Möbius transformations
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.
Visualization
Recall that we visualize a complex function by illustrating its effect on various domains in the complex plane. I’ve written two functions named rect_plot and disk_plot to make it relatively easy to visualize the image of rectangular domains and polar domains respectively.
Here, for example, is the image of the square \([-1,1]\times[-1,1]\) under the action of the Möbius transformation \[T(z) = i\frac{z+1}{2-z}.\]
And here’s the image of the unit disk under the transformation \[T(z) = i\frac{z+1}{1-z}.\]
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 \(LC\). First, let’s reproduce the textbook’s explanation of the fact that \(LC \leftrightarrow LC\) under multiplicative inversion. That is, the reciprocal function maps \(LC\) to \(LC\).
First, 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 \[\rho = \frac{1}{r} \text{ and } \varphi = -\theta.\] Plugging that into 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: