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Section3.5Möbius Transformations: A Closer Look

To visualize Möbius transformations it is helpful to focus on fixed points and, in the case of two fixed points, on two families of clines with respect to these points.

Given two points \(p\) and \(q\) in \(\mathbb{C}^+\text{,}\) a type I cline of \(\boldsymbol{p}\) and \(\boldsymbol{q}\) is a cline that goes through \(p\) and \(q\text{,}\) and a type II cline of \(\boldsymbol{p}\) and \(\boldsymbol{q}\) is a cline with respect to which \(p\) and \(q\) are symmetric. Type II clines are also called circles of Apollonius (see Exercise 3.3.2). Figure 3.5.1 shows some type I and type II clines of \(p\) and \(q\text{.}\) The type II clines of \(p\) and \(q\) are dashed.

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Figure3.5.1Type I clines (solid) and Type II clines (dashed) of \(p\) and \(q\text{.}\)

By Theorem 3.2.8, any type II cline of \(p\) and \(q\) intersects any type I cline of \(p\) and \(q\) at right angles. Furthermore, because Möbius transformations preserve clines and symmetry points, we can be assured that Möbius transformations preserve type I clines as well as type II clines. In particular, if \(C\) is a type I cline of \(p\) and \(q\text{,}\) then \(T(C)\) is a type I cline of \(T(p)\) and \(T(q)\text{.}\) Similarly, if \(C\) is a type II cline of \(p\) and \(q\text{,}\) then \(T(C)\) is a type II cline of \(T(p)\) and \(T(q)\text{.}\) We can use this to our advantage.

For instance, the type I clines of the points 0 and \(\infty\) are, precisely, lines through the origin, while the type II clines of 0 and \(\infty\) are circles centered at the origin. (Remember, inversion in a circle takes the center of the circle to \(\infty\text{.}\)) The type I clines in this case are clearly perpendicular to the type II clines, and they combine to create a coordinate system of the plane (polar coordinates), as demonstrated in Figure 3.5.2(a).

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Figure3.5.2(a) Type I clines (solid) and type II clines (dashed) of points 0 and \(\infty\) (solid); (b) A Möbius transformation sending \(0 \mapsto p\) and \(\infty \mapsto q\) sends type I and II clines of 0 and \(\infty\) to type I and II clines of \(p\) and \(q\text{,}\) respectively.

We can move this system of clines by considering a Möbius transformation that maps 0 to \(p\) and \(\infty\) to \(q\) (where \(p, q \neq \infty\)). Lines through the origin get mapped to type I clines of \(p\) and \(q\text{,}\) and circles through the origin get mapped to type II clines of \(p\) and \(q\text{.}\) The result is a system of clines that serves as a general coordinate system for the plane. Each point \(z\) in the plane is at the intersection of a single type I cline of \(p\) and \(q\) and a single type II cline of \(p\) and \(q\text{,}\) and these two clines intersect at right angles.

Let's get back to fixed points and how they can help us describe Möbius transformations. We consider the case that 0 and \(\infty\) are fixed before proceeding to the general case.

Example3.5.3Möbius transformations that fix 0 and \(\infty\)

Suppose \(T(z) = (az+b)/(cz+d)\) is a Möbius transformation that fixes 0 and \(\infty\text{.}\) In this case, the form of the Möbius transformation can be simplified. In particular, since \(T(0) = 0\text{,}\) it follows that \(b=0\text{.}\) And since \(T(\infty)=\infty\text{,}\) it follows that \(c = 0\text{.}\) Thus, \(T(z) = \frac{a}{d}z\) which may be written as \begin{equation*} T(z) = re^{i\theta}z. \end{equation*}

With \(T\) in this form, it is clear that if \(T\) fixes 0 and \(\infty\text{,}\) then \(T\) is a combination of a dilation (by factor \(r\)) and a rotation about the origin (by a factor \(\theta\)). We may assume that \(r > 0\) in the above equation, because if it is negative, we can turn it into a positive constant by adding \(\pi\) to the angle of rotation.

A dilation by \(r\) will push points along lines through the origin. These lines are precisely the type I clines of 0 and \(\infty\text{.}\) All points in the plane either head toward \(\infty\) (if \(r > 1\)) or they all head toward 0 (if \(0 \lt r \lt 1\)). Of course, if \(r = 1\) there is no dilation.

Meanwhile, rotation about 0 by \(\theta\) pushes points along circles centered at the origin. These circles are precisely the type II clines of 0 and \(\infty\text{.}\)

Now suppose \(T\) is a Möbius transformation that fixes two finite points \(p\) and \(q\) (neither is \(\infty\)). Let \begin{equation*} S(z) = \frac{z - p}{z-q} \end{equation*} be a Möbius transformation that takes \(p\) to 0 and \(q\) to \(\infty\text{.}\) Let \(U\) be the Möbius transformation determined by the composition equation \begin{equation*} U = S \circ T \circ S^{-1} \tag{1} \end{equation*}

Notice \begin{equation*} U(0) = S \circ T \circ S^{-1}(0)= S\circ T(p) = S(p) = 0, \end{equation*} and \begin{equation*} U(\infty) = S \circ T \circ S^{-1}(\infty) = S \circ T(q) = S(q) = \infty. \end{equation*}

That is, \(U\) is a Möbius transformation that fixes 0 and \(\infty\text{.}\) So, by Example 3.5.3, \(U\) is a rotation, a dilation, or some combination of those, and \(U\) looks like \(U(z) = re^{i\theta}z\text{.}\)

In any event, focusing on \(T\) again and using equation (1), which can be rewritten as \(S \circ T = U \circ S\text{,}\) we arrive at the following equation, called the normal form of the Möbius transformation in this case.

Normal form, two fixed points

The normal form of a Möbius transformation \(T\) fixing distinct points \(p\) and \(q\) (neither of which is \(\infty\)): \begin{equation*} \frac{T(z) - p}{T(z) - q} = re^{i\theta} \cdot \frac{z - p}{z -q} \end{equation*}

This normal form is much more illuminating than the standard \(a,b,c,d\) form because, although the map is still described in terms of four constants (\(p,q,r,\theta\)), each constant now has a simple geometric interpretation: \(p\) and \(q\) are fixed points, \(r\) is a dilation factor along type I clines of \(p\) and \(q\text{,}\) and \(\theta\) is a rotation factor around type II clines of \(p\) and \(q\text{.}\)

In particular, thanks to composition equation (1) we can view \(T\) as the composition \(T = S^{-1}\circ U \circ S.\) With this view, \(T\) moves points according to a three-leg journey. Think of a general point \(z\) clinging to the intersection of a single type II cline of \(p\) and \(q\) and a single type I cline of \(p\) and \(q\) (see Figure 3.5.4).

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Figure3.5.4Tracking the image of \(z\) if \(T\) fixes \(p\) and \(q\text{.}\)

First, \(z\) gets sent via \(S\) to \(S(z)\text{,}\) which is at the intersection of a line through the origin and a circle centered at the origin. Second, \(U\) (which has the form \(U(z) = re^{i\theta} z\)), sends \(S(z)\) along this line through the origin (by dilation factor \(r\)), and then around a new circle centered at the origin (by rotation factor \(\theta\)) to the point \(U(S(z))\text{.}\) Third, \(S^{-1}\) sends \(U(S(z))\) back to the intersection of a type I cline of \(p\) and \(q\) and a type II cline of \(p\) and \(q\text{.}\) This point of intersection is \(S^{-1}(U(S(z)))\) and is equivalent to \(T(z)\text{.}\)

Though fatigued, our well-traveled point realizes there's a shortcut. Why go through this complicated wash? We can understand \(T\) as follows: \(T\) will push points along type I clines of \(p\) and \(q\) (according to the dilation factor \(r\)) and along type II clines of \(p\) and \(q\) (according to the rotation factor \(\theta\)).

We emphasize two special cases of this normal form. If \(|re^{i\theta}| = 1\) there is no dilation, and points simply get rotated about type II clines of \(p\) and \(q\) as in Figure 3.5.5. Such a Möbius transformation is called an elliptic Möbius transformation.

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Figure3.5.5An elliptic Möbius transformation fixing \(p\) and \(q\) swirls points around type II clines of \(p\) and \(q\text{.}\)

The second special case occurs when \(\theta = 0\text{.}\) Here we have a dilation factor \(r\text{,}\) but no rotation. All points move along type I clines of \(p\) and \(q\text{,}\) as in Figure 3.5.6. A Möbius transformation of this variety is called a hyperbolic Möbius transformation. A hyperbolic Möbius transformation fixing \(p\) and \(q\) either sends all points away from \(p\) and toward \(q\) or vice versa, depending on the value of \(r\text{.}\)

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Figure3.5.6A hyperbolic Möbius transformation fixing \(p\) and \(q\) pushes points away from one fixed point and toward the other along type I clines of \(p\) and \(q\text{.}\)

If we are not in one of these special cases, then \(T\) is simply a combination of these two, and a Möbius transformation of this type is often called loxodromic.

If a Möbius transformation fixes two finite points, say \(p\) and \(q\text{,}\) and it is not the identity transformation, then some finite point gets sent to \(\infty\text{.}\) Moreover, \(\infty\) gets sent to some finite point. The point sent to infinity is called the pole of the transformation and is often denoted \(z_\infty\text{.}\) That is, \(T(z_\infty) = \infty\text{.}\) The inverse pole of \(\boldsymbol{T}\) is the image of \(\infty\) under the map, which is often denoted as \(w_\infty\text{.}\) That is, \(T(\infty) = w_\infty\text{.}\) There is a simple relationship between the four points \(p,q,z_\infty,\) and \(w_\infty\text{.}\)

Example3.5.9Find a Möbius transformation that fixes -1 and 1, and send \(i\) to \(\infty\)

By Lemma 3.5.7, the Möbius transformation sends \(\infty\) to \(-i\text{,}\) so by Theorem 3.5.8, \begin{equation*} T(z) = \frac{-iz-(1)(-1)}{z-i} = \frac{-iz+1}{z-i}. \end{equation*}

Example3.5.10Analyze and classify a Möbius transformation

Consider the Möbius transformation \begin{equation*} T(z) = \frac{(6+3i)z+(2-3i)}{z+3}. \end{equation*}

First we find the fixed points and the normal form of \(T\text{.}\) To find the fixed points we solve \(T(z) = z\) for \(z\text{.}\) \begin{align*} \frac{(6+3i)z+(2-3i)}{z+3}\amp = z\\ (6+3i)z + (2-3i)\amp = z^2+3z\\ z^2 - (3+3i)z - (2-3i) \amp = 0. \end{align*}

Hey! Wait a moment! This looks familiar. Let's see \(\ldots\) yes! We showed in Example 2.4.4 that this quadratic equation has solutions \(z = i\) and \(z = 3+2i\text{.}\)

So the map has these two fixed points, and the normal form of \(T\) is \begin{equation*} \frac{T(z)-i}{T(z)-(3+2i)}=\lambda\frac{z-i}{z-(3+2i)}. \end{equation*}

To find the value of \(\lambda\text{,}\) plug into the normal form a convenient value of \(z\text{.}\) For instance, \(T(-3)=\infty\text{,}\) so \begin{equation*} 1=\lambda\frac{-3-i}{-3-(3+2i)}. \end{equation*}

It follows that \(\lambda = 2\text{,}\) so \(T\) is a hyperbolic map that pushes points along clines through \(i\) and \(3+2i\text{.}\) Below is a schematic for how the map pushes points around in \(\mathbb{C}^+\text{.}\) Notice \(T(0) =\frac{2}{3}-i\text{,}\) \(T(1)=2\text{,}\) and \(T(4i) = 2.16+4.12i\text{.}\) Points are moving along type I clines of \(i\) and \(3+2i\) away from \(i\) and toward \(3 + 2i\text{.}\)

From the original description of \(T\) we observe that the pole of the map is \(z_\infty = -3\text{,}\) and the inverse pole of the map is \(w_{\infty} = 6+3i\text{.}\) Notice that \(z_\infty\text{,}\) \(w_\infty\text{,}\) and the two fixed points all lie on the same Euclidean line. This will always be the case for a hyperbolic Möbius transformation (Exercise 3.5.9).

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Example3.5.11Fix \(i\) and \(0\) and send \(1 \mapsto 2\)

The normal form of this map is \begin{equation*} \frac{T(z)-i}{T(z)-0}=\lambda\frac{z-i}{z-0}. \end{equation*}

Since \(T(1) = 2\) we know that \begin{equation*} \frac{2-i}{2}=\lambda(1-i). \end{equation*}

Solving for \(\lambda\) we have \begin{equation*} \lambda = \frac{3}{4}+\frac{1}{4}i, \end{equation*} and the map is loxodromic.

Expressing \(\lambda\) in polar form, \(\lambda = re^{i\theta}\text{,}\) gives \(r = \frac{\sqrt{10}}{4}\) and \(\theta=\arctan(1/3)\text{.}\) So \(T\) pushes points along type I clines of \(i\) and \(0\) according to the scale factor \(r\) and along type II clines of \(i\) and \(0\) according to the angle \(\theta\text{.}\)

Now we consider Möbius transformations that fix just one point. One such Möbius transformation comes to mind immediately. For any complex number \(d\text{,}\) the translation \(T(z) = z + d\) fixes just \(\infty\text{.}\) In the exercises, you prove that translations are the only Möbius transformations that fix \(\infty\) (and no other point).

Now suppose \(T\) fixes \(p \neq \infty\) (and no other point). Let \(S(z) = \frac{1}{z-p}\) be a Möbius transformation taking \(p\) to \(\infty\text{,}\) and let \(U = S \circ T \circ S^{-1}\text{.}\) Then \(U(\infty) = S(T(S^{-1}(\infty)))=S(T(p))=S(p)=\infty\text{,}\) and \(U\) fixes no other point. Thus, \(U(z) = z + d\) for some complex constant \(d\text{.}\)

The composition equation \(S \circ T = U \circ S\) gives the following equation called the normal form of a Möbius transformation \(\boldsymbol{T}\) fixing \(p \neq \infty\) (and no other point):

Normal form, one fixed point \(p\neq\infty\)

\begin{equation*} \frac{1}{T(z) - p} = \frac{1}{z-p} + d \end{equation*}

Observe that \(U(z)=z+d\) pushes points along lines parallel to one another in the direction of \(d\) (as in the right of Figure 3.5.12). All of these parallel lines meet at \(\infty\) and are mutually tangent at this point. The map \(S^{-1}\) takes this system of clines to a system of clines that meet just at \(p\text{,}\) and are tangent to one another at \(p\text{,}\) as pictured. The slope of the single line in this system depends on the value of the constant \(d\text{.}\) In fact, the single line in the system of clines is the line through \(p\) and \(T(\infty)\) (see Exercise 3.5.12 for details).

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Figure3.5.12A parabolic map fixing \(p\) pushes points along clines that are mutually tangent at \(p\text{.}\)

A map that fixes just \(p\) will push points along such a system of clines that are mutually tangent at \(p\text{.}\) Such a map is called parabolic. In a sense, a parabolic map sends points both toward and away from \(p\) along these clines, just as any translation pushes points along a line toward \(\infty\) and also away from \(\infty\text{.}\)

Example3.5.13Find the normal form of \(T(z) = (7z-12)/(3z-5)\)

As always, we start with fixed points. \begin{align*} z\amp =T(z)\\ z(3z-5) \amp = 7z-12\\ 3z^2-12z + 12 \amp = 0\\ z^2-4z + 4 \amp = 0\\ (z-2)^2 \amp = 0\\ z\amp = 2. \end{align*}

So \(T\) is parabolic and has normal form \begin{equation*} \frac{1}{T(z)-2} = \frac{1}{z-2}+d. \end{equation*}

To find \(d\) plug in the image of another point. Using the original description of the map, we know \(T(0)=2.4\) so \begin{equation*} \frac{1}{0.4}=\frac{1}{-2}+d \end{equation*} so that \(d = 3\text{.}\) The normal form is then \begin{equation*} \frac{1}{T(z)-2} = \frac{1}{z-2}+3. \end{equation*}

SubsectionExercises

2

Analyze each of the Möbius transformations below by finding the fixed points, finding the normal form, and sketching the appropriate coordinate system of clines, being sure to indicate the motion of the transformation.

a. \(T(z) = \frac{z}{2z-1}\)

b. \(T(z) = \frac{-z}{(1+i)z - i}\text{.}\)

c. \(T(z) = \frac{3iz-5}{z-i}\text{.}\)

3

A transformation \(T\) is called an involution if it is its own inverse. If this is the case, then \(T\circ T\) is the identity transformation. Prove that if a Möbius transformation \(T\) is an involution and not the identity transformation, it must be elliptic.

4

Suppose a Möbius transformation \(T\) has the following property: There are three points \(a, b, c\) in the complex plane \(\mathbb{C}\) such that \(T(a) = b, T(b) = c, T(c) = a\text{.}\)

a. What is the image of the unique cline through \(a, b,\) and \(c\) under \(T\text{?}\)

b. Explain why the triple composition \(T\circ T \circ T\) is the identity transformation.

c. Prove that \(T\) is elliptic.

5

Prove that if the Möbius transformation \(T\) fixes just \(\infty\text{,}\) then \(T(z) = z + d\) for some complex constant \(d\text{.}\)

6

Find the Möbius transformation that fixes 2 and 4 and sends \(2+i\) to \(\infty\text{.}\)

7

Use the normal form to build and classify a Möbius transformation that fixes 4 and 8 and sends \(i\) to 0.

8

Suppose \(T\) is an elliptic Möbius transformation that fixes the finite points \(p\) and \(q\text{.}\)

a. Prove that the points \(z_\infty\) and \(w_\infty\) as defined in Lemma 3.5.7 lie on the perpendicular bisector of segment \(pq\text{.}\)

b. Show that \(T\) is the composition of two inversions about clines that contain \(p\) and \(q\text{.}\) Hint: Think about which inversion fixes \(p\) and \(q\) and takes \(z_\infty\) to \(\infty\text{?}\)

9

Suppose the Möbius transformation \(T\) fixes the finite (and distinct) points \(p\) and \(q\) sends \(z_\infty\) to \(\infty\) and \(\infty\) to \(w_\infty\text{.}\) By Lemma 3.5.7 we know \(p+q=z_\infty+w_\infty\text{.}\) Use the normal form of \(T\) to prove the following facts:

a. If \(T\) is elliptic then the four points \(p\text{,}\) \(q\text{,}\) \(z_\infty\text{,}\) and \(w_\infty\) form a rhombus. Under what conditions is this rhombus actually a square?

b. If \(T\) is hyperbolic then these 4 points all lie on the same Euclidean line.

c. If \(T\) is loxodromic, then under what conditions do these four points determine a rectangle?

10

Prove that any pair of non-intersecting clines in \(\mathbb{C}\) may be mapped by a Möbius transformation to concentric circles. Hint: By Theorem 3.2.16 there are two points \(p\) and \(q\) in \(\mathbb{C}\) that are symmetric with respect to both clines. What happens if we apply a Möbius transformation that takes one of these points to \(\infty\text{?}\)

11

Suppose \(T = i_{C_1} \circ i_{C_2}\) where \(C_1\) and \(C_2\) are clines that do not intersect. Prove that \(T\) has two fixed points and these points are on all clines perpendicular to both \(C_1\) and \(C_2\text{.}\)

12

Suppose \(T(z)\) is parabolic with normal form \begin{equation*} \frac{1}{T(z)-p} = \frac{1}{z-p}+d. \end{equation*}

Prove that the line through \(p\) and \(p+\frac{1}{d}\) gets sent to itself by \(T\text{.}\)

13

Analyze \(T(z) = [(1+3i)z-9i]/[iz+(1-3i)]\) by finding the fixed points, finding the normal form, and sketching the appropriate system of clines indicating the motion of the transformation.

14

Find a parabolic transformation with fixed point \(2+i\) for which \(T(\infty)=8\text{.}\)

15

Given distinct points \(p\text{,}\) \(q\text{,}\) and \(z\) in \(\mathbb{C}\text{,}\) prove there exists a type II cline of \(p\) and \(q\) that goes through \(z\text{.}\)