The first principle is that you must not fool yourself – and you are the easiest person to fool, Richard Feynman

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Proposition. Möbius maps preserve generalized circles. A Möbius transformation maps generalized circles onto generalized circles.

Proof.

Step 1: Setting up the skeleton.
Let $\Gamma$ be a generalized circle in the Riemann sphere $\mathbb{C}_{\infty} = \mathbb{C} \cup \{ \infty \}$. Let S be a Möbius transformation. We claim that the image $S(\Gamma)$ is also a generalized circle.
Let $z_2, z_3, z_4$ be three distinct points on the circle $\Gamma$. When we apply our Möbius map S, those three points land at new locations: $w_k = S(z_k)$ for $k = 2, 3, 4$.
Since Möbius maps are bijective, these new points $w_2, w_3, w_4$ are also points. Since any three distinct points define a generalized circle, $w_2, w_3, w_4$ create a new target circle, which we’ll call $\Gamma_1$.

Step 2. The Forward Proof. $S(\Gamma) \subseteq \Gamma_1$
Now we need to prove that if we pick any random point z on the original circle $\Gamma$, its image S(z) will land exactly on the new circle $\Gamma_1$.
To do this, we use the Cross-Ratio. The cross-ratio has two magical properties:
1. The Geometry Property: If four points lie on the same circle, their cross-ratio is a purely real number (no imaginary part).
2. The invariance property. The cross-ratio is invariant under Möbius transformations: $(z, z_2, z_3, z_4) = (S(z), S(z_2), S(z_3), S(z_4))$

Let $z$ be any point on $\Gamma$. By the Cross-Ratio Theorem, four points are concyclic (a set of points is concyclic if all of them lie on a single circle or equivalently, there exists one unique circle that passes through every point in the set) if and only if their cross-ratio is real.

Since $z, z_2, z_3, z_4 \in \Gamma$, we have: $(z, z_2, z_3, z_4) \in \mathbb{R}$. Because Möbius maps preserve the cross-ratio, the mapped points $(S(z), w_2, w_3, w_4)$ output that exact same real number. Since the cross-ratio is real, the point $S(z)$ must lie on the circle determined by $w_2, w_3, w_4$, which is $\Gamma_1$. Thus, every point from the original circle $S(\Gamma)$ lands on the new circle $\Gamma_1$, so $S(\Gamma) \subseteq \Gamma_1$.

Step 3: The Backward Proof, $S(\Gamma) = \Gamma_1$

Step 2 proved that the original circle maps onto the new circle, but what if it only mapped to a small arc or half of the new circle? We have to prove it covers the whole thing.

Because S is a Möbius map, it is perfectly reversible. Its inverse, $S^{-1}$, is also a Möbius map. If we run the exact same logic from Step 2 backwards starting with $\Gamma_1$, we find that $S^{-1}$ maps all of $\Gamma_1$ entirely onto $\Gamma$.

Since the map perfectly covers the target space forward and backward, the image of the whole circle $\Gamma$ is exactly the whole circle $\Gamma_1$. Another reasoning is as follows. Since $S$ is a bijection and its inverse $S^{-1}$ is also a Möbius transformation, the reverse argument holds.

The points $w_2, w_3, w_4$ lie on $\Gamma_1$.
Therefore, their preimages $z_2, z_3, z_4$ lie on the generalized circle $S^{-1}(\Gamma_1)$.

But $z_2, z_3, z_4$ were chosen as three distinct points on $\Gamma$ and there is exactly one generalized circle passing through any three distinct points, thus $S^{-1}(\Gamma_1) \subseteq \Gamma \implies S(S^{-1}(\Gamma_1))\subseteq S(\Gamma) \implies \Gamma_1 \subseteq S(\Gamma) \implies[\text{Step 2, }S(\Gamma) \subseteq \Gamma_1] \Gamma_1 = S(\Gamma)$.

Proposition. You can map any circle to any other circle. For any given generalized circles $\Gamma$ and $\Gamma'$ in $\mathbb{C} \cup \{ \infty \}$, there is a Möbius transformation T such that $T(\Gamma) = \Gamma'$. Furthermore, there is a unique Möbius transformation taking three specified points on $\Gamma$ onto three specified points on $\Gamma'$.

Proof

Existence (Can we build this map?) Let $z_2, z_3, z_4$ be three distinct points on the source circle $\Gamma$ and $w_2, w_3, w_4$ be three distinct points on the target circle $\Gamma'$.

The Fundamental Theorem of Möbius Transformations guarantees that there is always a valid Möbius map T that will connect any three starting points to any three destination points. So, we build T to send $T(z_k) = w_k \quad \text{for } k=2,3,4$

A Möbius transformation maps circles onto circles, hence T maps the circle determined by $\{z_2, z_3, z_4\}$ to the circle determined by $\{w_2, w_3, w_4\}$. Since $\{z_k\}$ determines $\Gamma$ and $\{w_k\}$ determines $\Gamma'$, it follows that $T(\Gamma) = \Gamma'$.

Because we have proved in the previous Proposition that a Möbius map drags the rest of the circle along with its three skeleton points, the entire circle Γ is guaranteed to map perfectly onto Γ.

Uniqueness (Is there only one way to do this?)

Suppose $T_1$ and $T_2$ are Möbius transformations satisfying $T_1(z_k)=w_k=T_2(z_k)\quad (k=2,3,4)$.

By the 3‑point transitivity of Möbius transformations: A Möbius transformation is uniquely determined (100% locked) by the images of any three distinct points.

Thus, $T_1=T_2$.

Symmetry, Inversion, and Orientation in Möbius Geometry

Symmetry with Respect to a Line

Definition (Reflection in a Line). Two points $z,z^*\in \mathbb{C}$ are symmetric with respect to a line L if the line segment joining z and $z^*$ is perpendicular to L and L passes through the midpoint of the segment $[z, z^*]$. In other words, L is the perpendicular bisector of the segment connecting z and $z^*$.

If you stand in front of a flat mirror, your reflection is the same distance behind the mirror as you are in front of it. The segment connecting you and your reflection is perpendicular to the mirror, and the mirror cuts that segment exactly in half.

In coordinates, if $L = \mathbb{R}$ (the real axis), symmetry reduces to complex conjugation: $z = x + iy \quad \longmapsto \quad z^* = x - iy.$

Thus ,reflection in the real axis is simply $z \mapsto \bar z$.

Symmetry with Respect to a Circle (Inversion)

Definition (Inversion in a Circle). Let $C$ be a circle with center $O$ and radius $R$. Two points $z$ and $z^*$ are symmetric with respect to $C$ if they lie on the same ray from the center O and their distances to the center satisfy $|z - O| \cdot |z^* - O| = R^2$.

This transformation is called inversion in the circle $C$. Inversion (notice that being “symmetric with respect to C” means that one point is the inversion of the other in C) is a transformation I of the extended plane such that:

Points on the circle stay on the circle (they are their own reflection): if |z - O| = R, then I(z) is also on C.
Points inside go or are mapped outside and vice versa, in a way that depends only on distance to the center along each ray.
The center is special: the center $O$ maps to $\infty$, and $\infty$ maps to $O$.

In coordinates (for a circle centered at the origin), $I(z) = \frac{R^2}{\overline{z}}.$

Thus, inversion is a nonlinear reflection across a circle.

The Unified Analytic Definition

A generalized circle is either a Euclidean circle or a straight line (a circle passing through $\infty$). To treat lines and circles as the same species (generalized circles), we use the cross-ratio.

Definition (Cross‑Ratio Formulation). Let $\Gamma$ be a generalized circle passing through distinct points $z_2, z_3, z_4$. The points z and $z^*$ in $\mathbb{C} \cup \{ \infty \}$ are said to be symmetric with respect to $\Gamma$ if $\boxed{(z^*, z_2, z_3, z_4) = \overline{(z, z_2, z_3, z_4)}}$.

The cross-ratio map $S(\zeta) = (\zeta, z_2, z_3, z_4)$ maps the circle $\Gamma$ onto the real line $\mathbb{R} \cup \{\infty\}$. In the “target world” of the real line, symmetry is simply complex conjugation ($w \mapsto \bar{w}$). This definition forces $z$ and $z^*$ to mirror each other across $\Gamma$.

The Symmetry Principle

Symmetry principle. If a Möbius transformation $T$ maps a generalized circle $\Gamma_1$ onto another generalized circle $\Gamma_2$, then any pair of points $z, z^*$ symmetric with respect to $\Gamma_1$ are mapped to a pair of points $T(z), T(z^*)$ which are symmetric with respect to $\Gamma_2$.

Proof.

Choose three distinct points $z_2, z_3, z_4$ on $\Gamma_1$ and set $w_k = T(z_k).$
Since $T$ takes a generalized circle $\Gamma_1$ onto another generalized circle $\Gamma_2$, these points $w_k$ lie on the target circle $\Gamma_2$. By definition of symmetry (Cross‑Ratio Formulation), $(z^*, z_2, z_3, z_4) = \overline{(z, z_2, z_3, z_4)}.$
We want to prove that $T(z)$ and $T(z^*)$ satisfy the symmetry definition for $\Gamma_2$, that is, we must show: $(T(z^*), w_2, w_3, w_4) = \overline{(T(z), w_2, w_3, w_4)}$
We use the previously demonstrated property that the cross-ratio is invariant under Möbius transformations:

$$ \begin{aligned} (T(z^*), T(z_2), T(z_3), T(z_4)) &=(z^*, z_2, z_3, z_4) \\[2pt] &\text{Since z and z* are symmetric with respect to Γ₁:}\\[2pt] &=\overline{(z, z_2, z_3, z_4)} \\[2pt] &\text{Now, use the invariance property on the right-hand side (inside the conjugate):} \\[2pt] &=(\overline{T(z), T(z_2), T(z_3), T(z_4)}) \end{aligned} $$

Combining the steps, we have: $(T(z^*), w_2, w_3, w_4) = \overline{(T(z), w_2, w_3, w_4)}, \quad \blacksquare$.

Orientation and the “Left-Hand Rule”

A circle divides the Riemann sphere into two regions. To distinguish them, we “orient” the circle.

Definition (Orientation). An orientation for a generalized circle $\Gamma$ is an ordered triple of points $(z_1, z_2, z_3)$ on $\Gamma$ such that $z_1, z_2, z_3$ are distinct points on $\Gamma$.

📌 Geometric intuition. Imagine you are walking along the boundary of a country or region (the circle or line $\Gamma$). The orientation $(z_1, z_2, z_3)$ tells you the direction to travel: start at $z_1$, walk towards $z_2$, then continue toward $z_3$. As you walk, one region lies on your left and the other is on your right.

Definition (Right and Left Side). For an oriented circle $\Gamma$ determined by three distinct points $(z_1, z_2, z_3)$, the right side consists of all point $z$ such that $\text{Im}(z, z_1, z_2, z_3) > 0$. Similarly, the left side is is the set of points $z$ such that $\text{Im}(z, z_1, z_2, z_3) < 0$.

The cross-ratio detects which side of the circle a point lies on.

The Walking Rule: As you walk from $z_1 \to z_2 \to z_3$, the “Left Side” is the region physically to your left. For a circle, if the points are in counter-clockwise order, the “Left Side” is the interior (the disk).

Orientation principle. Möbius transformations preserve orientation. Let $\Gamma_1$ and $\Gamma_2$ be generalized circles in $\mathbb{C} \cup \{ \infty \}$ and let T be a Möbius transformation such that $T(\Gamma_1) = \Gamma_2$. Let $(z_1, z_2, z_3)$ be an orientation for $\Gamma_1$. Then, T takes the right and left side of $\Gamma_1$ onto the right and left side of $\Gamma_2$ with respect to the orientation of $\Gamma_2$ given by $(T_{z_1}, T_{z_2}, T_{z_3}).$

If you define “left” based on the direction you walk on the first circle, the map $T$ preserves this. If you walk along the image points $T(z_1) \to T(z_2) \to T(z_3)$ in the second world, the region that was on your left in the first world gets mapped to the region on your left in the second world. Möbius maps preserve angles and their orientation (they don’t flip images like a mirror unless there is a conjugate involved).

Proof:

Let $\Gamma_1$ be determined by $(z_1, z_2, z_3)$ and let its image $\Gamma_2 = T(\Gamma_1)$ be determined by $(w_1, w_2, w_3)$ where $w_k = T(z_k)$.

We know the cross-ratio is invariant under Möbius transformations: $(z, z_1, z_2, z_3) = (T(z), w_1, w_2, w_3)$

Let z be a point on the Right Side of $\Gamma_1$. By definition, $\text{Im}(z, z_1, z_2, z_3) > 0$. Because of the previous equality, the image point $T(z)$ satisfies: $\text{Im}(T(z), w_1, w_2, w_3) > 0$. Since the condition defining the “Right Side” is identical for the pre-image and the image, $T$ maps the Right Side of $\Gamma_1$ to the Right Side of $\Gamma_2$.

Example: Mapping the Disk to the Lower Half-Plane

The map $T(z) = \frac{(1+i)(z+1)}{z+i} = \frac{(1+i)z + (1+i)}{z+i}$ is a Möbius transformation that maps $i \to 1, -1 \to 0, -i \to \infty$. In other words, it sends the unit circle to the real line.

1. Define the Orientation (Source):
The trip $i \to -1 \to -i$ is a counter-clockwise walk along the unit circle.
Imagine you are a tiny person walking on the unit circle in the complex plane. You start at the top (i), walk to the left edge (−1), and continue to the bottom (−i). Because of the order of these points, you are walking counter-clockwise. If you stick your left arm out while walking counter-clockwise around a circle, your arm points toward the inside of the circle, the Unit Circle.

2. Find the Image Orientation (Target):
Calculate the images $w_k = T(z_k)$:
$T(i) = \frac{(1+i)(i+1)}{i+i} = \frac{(1+i)^2}{2i} = \frac{2i}{2i} = 1, T(-1) = \frac{(1+i)(0)}{-1+i} = 0, T(-i) = \frac{(1+i)(-i+1)}{-i+i} = \frac{(1+i)(-i+1)}{0} = \infty$

3. Determine the “Left Side” in the Target:
Now, teleport to the target plane (the w-plane). Your footprints must land in the exact same sequence:
$1 \to 0$. You are standing at 1 on the real number line and walk to 0. You are facing “backwards” (pointing towards the negative numbers).
$0 \to \infty$. In the extended complex plane, the real line is actually a giant circle that meets itself at a single point called ∞ (infinity). Because you are already facing left and moving left, to get from 0 to ∞, you just keep walking left! You walk past −1, −100 ,−1,000,000 all the way to −∞, which is the exact same destination as +∞.
So, your path is: Start at 1, walk left to 0, and keep walking left forever.
As you take this walk (1→0→left forever), stick your left arm out again. Because you are walking left along the horizontal axis, your left arm is pointing straight down. Therefore, the target region is the Lower Half-Plane.

4. The Sanity Check.
Because infinity can fry your brain, it is always a good idea to test a random point.
If the Orientation Principle works, any point inside the unit circle should map to a point inside the lower half-plane.
Let’s test the center point 0 which is obviously inside the circle. $T(0) = \frac{1+i}{i} = \frac{1}{i} + 1 = -i + 1 = 1 - i$.
The imaginary part is −1, which means it sits below the real axis. It successfully landed in the Lower Half-Plane, confirming our walking logic was perfectly correct!