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The maps $(az + b)/(cz + d)$ with non-zero determinant: bijections of the plane with one point added, built out of translations, scalings and inversion, conformal wherever defined, sending circles and lines to circles and lines, and having at most two fixed points unless they are the identity.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to evaluate a Möbius transformation including at the point at infinity, say which point it sends to infinity, decompose one into translations, scalings and an inversion, explain why it sends circles and lines to circles and lines, find its fixed points, and say why it can have at most two of them unless it is the identity.
Conformality, and the limit at infinity of a rational function from the limits lesson — which is where the point at infinity first appeared. This lesson takes that point seriously enough to add it to the plane, and a family of maps becomes a group of bijections as a result.
A Möbius transformation is $T(z) = \dfrac{az + b}{cz + d}$ with $ad - bc \ne 0$; that quantity is its determinant. The extended plane is the plane with one point at infinity added, and it is also called the Riemann sphere. A generalised circle is a circle or a line, a line counting as a circle through the point at infinity. A fixed point of $T$ is a point with $T(z) = z$.
The definition, and why the determinant. $T(z) = \dfrac{az + b}{cz + d}$. If $ad - bc = 0$ the two rows are proportional and $T$ is constant, so the condition $ad - bc \ne 0$ is exactly what makes $T$ a map worth having.
It is a bijection once infinity is added. Define $T(-d/c) = \infty$ and $T(\infty) = a/c$ when $c \ne 0$, and $T(\infty) = \infty$ when $c = 0$. Then $T$ is one-to-one and onto on the extended plane, with inverse $T^{-1}(w) = \dfrac{dw - b}{-cw + a}$ — another Möbius map. Composition corresponds to multiplying the matrices $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, so the Möbius maps form a group.
Everything is built from three kinds. Translation $z + \beta$; scaling with rotation $\alpha z$; and inversion $1/z$. When $c \ne 0$,
$$\frac{az + b}{cz + d} = \frac{a}{c} + \frac{bc - ad}{c}\cdot\frac{1}{cz + d},$$
which is a translation, then a scaling, then an inversion, then a scaling, then a translation. Any property that survives all three kinds survives every Möbius map, and that is how the next fact is proved.
Circles and lines go to circles and lines. Treating a line as a circle through the point at infinity, the family of generalised circles is preserved. Which of the two a particular one becomes depends on whether it passes through the point that is sent to infinity: $1/z$ sends the vertical line through $1$ to a circle, and sends the real axis to itself.
It is conformal everywhere it is defined, since $T'(z) = \dfrac{ad - bc}{(cz + d)^{2}}$ is never zero.
Fixed points. Solving $T(z) = z$ clears to a quadratic, so a Möbius map other than the identity has at most two fixed points. That single fact is what makes three points exactly the right amount of data to specify one, which is the next lesson.
Another way: picture
Put the plane on a sphere by stereographic projection: each point of the plane corresponds to a point of the sphere, and the north pole is the one point left over — the point at infinity. Circles on the sphere correspond to circles and lines in the plane, and a line is just a circle that happens to run through the pole. On the sphere, a Möbius map is a smooth motion taking circles to circles, and the distinction between a circle and a line stops existing.
Another way: steps
A handful of Möbius maps do almost all the work in practice, and it is worth knowing them by sight.
| Map | What it does |
|---|---|
| $z + \beta$ | translates; fixes only infinity |
| $\alpha z$ | rotates by $\arg\alpha$ and scales by $|\alpha|$; fixes $0$ and infinity |
| $1/z$ | exchanges $0$ and infinity, and the inside and outside of the unit circle |
| $\dfrac{z - i}{z + i}$ | the Cayley transform: upper half plane onto the unit disc |
| $\dfrac{z - \alpha}{1 - \bar\alpha z}$ | disc onto disc, sending $\alpha$ to $0$ |
The last two are the ones a boundary value problem actually needs. The Cayley transform is how a half plane problem becomes a disc problem, and the disc-to-disc family is how any chosen interior point is moved to the centre, where a formula is available.
It is worth checking the Cayley transform by hand once. A point $x$ on the real axis has $|x - i| = |x + i|$, so its image has modulus $1$: the real axis goes to the unit circle. And $i$ itself goes to $0$, which is inside — so the upper half plane, which contains $i$, goes to the inside of the circle rather than the outside. Two substitutions settle the whole map, because a Möbius map cannot do anything complicated in between.
The extended plane has exactly one point at infinity, not one in each direction. That is what makes a Möbius map a bijection: $1/z$ has to send $0$ somewhere, and it sends it to the single point that is the destination of every direction of escape. The sphere picture is the one to hold, because on the sphere the point at infinity is as ordinary as any other and there is visibly one of it.
Once it is a genuine point, the bookkeeping follows: $T(\infty)$ and $T^{-1}(\infty)$ are values to be computed, not exceptions to be apologised for. A solution that lists the domain of $T$ as everything except $-d/c$ has not yet made the move this lesson is about.
The second misreading is circles go to circles without the word generalised. A circle may become a line and a line may become a circle; which happens depends entirely on whether the original passes through the point that is sent to infinity. Treating the two as separate families makes a true theorem look like a list of exceptions.
And a third: preserving circles is not preserving centres. The image of a circle under $1/z$ is a circle, and the image of its centre is generally not the centre of the image. Nothing in the theorem promised it would be.
Where does $w = 1/z$ send the vertical line $x = 1$? Write $z = 1 + iy$.
Parametrise the line.
$w = \dfrac{1}{1 + iy} = \dfrac{1 - iy}{1 + y^{2}}$, whose real and imaginary parts satisfy $u^{2} + v^{2} = u$.
Substitute and simplify.
That is the circle of radius $\tfrac12$ centred at $\tfrac12$, passing through the origin — which is where the point at infinity went. A generalised circle became a genuine one.
The line passed through infinity; the circle passes through zero.
$T(z) = \dfrac{z}{z + 1}$: solving $\dfrac{z}{z + 1} = z$ gives $z = z^{2} + z$, so $z^{2} = 0$.
Clear the denominator.
One finite fixed point, $z = 0$, a repeated root. And $c = 1 \ne 0$, so infinity is not fixed.
Test infinity separately.
One fixed point in total: such a map is called parabolic, and it is the Möbius analogue of a translation.
Divide top and bottom by $z$: $\dfrac{2 + 1/z}{1 - 3/z}$.
The small terms vanish.
So the image is $2$, the ratio of the leading coefficients. And $z = 3$ is the point sent to infinity in exchange.
Let $T(z) = \dfrac{2z + 2}{z + 4}$. Fill in the table.
| Value | |
|---|---|
| The determinant of the map | |
| The image of zero | |
| The image of one | |
| The image of the point at infinity | |
| The point sent to infinity |
$T(z) = \dfrac{z + 1}{z + 4}$. What is $T(5)$? Give a fraction.
Answer:
Match each Möbius map to what it does to the extended plane.
| Translates the whole plane | Scales from the origin, fixing zero and infinity | Exchanges the inside and the outside of the unit circle | Sends the upper half plane onto the unit disc | |
|---|---|---|---|---|
| $z + 9$ | ||||
| $9z$ | ||||
| $\dfrac{1}{z}$ | ||||
| $\dfrac{z - i}{z + i}$ |
Put the steps of finding the fixed points of a Möbius transformation in order.
Number the steps in order (write the number in the box):
Select every statement that is true.
This task has no paper form; do it on a device.
A Möbius transformation other than the identity. At most how many fixed points can it have on the extended plane?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Let $T(z) = \dfrac{z + 5}{6z + 6}$. Fill in the table.
| Value | |
|---|---|
| The determinant of the map | |
| The image of zero | |
| The image of one | |
| The image of the point at infinity | |
| The point sent to infinity |
You can compute with a Möbius map on the extended plane and find its fixed points. Say in your own words why a line counts as a circle here. Next: the cross ratio, which turns the fixed-point count into a recipe for building the map you want.
9. Your turn: where does $T(z) = \dfrac{2z + 1}{z - 3}$ send the point at infinity?, step 2