Möbius Transformations

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Möbius Transformations

Definition. A linear fractional transformation is a function of the form $T(z) = \frac{az + b}{cz + d}$ (a fraction of two linear polynomials) where the coefficients $a, b, c, d \in \mathbb{C}$ are complex numbers. Definition. In addition, if $ad - bc \ne 0$, then T is called a Möbius transformation.

The quantity ad - bc is the determinant of the 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: it actually moves points around in a meaningful way, preserving the complex plane’s structure (angles, circles, lines).

If $ad-bc = 0$, the matrix is singular (non-invertible). This is the red flag 🚩: it means the transformation collapses everything.

If $ad-bc = 0$, then the two rows of the matrix are linearly dependent. In other words: $(a, b) =\lambda (c, d)$ for some scalar $\lambda$. So the numerator and denominator are proportional: $az+b=\lambda (cz+d), T(z)=\frac{az+b}{cz+d}=\frac{\lambda (cz+d)}{cz+d} = \lambda$ as long as $cz+d\neq 0$. That’s a constant function, every input z gets sent to the same output, so the entire plane collapses into a single point.

In conclusion. If $ad - bc = 0$, the function becomes “degenerate” (useless). We are left with a constant function $T(z) = \text{constant}$. That’s not a transformation; it collapses the whole plane into a single point.

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

Standard 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 space is called the Riemann Sphere or the Extended Complex Plane.

Handling the Singularities

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

1. If c = 0 (Linear Function), the formula simplifies to $T(z) = \frac{a}{d}z + \frac{b}{d}$. It never divides by zero. We define $T(\infin) := \infin$.
2. If $c \ne 0$ (Rational Function) there are two “problem” points we need to fix manually:
   (a) The Pole (dividing by zero). When $z = -d/c$, the denominator is zero. In the extended plane, $1/0 = \infty$, we define $T(\frac{-d}{c}) := \infin$.
   (b) The Limit at Infinity: What happens when z gets huge? The terms b and d become completely insignificant, $\lim_{z \to \infty} \frac{az+b}{cz+d} = \frac{az}{cz} = \frac{a}{c}$, then we define $T(\infin) := \frac{a}{c}$.
3. Then, with these definitions, T is a function from the extended plane to itself, $\mathbb{C} \setminus \{\frac{-d}{c} \} \to \mathbb{C} \setminus \{\frac{-d}{c} \}$.

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

Proof T is one to one (injective). We assume the outputs are equal ($T(z) = T(z_1)$) and prove the inputs must have been equal ($z = z_1$).

$$ \begin{aligned} \frac{az + b}{cz + d} &=\frac{az_1 + b}{cz_1 + d} \\[2pt] &\iff[\text{Cross multiply: }] (az + b)(cz_1 + d) = (az_1 + b)(cz + d) \\[2pt] &\iff[\text{Expand: }] aczz_1 + bd + bcz_1 + + adz = aczz_1 + bd +bcz + bd \\[2pt] &\iff[\text{Cancel common terms: }] adz + bcz_1 = adz_1 + bcz \\[2pt] &\iff[\text{Group terms: }] (ad -bc)z = (ad -bc)z_1 \\[2pt] &\iff[\text{Since we have defined } ad - bc \ne 0] z = z_1 \end{aligned} $$

This proves T is one-to-one (except potentially at the pole/infinity, but we defined those manually to be distinct).

Proof T is onto. 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 $c \neq 0$: If $w \neq a/c$, we use the formula above to find z. If $w = a/c$, we recall our definition that $T(\infty) = a/c$, so $z = \infty$. 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 $T(\infty) = a/c$).

   Möbius Mnemonic. Notice the coefficients of T are $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$. The coefficients of $T^{-1}$ are $\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$. This is exactly the pattern for the inverse of a $2\times2$ matrix!

Non-Uniqueness of Coefficients

The reader should notice that one can multiply the coefficients by a scalar $\lambda$ without changing the function (Since T(z) is a ratio, one can scale the numerator and the denominator by the same amount and the ratio stays the same).

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. However, we often choose a specific $\lambda$ to make the determinant equal to 1, ad - bc = 1. This is called 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 $a_1=\lambda a, b_1=\lambda b, c_1=\lambda c, d_1=\lambda d.$

Möbius transformations act on the Riemann sphere via matrices in $\mathrm{GL_{\mathnormal{2}}}(\mathbb{C})$: $\left( \begin{matrix}a&b\\ c&d\end{matrix}\right) \cdot z=\frac{az+b}{cz+d}.$ Scaling a matrix by $\lambda \neq 0$ does not change the induced map, so two matrices represent the same transformation iff they differ by a nonzero scalar.

The “Geometric Atoms” (Elementary Transformations)

Every Möbius transformation, no matter how complex, can be built by chaining together simpler transformations.

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. Translations preserve circles/lines.
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 length by $r = |a|$. If $r > 1$, it expands; if $r < 1$, it shrinks. (b) Rotation: Rotates the complex plane by the angle $\theta$. Notice that pure rotation is when |a| = 1. Pure dilation is when a is real and positive. 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).

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 -translations ($z \mapsto z + \beta$), dilations/rotations ($z \mapsto \alpha z$), and 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, $S_4 \circ S_3 \circ S_2 \circ S_1 (z) = T(z).$