Möbius transformations

complexPolarPlot(
    `1/z`, [0, 1], [0, 2*Math.PI],
    {xDomain: [-4,4], yDomain: [-4,4]}
)

Möbius transformations have the form \[T(z)=\frac{a z+b}{c z+d}.\] It’s not hard to show by direct computation that \[T^{-1}(z) = \frac{dz-b}{-cz+a}.\]

We do need to assume that \(ad-bc\neq0\) for this to be true, though. Otherwise, the function is a constant in disguise.

import {
    complexCartesianMark, 
    complexPolarMark,
    complexCartesianPlot, 
    complexPolarPlot
} from "../../components/complexImageMarks.js";

Visualization

Recall that we visualize a complex function by illustrating its effect on various domains in the complex plane. Here, for example, is the image of the square \([-1,1]\times[-1,1]\) under the action of the Möbius transformation \[T(z) = i\frac{z+1}{2-z}.\]

Here’s a little computer code to do it:

complexCartesianPlot(
    `i*(z+1)/(2-z)`, [-1, 1], [-1, 1],
    {xDomain: [-1.8,1.8], yDomain: [-0.1,2.1]}
)

And here’s the image of the unit disk under the function \[f(z) = i\frac{1+z}{1-z}.\]

complexPolarPlot(`i*(1+z)/(1-z)`, 
    [0,1], [0.0001, 2*Math.PI-0.0001],
    {xDomain: [-4.1,4.1], yDomain: [-1.1, 4.1]}
)

I guess we are curious to know how these types of images could come out of these types of functions??

Circles and lines map to circles and lines

We’ve discussed some of this before but it’s worth recalling some of the material as we go into it deeper this time.

Circles/Lines \(\leftrightarrow\) circles/lines: Let \(LC\) denote the set of all lines and circles in the complex plane. Any Möbius transformation maps \(LC\) to itself. That is, the image of any circle or line in the complex plane under a Möbius transformation is again a circle or line.

Our textbook calls the elements of \(LC\) “generalized circles” and we’ll see shortly why this might be a reasonable way to think of \(LC\).

To understand the ircles/Lines \(\leftrightarrow\) circles/lines phenomenon, let’s first reproduce the textbook’s explanation of the fact that \(LC \leftrightarrow LC\) under multiplicative inversion. That is, the reciprocal function maps \(LC\) to \(LC\).

To begin that process, we note that the equation of any circle or line can be written in the form \[ A(x^2+y^2) + Bx + Cy + D = 0, \] Where \(A,B,C,\) and \(D\) are real constants. We obtain a line precisely when \(A\neq0\). We can translate this polar form easily enough: \[ Ar^2+r(B\cos \theta +C\sin \theta) + D = 0\text{.} \tag{1}\] Now, suppose that \(z=re^{i\theta}\). Then, \[\frac{1}{z} = \frac{1}{r}e^{-i\theta}.\] Put another way, if we write \(\frac{1}{z} = \rho e^{i\varphi}\), then \[\rho = \frac{1}{r} \text{ and } \varphi = -\theta.\] Substituting for \(r\) and \(\theta\) in Equation 1, we get \[ A+\rho (B\cos \phi -C\sin \phi ) +D\rho ^2=0. \] This is again an equation of a line or circle, though which now depends on \(D\), rather than \(A\).

This much proves that

Inversion maps a circle or a line to a circle or line.

What about Möbius transformations more generally? Well, if \(c\neq0\), then we can write \[ \frac{az+b}{cz+d} = \frac{a}{c} - \frac{ad-bc}{c^2}\frac{1}{z+\frac{d}{c}}. \] Note that this is a composition of a shift, an inversion, a dialation/rotation with one more shift. Since each of those operations preserves \(LC\), so does the Möbius transformation.

If \(c=0\), it’s even easier: \[ \frac{az+b}{d} = \frac{a}{d}z + \frac{b}{d}. \]

Finding images under Möbius Transformations

Given a Möbius transformation and a circle or line, it’s natural to ask for the image of that set under the action of the Möbius transformation. Well, a line is well known to be determined by two points and a circle by three. More generally, three distinct points uniquely determine a line or circle; we get a line exactly when the three points are colinear.

Often, we are interested in the image of a solid disk bound by a circle. That image could be either

another disk,
the exterior of another disk, or
a half-plane.

We can find that image by simply finding the image of three points on the bounding circle. We can then examine the image of a point in the interior to characterize the image of the disk.

Example 1

Find the image of the unit disk under the Möbius transformation \[T(z) = \frac{2z+3}{4z+5}.\]

Solution: We simply apply the transformation to 3 points in the unit circle:

The first three of these points are clearly not colinear so they determine a circle, which is exactly the image of the unit circle under \(T\). The point \(0\) maps to \(3/5\), which lies between \(5/9\) and \(1\). Thus, the interior of the unit disk maps onto the interior of this circle:

complexPolarPlot(
    `(2*z + 3)/(4*z + 5)`, [0,1], [0.0001, 2*Math.PI-0.0001],
    {yDomain: [-0.31,0.31], xDomain: [-0.1, 1.1]}
)

Example 2

Find the image of the unit disk under the Möbius transformation \[T(z) = \frac{z-1}{z-i}.\]

We’ll again apply three points on the unit circle together with the point \(0\), which is in the interior.

Now, the first three points are collinear so the unit circle maps to a line. The origin maps to \(-i\) so the interior of the unit disk maps to the side of line that contains \(-i\).

complexPolarPlot(
    `(z - 1)/(z - I)`, [0,0.999], [0.0, 2*Math.PI],
    {yDomain: [-5.1,5.1], xDomain: [-5.1,5.1]}
)

Note that I didn’t plug \(z=i\) into the function since that throws an error in the Sage code. In fact, though, we often think of the point at \(\infty\) as lying on the line determined by the other two points. That’s really the easiest way to find that line.

Finding Möbius transformation for specified images

Given a disk or half-plane, suppose we’d like to find a Möbius that maps the unit disk on to that disk or half-plane. How might we do that?

The cross-ratio

Note that a Möbius transformation has four parameters but only three degrees of freedom. That is, given three parameters, we could normalize one of those to \(1\) by dividing top and bottom by the same number. We get the same Möbius in terms of three parameters.

As a result, the Möbius transformation should be uniquely determined by the image of three points. Thus, if we could find a Möbius transformation that fixes \(1\), maps \(-I\to0\), and sends \(-i\to\infty\), then we should have what we want.

Given fixed and distinct \(z_1\), \(z_2\), and \(z_3\), define the cross-ratio as \[ T(z) = \frac{(z-z_1) (z_2-z_3)}{(z-z_3)(z_2-z_1)}. \]

Note that the cross-ratio maps \[z_1\to0, \: z_2\to1, \text{ and } z_3\to\infty.\]

Thus, given three points, the cross-ratio maps the circle or line determined by those three points to the real axis.

Example 3

Let’s try to find a Möbius transformation that maps the unit circle to the upper half plane. To do so, we’ll apply the cross-ratio choosing \(z_1\), \(z_2\), and \(z_3\) to be three points on the unit circle. Then, we’ll check the image of the origin to see if it worked.

Choosing \(z_1 = 1\), \(z_2 = i\), and \(z_3 = -1\), we get

\[ T(z) = \frac{(z-1)(i+1)}{(z+1)(i-1)}. \]

3 points \(\to\) 3 points

The logical challenge is to see if, given two lists of three points each, can we find a Möbius transform mapping one list to the other? Suppose our lists are \(\{z_1,z_2,z_3\}\) and \(\{w_1,w_2,w_3\}\). Set up the following equation comparing two cross-ratios:

\[ T_1(z) = \frac{(z-z_1) (z_2-z_3)}{(z-z_3) (z_2-z_1)}=\frac{(w-w_1) (w_2-w_3)}{(w-w_3) (w_2-w_1)} = T_2(w) \]

The Möbius transformation on the left, \(T_1\), sends \[z_1\to0, z_2\to1, \text{ and } z_3\to\infty.\]
The inverse of the Möbius transformation on the right \(T_2^{-1}\) sends \[w_1\to0, w_2\to1, \text{ and } w_3\to\infty.\]
Thus, \(T_2^{-1}\circ T_1\) sends \[z_1\to w_1, z_2\to w_2, \text{ and } z_3\to w_3.\] We simply have to solve the equation for \(w\) to find it!

We can automate this procedure like so:

complexCartesianPlot(
    `-((I - 1)*z - 3*I + 1)/((I - 1)*z - I + 3)`,
    [-8,8], [0.0, 12], {xDomain: [-2.6,2.6], yDomain: [-1.5,1.5]}
)