Symmetry with Respect to a Line and Circle

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

Definition. Two points $z,z^*\in \mathbb{C}$ are symmetric with respect to a line L if the line through z and $z^*$ is perpendicular to L and L passes through the midpoint of the segment joining z and $z^*$. 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 line connects you and your reflection perpendicularly, and the mirror cuts that line in half.

For a circle with center O and radius R, two points $z$ and $z^*$ are symmetric if they lie on the same ray from the center and their distances to the center satisfy $|z - O| \cdot |z^* - O| = R^2$. 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:

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

Definition. 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 $(z^*, z_2, z_3, z_4) = \overline{(z, z_2, z_3, z_4)}$.

The cross-ratio maps $\Gamma$ to the real line $\mathbb{R}$. On the real line, symmetry is just complex conjugation, $x+iy \to x-iy$. This definition forces $z$ and $z^*$ to mirror each other relative to that real line.

Symmetry principle. If a Möbius transformation $T$ takes a generalized circle $\Gamma_1$ onto a 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.

Let $z, z^*$ be symmetric with respect to $\Gamma_1$, $z_2, z_3, z_4$ be three distinct points on the source circle $\Gamma_1$ and $w_k = T(z_k)$ for $k=2,3,4$. Since $T$ takes a generalized circle $\Gamma_1$ onto a generalized circle $\Gamma_2$, these points $w_k$ lie on the target circle $\Gamma_2$.

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: $(T(z^*), w_2, w_3, w_4) = \overline{(T(z), w_2, w_3, w_4)}$

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

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 walk: start at $z_1$, walk towards $z_2$, then to $z_3$. As you walk, one region is on your left and the other is on your right.

Definition. For an oriented circle $\Gamma$ determined by three distinct points $(z_1, z_2, z_3)$, the right side is the set of points $z$ 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: $\text{Im}(z, z_1, z_2, z_3) < 0$.

Orientation principle. Let $\Gamma_1$ and $\Gamma_2$ be two circles in $\mathbb{C} \cup \{ \infty \}$ and let T be a Möbius transformation such that $T(\Gamma_1) = T(\Gamma_2)$. 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:

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.

$T(i) = \frac{(1+i)(i+1)}{i+i} = \frac{(1+i)^2}{2i} = \frac{2i}{2i} = 1, T(-1) = \frac{(1+i)(0)}{-1+i} = 0, T(-i) = \frac{(1+i)(-i+1)}{-i+i} = \frac{(1+i)(-i+1)}{0} = \infty$

Source. The trip $i \to -1 \to -i$ is a counter-clockwise walk along the unit circle. If you walk counter-clockwise along a circle, its interior (the disk) is on the left, Left = Disk (Interior) Unit Circle.

Image. The trip $1 \to 0 \to \infty$ is a walk along the real axis moving to the left (from positive to negative, real line backwards). If we are standing on 1 and start walking towards 0, the left hand points down (negative imaginary axis, $\text{Im}(w) < 0$), Left = 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 point 1-i has a negative imaginary part. It is indeed in the Lower Half Plane.