11.8: Reflection and symmetry

Reflection and symmetry in a line

Example \(\PageIndex{1}\)

Suppose we have a line \(S\) and a point \(z_1\) not on \(S\). The reflection of \(z_1\) in \(S\) is the point \(z_2\) so that \(S\) is the perpendicular bisector to the line segment \(\overline{z_1 z_2}\). Since there is exactly one such point \(z_2\), the reflection of a point in a line is unique.

Definition

If \(z_2\) is the reflection of \(z_1\) in \(S\), we say that \(z_1\) and \(z_2\) are symmetric with respect to the line \(S\).

In the figure below the points \(z_1\) and \(z_2\) are symmetric in the \(x\)-axis. The points \(z_3\) and \(z_4\) are symmetric in the line \(S\).

In order to define the reflection of a point in a circle we need to work a little harder. Looking back at the previous example we can show the following.

Fact

If \(z_1\) and \(z_2\) are symmetric in the line \(S\), then any circle through \(z_1\) and \(z_2\) intersects \(S\) orthogonally.

Proof

Call the circle \(C\). Since \(S\) is the perpendicular bisector of a chord of \(C\), the center of \(C\) lies on \(S\). Therefore \(S\) is a radial line, i.e. it intersects \(C\) orthogonally.

Circles through symmetric points intersect the line at right angles.

Reflection and symmetry in a circle

We will adapt this for our definition of reflection in a circle. So that the logic flows correctly we need to start with the definition of symmetric pairs of points.

Definition

Suppose \(S\) is a line or circle. A pair of points \(z_1, z_2\) is called symmetric with respect to \(S\) if every line or circle through the two points intersects \(S\) orthogonally.

First we state an almost trivial fact.

Fact

Fractional linear transformations preserve symmetry. That is, if \(z_1, z_2\) are symmetric in a line or circle \(S\), then, for an FLT \(T\), \(T(z_1)\) and \(T(z_2)\) are symmetric in \(T(S)\).

Proof

The definition of symmetry is in terms of lines and circles, and angles. Fractional linear transformations map lines and circles to lines and circles and, being conformal, preserve angles.

Theorem \(\PageIndex{1}\)

Suppose \(S\) is a line or circle and \(z_1\) a point not on \(S\). There is a unique point \(z_2\) such that the pair \(z_1, z_2\) is symmetric in \(S\).

Proof

Let \(T\) be a fractional linear transformation that maps \(S\) to a line. We know that \(w_1 = T(z_1)\) has a unique reflection \(w_2\) in this line. Since \(T^{-1}\) preserves symmetry, \(z_1\) and \(z_2 = T^{-1} (w_2)\) are symmetric in \(S\). Since \(w_2\) is the unique point symmetric to \(w_1\) the same is true for \(z_2\) vis-a-vis \(z_1\). This is all shown in the figure below.

We can now define reflection in a circle.

Definition

The point \(z_2\) in the Theorem 11.8.1 is called the reflection of \(z_1\) in \(S\).

Reflection in the unit circle

Using the symmetry preserving feature of fractional linear transformations, we start with a line and transform to the circle. Let \(R\) be the real axis and \(C\) the unit circle. We know the FLT

\[T(z) = \dfrac{z - i}{z + i} \nonumber \]

maps \(R\) to \(C\). We also know that the points \(z\) and \(\overline{z}\) are symmetric in \(R\). Therefore

\[w_1 = T(z) = \dfrac{z - i}{z + i} \ \ \ \text{and} \ \ \ w_2 = T(\overline{z}) = \dfrac{\overline{z} - i}{\overline{z} + i} \nonumber \]

are symmetric in \(D\). Looking at the formulas, it is clear that \(w_2 = 1/\overline{w_1}\). This is important enough that we highlight it as a theorem.

Theorem \(\PageIndex{2}\) Reflection in the unit circle

The reflection of \(z = x + iy = re^{i \theta}\) in the unit circle is

\[\dfrac{1}{\overline{z}} = \dfrac{z}{|z|^2} = \dfrac{x + iy}{x^2 + y^2} = \dfrac{e^{i \theta}}{r}. \nonumber \]

The calculations from \(1/\overline{z}\) are all trivial.

Note

1. It is possible, but more tedious and less insightful, to arrive at this theorem by direct calcula- tion.
2. If \(z\) is on the unit circle then \(1/\overline{z} = z\). That is, \(z\) is its own reflection in the unit circle –as it should be.
3. The center of the circle 0 is symmetric to the point at \(\infty\).

The figure below shows three pairs of points symmetric in the unit circle:

\(z_1 = 2; \ w_1 = \dfrac{1}{2}, \ z_2 = 1 + i; \ w_2 = \dfrac{1 + i}{2},\ z_3 = -2 + i; \ w_3 = \dfrac{-2 + i}{5}.\)

Pairs of points \(z_j\): \(w_j\) symmetric in the unit circle.

Example \(\PageIndex{2}\) Reflection in the circle of radius \(R\)

Suppose \(S\) is the circle \(|z| = R\) and \(z_1\) is a pint not on \(S\). Find the reflection of \(z_1\) in S.

Solution

Our strategy is to map \(S\) to the unit circle, find the reflection and then map the unit circle back to \(S\).

Start with the map \(T(z) = w = z/R\). Clearly \(T\) maps \(S\) to the unit circle and

\[w_1 = T(z_1) = z_1/R. \nonumber \]

The reflection of \(w_1\) is

\[w_2 = 1/\overline{w_1} = R/\overline{z}_1. \nonumber \]

Mapping back from the unit circle by \(T^{-1}\) we have

\[z_2 = T^{-1} (w_2) = Rw_2 = R^2/\overline{z}_1. \nonumber \]

Therefore the reflection of \(z_1\) is \(R^2/\overline{z}_1.\)

Here are three pairs of points symmetric in the circle of radius 2. Note, that this is the same figure as the one above with everything doubled.

\(z_1 = 4;\ w_1 = 1,\ z_2 = 2 + 2i;\ w_2 = 1 + i, \ z_3 = -4 + 2i;\ w_3 = \dfrac{-4 + 2i}{5}.\)

Pairs of points \(z_j\); \(w_j\) symmetric in the circle of radius 2.

Example \(\PageIndex{3}\)

Find the reflection of \(z_1\) in the circle of radius \(R\) centered at \(c\).

Solution

Let \(T(z) = (z - c)/R\). \(T\) maps the circle centered at \(c\) to the unit circle. The inverse map is

\[T^{-1} (w) = Rw + c. \nonumber \]

So, the reflection of \(z_1\) is given by mapping \(z\) to \(T(z)\), reflecting this in the unit circle, and mapping back to the original geometry with \(T^{-1}\). That is, the reflection \(z_2\) is

\[z_1 \to \dfrac{z_1 - c}{R} \to \dfrac{R}{\overline{z_1 - c}} \to z_2 = \dfrac{R^2}{\overline{z_1 - c}} + c. \nonumber \]

We can now record the following important fact.

(Reflection of the center)

For a circle \(S\) with center \(c\) the pair \(c\), \(\infty\) is symmetric with respect to the circle.

Proof

This is an immediate consequence of the formula for the reflection of a point in a circle. For example, the reflection of \(z\) in the unit circle is \(1/\overline{z}\). So, the reflection of 0 is infinity.

Example \(\PageIndex{4}\)

Show that if a circle and a line don’t intersect then there is a pair of points \(z_1, z_2\) that is symmetric with respect to both the line and circle.

Solution

By shifting, scaling and rotating we can find a fractional linear transformation \(T\) that maps the circle and line to the following configuration: The circle is mapped to the unit circle and the line to the vertical line \(x = a > 1\).

For any real \(r\), \(w_1 = r\) and \(w_2 = 1/r\) are symmetric in the unit circle. We can choose a specific \(r\) so that \(r\) and \(1/r\) are equidistant from \(a\), i.e. also symmetric in the line \(x = a\). It is clear geometrically that this can be done. Algebraically we solve the equation

\[\dfrac{r + 1/r}{2} = a \ \ \Rightarrow \ \ r^2 - 2ar + 1 = 0 \ \ \Rightarrow \ \ r = a + \sqrt{a^2 - 1} \ \ \Rightarrow \ \ \dfrac{1}{r} = a - \sqrt{a^2 - 1}. \nonumber \]

Thus \(z_1 = T^{-1} (a + \sqrt{a^2 - 1})\) and \(z_2 = T^{-1} (a - \sqrt{a^2 - 1})\) are the required points.

Example \(\PageIndex{5}\)

Show that if two circles don’t intersect then there is a pair of points \(z_1, z_2\) that is symmetric with respect to both circles.

Solution

Using a fractional linear transformation that maps one of the circles to a line (and the other to a circle) we can reduce the problem to that in the previous example.

Example \(\PageIndex{6}\)

Show that any two circles that don’t intersect can be mapped conformally to con- centric circles.

Solution

Call the circles \(S_1\) and \(S_2\). Using the previous example start with a pair of points \(z_1, z_2\) which are symmetric in both circles. Next, pick a fractional linear transformation \(T\) that maps \(z_1\) to 0 and \(z_2\) to infinity. For example,

\[T(z) = \dfrac{z - z_1}{z - z_2}. \nonumber \]

Since \(T\) preserves symmetry 0 and \(\infty\) are symmetric in the circle \(T(S_1)\). This implies that 0 is the center of \(T(S_1)\). Likewise 0 is the center of \(T(S_2)\). Thus, \(T(S_1)\) and \(T(S_2)\) are concentric.