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Section3.2Inversion

Inversion offers a way to reflect points across a circle. This transformation plays a central role in visualizing the transformations of non-Euclidean geometry, and this section is the foundation of much of what follows.

Suppose \(C\) is a circle with radius \(r\) and center \(z_0\text{.}\) Inversion in the circle \(C\) sends a point \(z \neq z_0\) to the point \(z^*\) defined as follows: First, construct the ray from \(z_0\) through \(z\text{.}\) Then, let \(z^*\) be the unique point on this ray that satisfies the equation \begin{equation*} |z-z_0|\cdot|z^*-z_0| = r^2. \end{equation*} The point \(z^*\) is called the symmetric point to \(z\) with respect to \(C\text{.}\)

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Inversion in a circle centered at \(z_0\) is a transformation on the set \(\mathbb{C}-{z_0}\text{.}\) We usually denote inversion in the circle \(C\) by by \(i_C(z) = z^*\text{.}\) In the next section we will discuss how to extend this transformation in a way to include the center \(z_0\text{.}\)

You will work through several features of circle inversions in the exercises, including how to construct symmetry points with compass and ruler (see Figure 3.2.18). We note here that \(i_C\) fixes all the points on the circle \(C\text{,}\) and points inside the circle get mapped to points outside the circle and vice versa. The closer \(z\) gets to the center of the circle, the further \(i_C(z)\) gets from the circle.

Inversion in the circle \(C\) centered at \(z_0\) with radius \(r\)

Inversion in the circle \(C\) centered at \(z_0\) with radius \(r\) is given by \begin{align*} i_C(z) \amp = \frac{r^2}{(\overline{z-z_0})} + z_0. \end{align*} The formula can be obtained by first considering inversion in the unit circle (see Example 3.2.1) and composing this with some general linear transformations. The details are left to Exercise 3.2.1.

Example3.2.1Inversion in the unit circle

The unit circle in \(\mathbb{C}\text{,}\) denoted \(\mathbb{S}^1\text{,}\) is the circle with center \(z_0 = 0\) and radius \(r = 1\text{.}\) The equation for the point \(z^*\) symmetric to a point \(z \neq 0\) thus reduces from \(|z-z_0|\cdot|z^*-z_0| = r^2\) to \begin{equation*} |z|\cdot|z^*| = 1. \end{equation*}

Moreover, \(z^*\) is just a scaled version of \(z\) since they are on the same ray through the origin. That is, \(z^* = kz\) for some positive real number \(k\text{.}\) Take the absolute value of both sides of this equation to obtain \(k = |z^*|/|z|\text{.}\) Since \(|z^*|=1/|z|\) by the symmetry point equation, we see that \(k = 1/|z|^2\text{,}\) so \(z^*=(1/|z|^2)z\text{.}\) Moreover, \(|z|^2 = z \cdot \overline{z}\text{,}\) so inversion in the unit circle \(\mathbb{S}^1\) may be written as \begin{equation*} i_{\mathbb{S}^1}(z) = 1/\overline{z}. \end{equation*}

Example3.2.2Inverting some figures in a circle

Below we have inverted a circle, the letter ‘M,’ and a small grid across the circle \(C\text{.}\) It looks as if the image of the circle is another circle, which we will soon prove is indeed the case. We will also prove that lines not intersecting the center \(z_0\) of the circle of inversion also get inverted into circles. It follows that the line segments in the ‘M’ get mapped to arcs of circles.

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As Example 3.2.2 suggests, the distinction between lines and circles gets muddied a bit by inversion. A line can get mapped to a circle and vice versa. In what follows, it will be helpful to view reflection in a line and inversion in a circle as special cases of the same general map. To arrive at this view we first make lines and circles special cases of the same general type of figure.

Definition3.2.3

A cline is a Euclidean circle or line. Any cline can be described algebraically by an equation of the form \begin{equation*} cz\overline{z} + \alpha z + \overline{\alpha}\overline{z} + d = 0 \end{equation*} where \(z = x + yi\) is a complex variable, \(\alpha\) is a complex constant, and \(c, d\) are real numbers. If \(c = 0\) the equation describes a line, and if \(c \neq 0\) and \(|\alpha|^2 > cd\) the equation describes a circle.

The word “cline” (pronounced ‘Klein’) might seem a bit forced, but it represents the shift in thinking we aim to achieve. We need to start thinking of lines and circles as different manifestations of the same general class of objects. What class? The class of clines.

Letting \(\alpha = a + bi\) and \(z = x + yi\text{,}\) the cline equation \(cz\overline{z} + \alpha z + \overline{\alpha} \overline{z} + d = 0\) can be written as \begin{equation*} c(x^2+y^2)+[ax - by + (ay+bx)i]+[ax-by - (ay+bx)i] + d = 0 \end{equation*} which simplifies to \begin{equation*} c(x^2+y^2) + 2(ax - by) + d = 0. \end{equation*}

If \(c = 0\) then we have the equation of a line, and if \(c \neq 0\) we have the equation of a circle, so long as \(a^2 + b^2 > cd\text{.}\) In this case, the equation can be put into standard form by completing the square. Let's run through this.

If \(c \neq 0\text{,}\) \begin{align*} c(x^2+y^2) + 2ax - 2by + d \amp = 0\\ x^2 + \frac{2a}{c}x + y^2 - \frac{2b}{c}y \amp = -\frac{d}{c}\\ x^2 + \frac{2a}{c}x + \bigg(\frac{a}{c}\bigg)^2 + y^2 - \frac{2b}{c}y + \bigg(\frac{b}{c}\bigg)^2 \amp = -\frac{d}{c}+\bigg(\frac{a}{c}\bigg)^2+\bigg(\frac{b}{c}\bigg)^2\\ \bigg(x+\frac{a}{c}\bigg)^2+\bigg(y-\frac{b}{c}\bigg)^2\amp =\frac{ a^2 + b^2-cd}{c^2} \end{align*} and we have the equation of a circle so long as the right-hand side (the radius term) is positive. In other words, we have the equation of a circle so long as \(a^2 + b^2>cd\text{.}\)

The cline equation

To summarize, if \(c \neq 0\text{,}\) the cline equation \begin{equation*} cz\overline{z} + \alpha z + \overline{\alpha} \overline{z} + d = 0 \end{equation*} gives a circle with center \(z_0\) and radius \(r\text{,}\) where \begin{equation*} z_0=\bigg(-\frac{\text{Re}(\alpha)}{c}, \frac{\text{Im}(\alpha)}{c}\bigg)~~~\text{and}~~~ r = \sqrt{\frac{|\alpha|^2-cd}{c^2}}, \end{equation*} so long as \(|\alpha|^2 > cd\text{.}\) If \(c = 0\text{,}\) the cline equation gives a line.

From now on, if you read the phrase “inversion in a cline,” know that this means inversion in a circle or reflection about a line, and if someone hands you a cline \(C\text{,}\) you might say, “Thanks! By the way, is this a line or a circle?”

We note here the construction of a cline through three points in \(\mathbb{C}\text{.}\) This construction is used often in later chapters to generate figures in non-Euclidean geoemtry.

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Figure3.2.5Constructing the unique circle through three points.

We will call two clines orthogonal if they intersect at right angles. For instance, a line is orthogonal to a circle if and only if it goes through the center of the circle. One very important feature of inversion in \(C\) is that clines orthogonal to \(C\) get inverted to themselves. To prove this fact, we first prove the following result, which can be found in Euclid's Elements (Book III, Proposition 36).

We note that the quantity \(s^2-r^2\) in the previous lemma is often called the power of the point \(p\) with respect to the circle \(C\text{.}\) That is, if circle \(C\) has radius \(r\) and a point \(p\) is a distance \(s\) from the center of \(C\) then the quantity \(s^2-r^2\) is called the power of the point \(p\text{.}\)

Another important feature of inversion in a cline is that it preserves symmetry points.

We close the section with two applications of inversion.

As we let \(k\) run through all positive real numbers, we obtain a family of clines, called the circles of Apollonius of the points \(\boldsymbol{p}\) and \(\boldsymbol{q}\). We note that \(p\) and \(q\) are symmetric with respect to each cline in this family (see Exercise 3.3.2).

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Figure3.2.15Finding two points symmetric with respect to a line and circle.

SubsectionExercises

1

Prove the general formula for inversion in a circle \(C\) centered at \(z_0\) with radius \(r\text{.}\) In particular, show in this case that \begin{equation*} i_C(z) = \frac{r^2}{(\overline{z-z_0})} + z_0. \end{equation*}

2

Constructing the symmetric point to \(z\) when \(z\) is inside the circle of inversion.

Prove that for a point \(z\) inside the circle \(C\) with center \(z_0\) (Figure 3.2.18(a)), the following construction finds the symmetry point of \(z\text{.}\) (1) Draw the ray from \(z_0\) through \(z\text{.}\) (2) Construct the perpendicular to this ray at \(z\text{.}\) Let \(t\) be a point of intersection of this perpendicular and \(C\text{.}\) (3) Construct the radius \(z_0t\text{.}\) (4) Construct the perpendicular to this radius at \(t\text{.}\) The symmetric point \(z^*\) is the point of intersection of this perpendicular and ray \(\overrightarrow{z_oz}\text{.}\)

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Figure3.2.18Constructing the symmetric point (a) if \(z\) is inside the circle of inversion; (b) if \(z\) is outside the circle of inversion.
3

Constructing the symmetric point to \(z\) when \(z\) is outside the circle of inversion.

Prove that for a point \(z\) outside the circle \(C\) with center \(z_0\) (Figure 3.2.18(b)), the following construction finds the symmetry point of \(z\text{.}\) (1) Construct the circle having diameter \(z_0z\text{.}\) Let \(t\) be a point of intersection of the two circles. (2) Construct the perpendicular to \(z_0z\) through \(t\text{.}\) Let \(z^*\) be the intersection of this perpendicular with segment \(z_0z\text{.}\)

4

Suppose \(T_1\) is inversion in the circle \(|z| = r_1\text{,}\) and \(T_2\) is inversion in the circle \(|z| = r_2\text{,}\) where \(r_1, r_2 > 0\text{.}\) Prove that \(T_2 \circ T_1\) is a dilation. Conversely, show any dilation is the composition of two inversions.

5

Determine the image of the line \(y = mx + b\) (when \(b \neq 0)\) under inversion in the unit circle. In particular, show that the image is a circle with center \((-m/2b, 1/2b)\) and radius \(\sqrt{(m^2 + 1)/4b^2}\text{.}\) Hint: Refer to Exercise 2.4.1.

6

Determine the image of the line \(L\) given by \(y = 3x + 4\) under inversion in the unit circle. Give a careful plot of the unit circle, the line \(L\text{,}\) and the image of \(L\) under the inversion.

7

Prove that inversion in the unit circle maps the circle \((x-a)^2 + (y-b)^2 = r^2\) to the circle \begin{equation*} \bigg(x-\frac{a}{d}\bigg)^2 + \bigg(y-\frac{b}{d}\bigg)^2 = \bigg(\frac{r}{d}\bigg)^2 \end{equation*} where \(d = a^2+b^2-r^2\text{,}\) provided that \(d \neq 0\text{.}\)

8

Determine in standard form the image of the circle \(C\) given by \((x-1)^2 + y^2 = 4\) under inversion in the unit circle. Give a careful plot of the unit circle, the circle \(C\text{,}\) and the image of \(C\) under the inversion.

9

True or False? If a circle \(C\) gets mapped to another circle under inversion in the unit circle, then the center of the \(C\) gets mapped to the center of the image circle, \(i_{\mathbb{S}^1}(C)\text{.}\) If the statement is true, prove it; if it is false, provide a counterexample.

10

Suppose \(C\) and \(D\) are orthogonal circles. Corollary 3.2.9 tells us that inversion in \(C\) maps \(D\) to itself. Prove that this inversion also takes the interior of \(D\) to itself.

11

Finish the proof of Theorem 3.2.10 by showing that the angle of intersection at \(z^*\) equals the angle of intersection at \(z\) in Figure 3.2.11.

12

Suppose \(C\) is the circle \(|z - z_0| = r\) and \(C^\prime\) is the circle \(|z - z_0| = r^\prime\text{.}\) Find the stretch factor \(k\) in the dilation \(S(z) = k(z-z_0) + z_0\) so that \(i_C = S \circ i_{C^\prime}.\)

13

Complete the proof of Lemma 3.2.7 by proving the case in which the line through \(p\) passes through the center of \(C\text{.}\)