive mathematical red flag 🚩. It means the two rows of the matrix are linearly dependent: $(a,b) = \lambda(c,d)$ for some scalar $\lambda$.
  Consequently, the numerator and denominator are strictly proportional: $T(z) = \frac{\lambda(cz + d)}{cz + d} = \lambda \quad (\text{for } cz+d \neq 0) $
  The function reduces to a flat constant. Every single input $z$ is crushed into the exact same output $\lambda$. The entire complex plane collapses into a single point. This is useless for geometry, which is why Möbius transformations strictly require $ad - bc \neq 0$.

The Extended Complex Plane ($\mathbb{C} \cup \{\infty\}$)

Standard rational functions often break when you divide by zero. Möbius transformations are designed to handle this gracefully by including Infinity ($\infty$) as an actual point. This new topological space, denoted $\hat{\mathbb{C}} = \mathbb{C} \cup \{\infty\}$, is called the Extended Complex Plane (or the Riemann Sphere).

Handling the Singularities

By defining how $T(z)$ behaves at infinity and at its pole, we upgrade it from a flawed function on $\mathbb{C}$ to a perfect, unbroken bijection on $\hat{\mathbb{C}}$.

Consider $T(z) = \frac{az + b}{cz + d}$ where $a, b, c, d \in \mathbb{C}, \qquad ad-bc\neq 0$.

1. If c = 0 (The Linear Case), the formula simplifies to a linear polynomial: $T(z) = \frac{a}{d}z + \frac{b}{d}$.
   Because it never divides by zero, it maps finite points to finite points. As $z$ grows infinitely large, the output grows infinitely large.
   Thus, we define: $ \mathbf{T(\infty) := \infty} $.
2. If $c \ne 0$ (The Rational Case) the variable $z$ is in the denominator. We must manually fix two “problem” points:
   (a) The Pole (dividing by zero). When $z = -d/c$, the denominator becomes exactly zero.
   Because $1/0 \to \infty$ on the Riemann Sphere, we define $ \mathbf{T\left(\frac{-d}{c}\right) := \infty} $.
   (b) The Limit at Infinity: What happens when the imput z gets huge? We look at the limit of the function as $z \to \infty$.
   The terms b and d become mathematically insignificant compared to the massive $z$ term:
   $ \lim_{z \to \infty} \frac{az + b}{cz + d} = \lim_{z \to \infty} \frac{z(a + b/z)}{z(c + d/z)} = \frac{a + 0}{c + 0} = \frac{a}{c} $
   Therefore, we define: $ \mathbf{T(\infty) := \frac{a}{c}} $
3. With these definitions securely in place, $T$ is a mathematically flawless function from the extended plane to itself: $T: \hat{\mathbb{C}} \to \hat{\mathbb{C}}$.

Möbius Transformations are Bijections f the sphere

Every Möbius transformation is a bijection of the extended complex plane.

T is well defined, one-to-one, and onto.

Proof of Injectivity (One-to-One)

We assume two finite outputs are equal, $T(z) = T(w)$ and prove the inputs must have been equal $z = w$.

$$ \begin{aligned} \frac{az + b}{cz + d} &= \frac{aw + b}{cw + d} \\[2pt] &\iff[\text{Cross multiply: }] (az + b)(cw + d) = (aw + b)(cz + d) \\[2pt] &\iff[\text{Expand: }] aczw + adz + bcw + bd = aczw + adw + bcz + bd \\[2pt] &\iff[\text{Cancel common terms: }] adz + bcw = adw + bcz \\[2pt] &\iff[\text{Group terms: }] (ad -bc)z = (ad -bc)w \\[2pt] &\iff[\text{Since we have stricly defined } ad - bc \ne 0 \text{, we can divide both sides by it, leaving exactly}] z = w \end{aligned} $$

This proves T is one-to-one on finite points or except potentially at the pole/infinity, but we have defined those manually to be distinct.

Proof of Surjectivity (Onto) & The Inverse Function

We pick any target value w and show we can find a z that maps to it. We do this by solving the equation for z.

$$ \begin{aligned} w &=\frac{az + b}{cz + d} \\[2pt] &\iff w(cz + d) = az + b \\[2pt] &\iff wcz + wd = az + b \\[2pt] &\iff wcz - az = b - wd \\[2pt] &\iff z(wc - a) = b - wd \\[2pt] &\iff z = \frac{b - dw}{wc - a} = \frac{-dw + b}{cw - a} \end{aligned} $$

1. If $c \neq 0$, there are three cases:
   (a) If $w \neq a/c$, we use the formula above to find z.
   (b) If $w = a/c$, we recall our definition that $T(\infty) = a/c$, so $z = \infty$.
   (c) If $w = \infty$, we recall our definition that $T(-d/c) = \infty$, so $z = -d/c$.
2. If $c = 0$: The equation is linear, which is trivially solvable for all finite w. Besides, we recall our definition that $T(\infty) = \infty$

Therefore, since we can find a “parent” z for every possible child $w$ (including $\infty$), the function is Onto. Because $T$ is both One-to-One and Onto on the domain $\mathbb{C} \cup \{\infty\}$, it is a Bijection and $T^{-1}: \mathbb{C} \cup \{ \infin \} \to \mathbb{C} \cup \{ \infin \}$ is well-defined.

1. If c = 0 (Linear Transformation), then $T^{-1}(w)=\frac{dw-b}{a}, \forall w \ne \infin$. In the original function, $T(\infty) = \infty$. Therefore, the inverse must map infinity back to infinity: $T^{-1}(\infin) = \infin$.
2. If $c \ne 0$ (Rational Transformation), then $T^{-1}(w)=\frac{dw-b}{-cw + a}, \forall w \ne \infin, w \ne \frac{-d}{c}$. $T^{-1}(\infin) = \frac{-d}{c}$ (this matches the original definition where $T(-d/c) = \infty$). $T^{-1}(\frac{a}{c}) = \infty$ (this matches the original definition where $T(\infty) = a/c$).

   Notice the coefficients of $T(z)$ form the matrix $M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$. The coefficients of $T^{-1}(w)$ form the matrix $\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$. This is exactly the classical adjugate pattern for finding the inverse of a $2 \times 2$ matrix! Because multiplying a fractional transformation by a scalar (like the determinant) doesn’t change the fraction, $M^{-1}$ perfectly encodes $T^{-1}$.

Non-Uniqueness and Normalization

A fractional transformation is essentially a ratio. If you multiply the numerator and denominator by the exact same non-zero complex scalar $\lambda$, the function does not change at all: T(z) = $\frac{az+b}{cz+d} =\frac{a\lambda z + b\lambda}{c\lambda z + d\lambda}$ where $\lambda$ is a non-zero complex number.

Because of this redundancy, the coefficients a, b, c, and d are not unique. For any non-zero $\lambda$, the sets $\{a,b,c,d\}$ and $\{\lambda a, \lambda b, \lambda c, \lambda d\}$ represent the exact same geometric transformation. In the language of linear algebra, scaling a matrix by $\lambda \neq 0$ does not change the map it induces on the Riemann sphere.

So the matrices $\begin{pmatrix}a&b\\ c&d\end{pmatrix} \quad\text{and} \quad \lambda\begin{pmatrix}a&b\\ c&d\end{pmatrix}$ define exactly the same Möbius transformation. Thus, Möbius transformations correspond not to individual invertible matrices, but to invertible matrices up to nonzero scalar multiples.

This is why the natural matrix group for Möbius transformations is $\mathrm{PGL}_2(\mathbb C)=\mathrm{GL}_2(\mathbb C)/\mathbb C^\times.$

Normalization: To remove this redundancy and make calculations elegant, we often choose a specific scalar $\lambda$ to force the determinant to equal exactly $1$: $ (\lambda a)(\lambda d) - (\lambda b)(\lambda c) = \lambda^2(ad - bc) = 1 \implies \lambda = \frac{1}{\pm\sqrt{ad-bc}} $

When $ad - bc = 1$, we call it a normalized Möbius transformation. It makes calculations cleaner and easier.

Möbius transformation coefficients are proportional

Let T be a Möbius transformation with two representations $T(z)=\frac{az+b}{cz+d}=\frac{a_1z+b_1}{c_1z+d_1}$, where not all of a, b, c, d are zero and likewise for $a_1, b_1, c_1, d_1$, and at least one determinant is nonzero: $ad-bc \neq 0$ (so T is non-constant).

Then, there exists $\lambda \in \mathbb{C^{\mathnormal{\times }}}$ such that $a_1=\lambda a, b_1=\lambda b, c_1=\lambda c, d_1=\lambda d.$

$\frac{az+b}{cz+d}=\frac{a_1z+b_1}{c_1z+d_1}$.

Cross-multiplying: $(az+b)(c_1z+d_1)=(a_1z+b_1)(cz+d).$

Expand both sides: $ac_1z^2+(ad_1+bc_1)z+bd_1=a_1cz^2+(a_1d+b_1c)z+b_1d.$ Since these are polynomials in z that agree for infinitely many z, their coefficients must match:

• $z^2$-coefficient: $ac_1=a_1c$.
• z-coefficient: $ad_1+bc_1=a_1d+b_1c$.
• constant term: $bd_1=b_1d.$

Extracting the scalar $\lambda$. We want to show there exists $\lambda \neq 0$ with $a_1=\lambda a,\quad b_1=\lambda b,\quad c_1=\lambda c,\quad d_1=\lambda d$.

Because $ad-bc\neq 0$, at least one of a, c is nonzero and at least one of b, d is nonzero.

Suppose $c\neq 0$. From $ac_1=a_1c$, we get $\frac{a_1}{a}=\frac{c_1}{c}$ whenever $a\neq 0$. If a = 0, then $ac_1=a_1c$ becomes $0=a_1c$, so $a_1=0$ (since $c\neq 0$), and we can instead use the other coefficients; but generically, this suggests a common ratio [… The reader should continue with the different cases]

Associate to T the matrix $M=\left( \begin{matrix}a&b\\ c&d\end{matrix}\right) \in \mathrm{GL_{\mathnormal{2}}}(\mathbb{C})$, and define its action on the Riemann sphere by $M\cdot z=\frac{az+b}{cz+d}.$

If we scale M by $\lambda \neq 0$, $\lambda M=\left( \begin{matrix}\lambda a&\lambda b\\ \lambda c&\lambda d\end{matrix}\right)$, then $(\lambda M)\cdot z=\frac{\lambda az+\lambda b}{\lambda cz+\lambda d}=\frac{\lambda (az+b)}{\lambda (cz+d)}=\frac{az+b}{cz+d}=M\cdot z$.

So scaling by a nonzero scalar does not change the induced Möbius transformation.

Conversely, we just proved: if two matrices induce the same Möbius transformation and at least one has nonzero determinant (so the map is non-constant), then the matrices differ by a nonzero scalar.

Möbius transformations are represented by matrices in $\mathrm{GL_{\mathnormal{2}}}(\mathbb{C})$, and two such matrices represent the same transformation if and only if they differ by a nonzero scalar. Equivalently, the group of Möbius transformations is $\mathrm{PGL_{\mathnormal{2}}}(\mathbb{C})=\mathrm{GL_{\mathnormal{2}}}(\mathbb{C})/\mathbb{C^{\mathnormal{\times }}}$.

The “Geometric Atoms” (Elementary Transformations)

The most powerful geometric feature of Möbius transformations is that every Möbius transformation, no matter how complex, can be built by chaining together simpler transformations, mainly three elementary “atoms” of geometry.

1. Translation. T(z) = z + a. This shifts the entire complex plane by the vector a. It does not rotate or stretch the plane, just slides it. It preserves all shapes, angles, and sizes.
2. Dilation & Rotation. T(z) = az, $a \ne 0$. This combines two effects based on the polar form of the constant $a = re^{i\theta}$:
   (a) Scaling (Dilation): Multiplies every length by $r = |a|$. If $r > 1$, it expands; if $r < 1$, it shrinks.
   (b) Rotation: Rotates the complex plane around the origin by the angle $\theta$. Notice that pure rotation is when |a| = 1. Pure dilation is when a is real and positive (a = r, $r \in \mathbb{R}$). Dilations preserve circles/lines.
3. Inversion. $T(z) = \frac{1}{z}$. This turns the complex plane “inside out.”
   Points inside the unit circle map to the outside (if |z|<1, then |1/z|>1). Points outside the unit circle map to the inside (If |z|>1, then |1/z|<1).
   It swaps the origin and infinity: $0 \leftrightarrow \infty$ (At z = 0, we define $T(0) = \infty$. At $z = \infty$, we define $T(\infty) = 0$). Inversion maps circles/lines to circles/lines (with the subtlety that circles through the origin become lines not passing through the origin, and lines not through the origin become circles passing through the origin).

Proof Sketch:

Any circle or line in $\mathbb{C}\cong \mathbb{R^{\mathnormal{2}}}$ can be written as: $\boxed{A|z|^2+Bz+\overline{B}\, \overline{z}+C=0}$ with $A\in \mathbb{R}, B\in \mathbb{C}, C\in \mathbb{R}$.

• If $A \neq 0$, this is a circle.
• If $A = 0$, this is a line.

This is the algebraic description of a “generalized circle.”

Apply inversion. Let w = T(z) = 1/z, so z = 1/w.

Plug into the equation: $A\left| \frac{1}{w}\right| ^2+B\left( \frac{1}{w}\right) +\overline{B}\left( \frac{1}{\overline{w}}\right) +C=0.$

Multiply both sides by $|w|^2$ to clear denominators: $A+B\overline{w}+\overline{B}w+C|w|^2=0.$

This is again a real quadratic equation in w and $\overline{w}$. And therefore: the image of a circle/line under inversion is again a circle/line.

If the original circle passes through 0, then in its equation the constant term C must be 0 (Plug z = 0: you get C = 0.)

Then, the transformed equation becomes: $A+B\overline{w}+\overline{B}w=0$, which is linear in w. A linear real equation describes a line. So, circles through the origin become lines not passing through the origin.

If the original line does not pass through 0, then its equation has $C\neq 0$. After inversion, the term $C|w|^2$ appears, making the new equation quadratic and so we get a circle.

Proposition. Decomposition of Möbius Transformations. Every Möbius transformation $T(z) = \frac{az+b}{cz+d}$ can be expressed or written as a composition of simpler transformations or building blocks, more specifically as a composition of simple translations ($z \mapsto z + \beta$), dilations/rotations ($z \mapsto \alpha z$), and at most one inversion ($z \mapsto \frac{1}{z}$).

This means that geometrically, a Möbius transformation is nothing more than a sequence of shifting, spinning, scaling, and flipping the plane. It gives us a powerful geometric result. If we want to know what a Möbius transformation does to a shape (like a circle or a line), we don’t have to calculate the full messy formula. We just need to now that translations and dilations/rotations preserve circles and lines, and inversions map circles/lines to circles/lines (with the subtlety that circles through the origin become lines not passing through the origin, and lines not through the origin become circles passing through the origin). Since Möbius transformations are compositions of these, they must map circles/lines to circles/lines.

Proof.

Suppose c = 0 (the linear case), then $d \neq 0$ (otherwise $ad-bc=0$). The function simplifies to a linear polynomial: $T(z) =\frac{az+b}{d}=\frac{a}{d}z + \frac{b}{d}$.

We can break this into two steps: $S_1(z) = \frac{a}{d}z, S_2(z) = z + \frac{b}{d}$ a dilation/rotation and translation respectively. Then, the composition matches T(z), $T(z) = (S_2 \circ S_1)(z)$.

If $c \ne 0$ (the rational case), then we use algebraic manipulation to isolate the $z$ term in the denominator: $T(z) = \frac{az+b}{cz+d}=\frac{\frac{a}{c}(cz+d)-\frac{ad}{c}+b}{cz+d} = \frac{a}{c} + \frac{-ad + bc}{c^2z + dc} = \frac{a}{c} + \frac{1}{(\frac{c^2}{bc-ad})z + \frac{dc}{bc-ad}}$

We can identify the operations in order of application (from inside out, starting at z): $S_1(z) = (\frac{c^2}{bc-ad})z, S_2(z) = z + \frac{dc}{bc-ad}, S_3(z) = \frac{1}{z}, S_4 = z + \frac{a}{c}$.

Translations ($S_4, S_2$): Shift z by $a/c$ and $\frac{dc}{bc-ad}$ respectively. Inversion ($S_3$): Flip the result. Dilation/Rotation ($S_1$): Scale by the constant factor $\frac{c^2}{bc-ad}$.

Then, $S_4 \circ S_3 \circ S_2 \circ S_1 (z) = T(z).$

Every Möbius transformation is simply a sequence of shifting, flipping, spinning, and shifting again! $\blacksquare$