3.4: Möbius Transformations
Consider the function defined on \(\mathbb{C}^+\) by \(T(z) = \dfrac{(az + b)}{(cz + d)}\) where \(a, b, c\) and \(d\) are complex constants. Such a function is called a Möbius transformation if \(ad - bc \neq 0\text{.}\) Transformations of this form are also called fractional linear transformations. The complex number \(ad - bc\) is called the determinant of \(T(z) = \dfrac{(az+b)}{(cz+d)}\text{,}\) and is denoted as Det\((T)\text{.}\)
The function
\[ T(z) = \dfrac{az + b}{cz + d} \]
is a transformation of \(\mathbb{C}^+\) if and only if \(ad - bc \neq 0\text{.}\)
- Proof
-
First, suppose \(T(z) = \dfrac{(az+b)}{(cz+d)}\) and \(ad - bc \neq 0\text{.}\) We must show that \(T\) is a transformation. To show \(T\) is one-to-one, assume \(T(z_1) = T(z_2)\text{.}\) Then
\[ \dfrac{az_1 + b}{cz_1 + d} = \dfrac{az_2 + b}{cz_2 + d}\text{.} \]
Cross multiply this expression and simplify to obtain
\[ (ad - bc)z_1 = (ad - bc)z_2\text{.} \]
Since \(ad - bc \neq 0\) we may divide this term out of the expression to see \(z_1 = z_2\text{,}\) so \(T\) is 1-1 and it remains to show that \(T\) is onto.
Suppose \(w\) in \(\mathbb{C}^+\) is given. We must find \(z \in \mathbb{C}^+\) such that \(T(z) = w\text{.}\) If \(w = \infty\text{,}\) then \(z = \dfrac{-d}{c}\) (which is \(\infty\) if \(c = 0\)) does the trick, so assume \(w \neq \infty\text{.}\) To find \(z\) such that \(T(z) = w\) we solve the equation
\[ \dfrac{az + b}{cz+d} = w \]
for \(z\text{,}\) which is possible so long as \(a\) and \(c\) are not both 0 (causing the \(z\) terms to vanish). Since \(ad-bc \neq 0\text{,}\) we can be assured that this is the case, and solving for \(z\) we obtain
\[ z = \dfrac{-dw + b}{cw-a}\text{.} \]
Thus \(T\) is onto, and \(T\) is a transformation.
To prove the converse we show the contrapositive. We suppose \(ad-bc=0\) and show \(T(z)=\dfrac{(az+b)}{(cz+d)}\) is not a transformation by tackling two cases.
Case 1: \(ad = 0\text{.}\) In this case, \(bc = 0\) as well, so \(a\) or \(d\) is zero, and \(b\) or \(c\) is zero. In all four scenarios, one can check immediately that \(T\) is not a transformation of \(\mathbb{C}^+\text{.}\) For instance, if \(a = c = 0\) then \(T(z) = \dfrac{b}{d}\) is neither 1-1 nor onto \(\mathbb{C}^+\text{.}\)
Case 2: \(ad \neq 0\text{.}\) In this case, all four constants are non-zero, and \(\dfrac{a}{c} = \dfrac{b}{d}\text{.}\) Since \(T(0)=\dfrac{b}{d}\) and \(T(\infty)= \dfrac{a}{c}\text{,}\) \(T\) is not 1-1, and hence not a transformation of \(\mathbb{C}^+\text{.}\)
Note that in the preceeding proof we found the inverse transformation of a Möbius transformation. This inverse transformation is itself a Möbius transformation since its determinant is not \(0\). In fact, its determinant equals the determinant of the original Möbius transformation. We summarize this fact as follows.
The Möbius transformation
\[ T(z) = \dfrac{az + b}{cz+d} \]
has the inverse transformation
\[ T^{-1}(z) = \dfrac{-dz + b}{cz - a}\text{.} \]
In particular, the inverse of a Möbius transformation is itself a Möbius transformation.
If we compose two Möbius transformations, the result is another Möbius transformation. Proof of this fact is left as an exercise.
The composition of two Möbius transformations is again a Möbius transformation.
Just as translations and rotations of the plane can be constructed from reflections across lines, the general Möbius transformation can be constructed from inversions about clines.
A transformation of \(\mathbb{C}^+\) is a Möbius transformation if and only if it is the composition of an even number of inversions.
- Proof
-
We first observe that any general linear transformation \(T(z)=az+b\) is the composition of an even number of inversions. Indeed, such a map is a dilation and rotation followed by a translation. Rotations and translations are each compositions of two reflections (Theorem \(3.1.4\)), and a dilation is the composition of two inversions about concentric circles (Exericise \(3.2.4\)). So, in total, we have that \(T(z) = az+ b\) is the composition of an even number of inversions.
Now suppose \(T\) is the Möbius transformation \(T(z) = \dfrac{(az+b)}{(cz+d)}\text{.}\) If \(c = 0\) then \(T\) is a general linear transformation of the form \(T(z) = \dfrac{a}{d}z + \dfrac{b}{d}\text{,}\) and we have nothing to show.
So we assume \(c \neq 0\text{.}\) By doing some long division, the Möbius transformation can be rewritten as
\[ T(z) = \dfrac{az+b}{cz+d} = \dfrac{a}{c} + \dfrac{\dfrac{(bc-ad)}{c}}{cz+d}\text{,} \]
which can be viewed as the composition \(T_3 \circ T_2 \circ T_1(z)\text{,}\) where \(T_1(z) = cz + d\text{,}\) \(T_2(z) = \dfrac{1}{z}\) and \(T_3(z) = \dfrac{bc-ad}{c}z + \dfrac{a}{c}\text{.}\) Note that \(T_1\) and \(T_3\) are general linear transformations, and
\[ T_2(z) = \dfrac{1}{z} = \overline{\bigg[\dfrac{1}{\overline{z}}\bigg]} \]
is inversion in the unit circle followed by reflection about the real axis. Thus, each \(T_i\) is the composition of an even number of inversions, and the general Möbius transformation \(T\) is as well.
To prove the other direction, we show that if \(T\) is the composition of two inversions then it is a Möbius transformation. Then, if \(T\) is the composition of any even number of inversions, it is the composition of half as many Möbius transformations and is itself a Möbius transformation by Theorem \(3.4.3\).
Case 1: \(T\) is the composition of two circle inversions. Suppose \(T = i_{C_1} \circ i_{C_2}\) where \(C_1\) is the circle \(|z - z_1| = r_1\) and \(C_2\) is the circle \(|z-z_2| = r_2\text{.}\) For \(i = 1, 2\) the inversion may be described by
\[ i_{C_i} = \dfrac{r_i^2}{\overline{z-z_i}}+z_i\text{,} \]
and if we compose these two inversions we do in fact obtain a Möbius transformation. We leave the details of this computation to the reader but note that the determinant of the resulting Möbius transformation is \(r_1^2r_2^2\text{.}\)
Case 2: \(T\) is the composition of one circle inversion and one line reflection. Reflection in the line can be given by \(r_L(z) = e^{i\theta}\overline{z}+b\) and inversion in the circle \(C\) is given by \(i_{C} = \dfrac{r^2}{\overline{z-z_o}}+z_o\) where \(z_0\) and \(r\) are the center and radius of the circle, as usual. Work out the composition and you'll see that we have a Möbius transformation with determinant \(e^{i\theta}r^2\) (which is non-zero). Its inverse, the composition \(i_C \circ r_L\text{,}\) is also a Möbius transformation.
Case 3: \(T\) is the composition of two reflections. Either the two lines of reflection are parallel, in which case the composition gives a translation, or they intersect, in which case we have a rotation about the point of intersection (Theorem \(3.1.4\)). In either case we have a Möbius transformation.
It follows that the composition of any even number of inversions yields a Möbius transformation.
Since Möbius transformations are composed of inversions, they will embrace the finer qualities of inversions. For instance, since inversion preserves clines, so do Möbius transformations, and since inversion preserves angle magnitudes, Möbius transformations preserve angles (as an even number of inversions).
Möbius transformations take clines to clines and preserves angles.
The following fixed point theorem is useful for understanding Möbius transformations.
Any Möbius transformation \(T: \mathbb{C}^+ \to \mathbb{C}^+\) fixes \(1\), \(2\), or all points of \(\mathbb{C}^+\text{.}\)
- Proof
-
To find fixed points of \(T(z) = \dfrac{(az+b)}{(cz + d)}\) we want to solve
\[ \dfrac{az + b}{cz + d} = z \]
for \(z\text{,}\) which gives the quadratic equation
\[ cz^2 + (d-a)z - b = 0 \text{.} \label{3.4.1}\]
If \(c \neq 0\) then, as discussed in Example \(2.4.2\), equation (\(\ref{3.4.1}\)) must have \(1\) or \(2\) solutions, and there are \(1\) or \(2\) fixed points in this case.
If \(c = 0\) and \(a \neq d\text{,}\) then the transformation has the form \(T(z) = \dfrac{(az + b)}{d}\), which fixes \(\infty\text{.}\) From equation (1), \(z = \dfrac{b}{(d-a)} \neq \infty\) is a fixed point as well. So we have \(2\) fixed points in this case.
If \(c = 0\) and \(a = d\text{,}\) then equation (\(\ref{3.4.1}\)) reduces to \(0 = -b\text{,}\) so \(b = 0\) too, and the transformation is the identity transformation \(T(z) = \dfrac{(az+0)}{(0z + a)} = z\text{.}\) This transformation fixes every point.
With this fixed point theorem in hand, we can now prove the Fundamental Theorem of Möbius Transformations, which says that if we want to induce a one-to-one and onto motion of the entire extended plane that sends my favorite three points (\(z_1, z_2, z_3\)) to your favorite three points (\(w_1, w_2, w_3)\text{,}\) as dramatized below, then there is a Möbius transformation that will do the trick, and there's only one.
There is a unique Möbius transformation taking any three distinct points of \(\mathbb{C}^+\) to any three distinct points of \(\mathbb{C}^+\text{.}\)
- Proof
-
Suppose \(z_1, z_2,\) and \(z_3\) are distinct points in \(\mathbb{C}^+\text{,}\) and \(w_1, w_2\text{,}\) and \(w_3\) are distinct points in \(\mathbb{C}^+\text{.}\) We show there exists a unique Möbius transformation that maps \(z_i \mapsto w_i\) for \(i = 1,2,3\text{.}\) To start, we show there exists a map, built from inversions, that maps \(z_1 \mapsto 1\text{,}\) \(z_2 \mapsto 0\) and \(z_3 \mapsto \infty\text{.}\) We do so in the case that \(z_3 \neq \infty\text{.}\) This special case is left to the exercises.
First, invert about any circle centered at \(z_3\text{.}\) This takes \(z_3\) to \(\infty\) as desired. Points \(z_1\) and \(z_2\) no doubt get moved, say to \(z_1^\prime\) and \(z_2^\prime\text{,}\) respectively, neither of which is \(\infty\text{.}\) Second, do a translation that takes \(z_2^\prime\) to 0. Such a translation will keep \(\infty\) fixed, and take \(z_1^\prime\) to some new spot \(z_1^{\prime\prime}\) in \(\mathbb{C}\text{.}\) Third, rotate and dilate about the origin, (which keeps 0 and \(\infty\) fixed) so that \(z_1^{\prime\prime}\) moves to 1. This process yields a composition of inversions that maps \(z_1 \mapsto 1\text{,}\) \(z_2 \mapsto 0\text{,}\) and \(z_3 \mapsto \infty\text{.}\) However, this composition actually involves an odd number of inversions, so it's not a Möbius transformation. To make a Möbius transformation out of this composition, we do one last inversion: reflect across the real axis. This keeps 1, 0, and \(\infty\) fixed. Thus, there is a Möbius transformation taking any three distinct points to the points \(1, 0,\) and \(\infty\text{.}\) For now, we let \(T\) denote the Möbius transformation that maps \(z_1 \mapsto 1\text{,}\) \(z_2 \mapsto 0\) and \(z_3 \mapsto \infty\text{.}\)
One can similarly construct a Möbius transformation, call it \(S\text{,}\) that takes \(w_1 \mapsto 1\text{,}\) \(w_2 \mapsto 0\text{,}\) and \(w_3 \mapsto \infty\text{.}\)
If we let \(S^{-1}\) denote the inverse transformation of \(S\text{,}\) then the composition \(S^{-1} \circ T\) is a Möbius transformation, and this transformation does what we set out to accomplish as suggested by Figure \(3.4.2\). In particular,
\begin{align*} S^{-1} \circ T(z_1) & = S^{-1}(1) = w_1\\ S^{-1} \circ T(z_2) & = S^{-1}(0) = w_2\\ S^{-1} \circ T(z_3) & = S^{-1}(\infty) = w_3\text{.} \end{align*}
Finally, to prove that this Möbius transformation is unique, assume that there are two Möbius transformations \(U\) and \(V\) that map \(z_1 \mapsto w_1\text{,}\) \(z_2 \mapsto w_2\) and \(z_3 \mapsto w_3\text{.}\) Then \(V^{-1} \circ U\) is a Möbius transformation that fixes \(z_1, z_2\text{,}\) and \(z_3\text{.}\) According to Theorem \(3.4.6\) there is only one Möbius transformation that fixes more than two points, and this is the identity transformation. Thus \(V^{-1} \circ U(z) = z\) for all \(z \in \mathbb{C}^+\text{.}\) Similarly, \(U \circ V^{-1}(z) = z\text{,}\) and it follows that \(U(z) = V(z)\) for all \(z \in \mathbb{C}^+\text{.}\) That is, \(U\) and \(V\) are the same map.
There is an algebraic description of the very useful Möbius transformation mapping \(z_1 \mapsto 1\text{,}\) \(z_2 \mapsto 0\) and \(z_3 \mapsto \infty\) that arose in the proof of Theorem \(3.4.7\):
\[ T(z) = \dfrac{(z-z_2)}{(z-z_3)}\cdot \dfrac{(z_1 - z_3)}{(z_1-z_2)}\text{.} \]
The reader can check that the map works as advertised and that it is indeed a Möbius transformation. (While it is clear that the transformation has the form \(\dfrac{(az + b)}{(cz + d)}\text{,}\) it might not be clear that the determinant is nonzero. It is, since the \(z_i\) are distinct.) We also note that if one of the \(z_i\) is \(\infty\text{,}\) the form of the map reduces by cancellation of the terms with \(\infty\) in them. For instance, if \(z_2 = \infty\text{,}\) the map that sends \(z_1 \mapsto 1\text{,}\) \(\infty \mapsto 0\) and \(z_3 \mapsto \infty\) is \(T(z) = \dfrac{(z_1-z_3)}{(z-z_3)} \text{.}\)
The Möbius transformation that sends any three distinct points to \(1\), \(0\), and \(\infty\) is so useful that it gets its own name and special notation.
The cross ratio of 4 complex numbers \(z,w,u,\) and \(v\text{,}\) where \(w,u,\) and \(v\) are distinct, is denoted \((z,w;u,v)\text{,}\) and
\[ (z,w;u,v) = \dfrac{z-u}{z-v}\cdot\dfrac{w-v}{w-u}\text{.} \]
If \(z\) is a variable, and \(w, u,\) and \(v\) are distinct complex constants, then \(T(z) = (z,w;u,v)\) is the (unique!) Möbius transformation that sends \(w \mapsto 1\text{,}\) \(u \mapsto 0\text{,}\) and \(v \mapsto \infty\text{.}\)
Find the unique Möbius transformation that sends \(1 \mapsto 3\text{,}\) \(i \mapsto 0\text{,}\) and \(2 \mapsto -1\text{.}\)
One approach: Find \(T(z) = (z,1;i,2)\) and \(S(w) = (w,3;0,-1)\text{.}\) In this case, the transformation we want is \(S^{-1} \circ T\text{.}\)
To find this transformation, we set the cross ratios equal:
\begin{align*} (z,1;i,2) & = (w,3;0,-1)\\ \dfrac{z-i}{z-2}\cdot\dfrac{1-2}{1-i} & = \dfrac{w-0}{w+1}\cdot\dfrac{3+1}{3-0}\\ \dfrac{-z+i}{(1-i)z-2+2i} & = \dfrac{4w}{3w+3}\text{.} \end{align*}
Then solve for \(w\text{:}\)
\begin{align*} -3zw + 3iw + 3i - 3z & = 4[(1-i)z-2+2i]w\\ -3z+3i & = [3z-3i+4[(1-i)z-2+2i]]w\\ w & = \dfrac{-3z + 3i}{(7-4i)z + (-8+5i)}\text{.} \end{align*}
Thus, our Möbius transformation is
\[ V(z) = \dfrac{-3z + 3i}{(7-4i)z + (-8+5i)}\text{.} \]
It's quite easy to check our answer here. Since there is exactly one Möbius transformation that does the trick, all we need to do is check whether \(V(1) = 3, V(i) = 0\) and \(V(2) = -1\text{.}\) Ok... yes... yes... yep! We've got our map!
The probabilities assigned to events by a distribution function on a sample space are given by.
- Proof
-
Suppose \(z_0, z_1, z_2,\) and \(z_3\) are four distinct points in \(\mathbb{C}^+\text{,}\) and \(T\) is any Möbius transformation. Then
\[ (z_0,z_1;z_2,z_3) = (T(z_0), T(z_1); T(z_2), T(z_3))\text{.} \]
In addition to defining maps that send points to \(1\), \(0\), and \(\infty\text{,}\) the cross ratio can proclaim whether four points lie on the same cline: If \((z,w;u,v)\) is a real number then the points are all on the same cline; if \((z,w;u,v)\) is complex, then they aren't. The proof of this fact is left as an exercise.
Take the points \(1, i, -1, -i\text{.}\) We know these four points lie on the circle \(|z| = 1\text{,}\) so according to the statement above, \((1,i;-1,-i)\) is a real number. Let's check:
\begin{align*} (1,i;-1,-i) & = \dfrac{1+1}{1+i}\cdot\dfrac{i+i}{i+1}\\ & = \dfrac{2}{1+i}\dfrac{2i}{1+i}\\ & = \dfrac{4i}{(1-1)+2i}\\ & = \dfrac{4i}{2i}\\ & = 2 \tag{Yep!}\text{.} \end{align*}
Another important feature of inversion that gets passed on to Möbius transformations is the preservation of symmetry points. The following result is a corollary to Theorem \(3.2.8\).
If \(z\) and \(z^*\) are symmetric with respect to the cline \(C\text{,}\) and \(T\) is any Möbius transformation, then \(T(z)\) and \(T(z^*)\) are symmetric with respect to the cline \(T(C)\text{.}\)
We close the section with one more theorem about Möbius transformations.
Given any two clines \(C_1\) and \(C_2\text{,}\) there exists a Möbius transformation \(T\) that maps \(C_1\) onto \(C_2\text{.}\) That is, \(T(C_1) = C_2\text{.}\)
- Proof
-
Let \(p_1\) be a point on \(C_1\) and \(q_1\) and \(q_1^*\) symmetric with respect to \(C_1\text{.}\) Similarly, let \(p_2\) be a point on \(C_2\) and \(q_2\) and \(q_2^*\) be symmetric with respect to \(C_2\text{.}\) Build the Möbius transformation that sends \(p_1 \mapsto p_2\text{,}\) \(q_1 \mapsto q_2\) and \(q_1^* \mapsto q_2^*\text{.}\) Then \(T(C_1)=C_2\text{.}\)
Exercises
Find a transformation of \(\mathbb{C}^+\) that rotates points about \(2i\) by an angle \(\dfrac{\pi}{4}\text{.}\) Show that this transformation has the form of a Möbius transformation.
Find the inverse transformation of \(T(z) = \dfrac{3z + i}{2z + 1}\text{.}\)
Prove Theorem \(3.4.3\). That is, suppose \(T\) and \(S\) are two Möbius transformations and prove that the composition \(T\circ S\) is again a Möbius transformation.
Prove that any Möbius transformation can be written in a form with determinant \(1\), and that this form is unique up to sign.
- Hint
-
How does the determinant of \(T(z) = \dfrac{(az+b)}{(cz+d)}\) change if we multiply top and bottom of the map by some constant \(k\text{?}\)
Find the unique Möbius transformation that sends \(1 \mapsto i\text{,}\) \(i \mapsto -1\text{,}\) and \(-1 \mapsto -i\text{.}\) What are the fixed points of this transformation? What is \(T(0)\text{?}\) What is \(T(\infty)\text{?}\)
Repeat the previous exercise, but send \(2 \to 0\text{,}\) \(1 \to 3\) and \(4 \to 4\text{.}\)
Prove this feature of the cross ratio: \(\overline{(z, z_1; z_2, z_3)} = (\overline{z},\overline{z_1};\overline{z_2},\overline{z_3})\text{.}\)
Prove that the cross ratio of four distinct real numbers is a real number.
Prove that the cross ratio of four distinct complex numbers is a real number if and only if the four points lie on the same cline.
- Hint
-
Prove that the cross ratio of four distinct complex numbers is a real number if and only if the four points lie on the same cline.
Do the points \(2+i, 3, 5,\) and \(6 + i\) lie on a single cline?
More on Möbius transformations.
- Give an example of a Möbius transformation \(T\) such that \(\overline{T(z)} \neq T(\overline{z})\) for some \(z\) in \(\mathbb{C}^+\text{.}\)
- Suppose \(T\) is a Möbius transformation that sends the real axis onto itself. Prove that in this case, \(\overline{T(z)} = T(\overline{z})\) for all \(z\) in \(\mathbb{C}\text{.}\)
Is there a Möbius transformation that sends \(1\) to \(3\), \(i\) to \(4\), \(-1\) to \(2 + i\) and \(-i\) to \(4 + i\text{?}\)
- Hint
-
It may help to observe that the input points are on a single cline.
Find the fixed points of these transformations on \(\mathbb{C}^+\text{.}\) Remember that \(\infty\) can be a fixed point of such a transformation.
- \(\displaystyle T(z) = \dfrac{2z}{3z-1}\)
- \(\displaystyle T(z) = iz\)
- \(\displaystyle T(z) = \dfrac{-iz}{(1-i)z - 1}\)
Find a Möbius transformation that takes the circle \(|z| = 4\) to the straight line \(3x + y = 4\text{.}\)
- Hint
-
Track the progress of three points, and the rest will follow.
Find a non-trivial Möbius transformation that fixes the points \(-1\) and \(1\), and call this transformation \(T\text{.}\) Then, let \(C\) be the imaginary axis. What is the image of \(C\) under this map. That is, what cline is \(T(C)\text{?}\)
Suppose \(z_1, z_2, z_3\) are distinct points in \(\mathbb{C}^+\text{.}\) Show that by an even number of inversions we can map \(z_1 \mapsto 1\text{,}\) \(z_2 \mapsto 0\text{,}\) and \(z_3 \mapsto \infty\) in the case that \(z_3 = \infty\text{.}\)