##### Theorem3.4.1

The function \begin{equation*} T(z) = \frac{az + b}{cz + d} \end{equation*} is a transformation of \(\mathbb{C}^+\) if and only if \(ad - bc \neq 0\text{.}\)

Consider the function defined on \(\mathbb{C}^+\) by \(T(z) = (az + b)/(cz + d)\) 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 \(T(z) = (az+b)/(cz+d)\text{,}\) and is denoted as Det\((T)\text{.}\)

The function \begin{equation*} T(z) = \frac{az + b}{cz + d} \end{equation*} is a transformation of \(\mathbb{C}^+\) if and only if \(ad - bc \neq 0\text{.}\)

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 \begin{equation*} T(z) = \frac{az + b}{cz+d} \end{equation*} has the inverse transformation \begin{equation*} T^{-1}(z) = \frac{-dz + b}{cz - a}. \end{equation*} 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.

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 \(\mathbb{C}^+\text{.}\)

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 \(\mathbb{C}^+\text{.}\)

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 3.4.8: \begin{equation*} T(z) = \frac{(z-z_2)}{(z-z_3)}\cdot \frac{(z_1 - z_3)}{(z_1-z_2)}. \end{equation*}

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 \((az + b)/(cz + d)\text{,}\) 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 \(T(z) = (z_1-z_3)/(z-z_3)\text{.}\)

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
\begin{equation*}
(z,w;u,v) = \frac{z-u}{z-v}\cdot\frac{w-v}{w-u}.
\end{equation*}

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) \amp = (w,3;0,-1)\\ \frac{z-i}{z-2}\cdot\frac{1-2}{1-i} \amp = \frac{w-0}{w+1}\cdot\frac{3+1}{3-0}\\ \frac{-z+i}{(1-i)z-2+2i} \amp = \frac{4w}{3w+3}. \end{align*}

Then solve for \(w\text{:}\) \begin{align*} -3zw + 3iw + 3i - 3z \amp = 4[(1-i)z-2+2i]w\\ -3z+3i \amp = [3z-3i+4[(1-i)z-2+2i]]w\\ w \amp = \frac{-3z + 3i}{(7-4i)z + (-8+5i)}. \end{align*}

Thus, our Möbius transformation is \begin{equation*} V(z) = \frac{-3z + 3i}{(7-4i)z + (-8+5i)}. \end{equation*}

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!

Suppose \(z_0, z_1, z_2,\) and \(z_3\) are four distinct points in \(\mathbb{C}^+\text{.}\) Then \((z_0,z_1;z_2,z_3) = (T(z_0), T(z_1); T(z_2), T(z_3))\) for any Möbius transformation \(T\text{.}\)

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) \amp = \frac{1+1}{1+i}\cdot\frac{i+i}{i+1}\\ \amp = \frac{2}{1+i}\frac{2i}{1+i}\\ \amp = \frac{4i}{(1-1)+2i}\\ \amp = \frac{4i}{2i}\\ \amp = 2. \tag{Yep!} \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 3.2.12.

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, \(T(C_1) = C_2\text{.}\)

Find a transformation of \(\mathbb{C}^+\) that rotates points about \(2i\) by an angle \(\pi/4\text{.}\) Show that this transformation has the form of a Möbius transformation.

Find the inverse transformation of \(T(z) = \frac{3z + i}{2z + 1}\text{.}\)

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 \(T(z) = (az+b)/(cz+d)\) 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: Use the previous exercise and the invariance of the cross ratio.

Do the points \(2+i, 3, 5,\) and \(6 + i\) lie on a single cline?

More on Möbius transformations.

a. Give an example of a Möbius transformation \(T\) such that \(\overline{T(z)} \neq T(\overline{z})\) for some \(z\) in \(\mathbb{C}^+\text{.}\)

b. 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.

a. \(T(z) = \frac{2z}{3z-1}\)

b. \(T(z) = iz\)

c. \(T(z) = \frac{-iz}{(1-i)z - 1}\)

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 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 \(z_3 \mapsto \infty\text{,}\) in three special cases:

a. When \(z_1 = \infty\text{.}\)

b. When \(z_2 = \infty\text{.}\)

c. When \(z_3 = \infty\text{.}\)