13. Inversion and complex geometry¶

Some problems about circles are hard only because circles are round. Find a circle that touches three given circles; show that a ring of circles squeezed between two others always closes up, wherever it starts; prove Ptolemy's theorem about four points on a circle. This chapter gives you two tools that make such problems easy. The first is to write points as complex numbers, so that turning and scaling become multiplication. The second is inversion, a map that turns a circle inside out and can change circles into straight lines. Together they lead to the Möbius transformations, the cross-ratio and the Riemann sphere, where the plane is closed up by one extra point at infinity.

In this chapter

You will learn:

how to do plane geometry with complex numbers: distance, angle, rotation, similarity, and a test for equilateral triangles
what inversion in a circle is, and why it sends lines and circles to lines and circles (two proofs)
how inversion solves Apollonius' problem and proves Steiner's porism about chains of circles
Möbius transformations, the cross-ratio, and the test for four points on one circle
Ptolemy's theorem and inequality, proved twice (by inversion and by a complex identity)
the Riemann sphere and stereographic projection, and why circles on the sphere become circles in the plane

You need: Chapter 6 (inscribed angles, tangents, power of a point), Chapter 10 (isometries and similarities) and Chapter 12 (where Ptolemy's theorem first appears). From the Algebra book, Complex numbers: the complex plane, modulus, argument and polar form.

Levels: 13.1 and 13.2 are advanced but concrete; 13.3 and 13.5 are advanced; 13.4 and 13.6 are advanced to expert.

13.1 Geometry with complex numbers¶

Level: advanced. Here is a problem. On each side of any triangle, build an equilateral triangle pointing outward, and mark its centre. Join the three centres. The result is always an equilateral triangle, whatever triangle you began with. This is called Napoleon's theorem (the name is a tradition; there is no evidence that Napoleon Bonaparte found or proved it). A proof with angles and congruent triangles is possible but fiddly. With complex numbers it is a few lines of algebra. First we set up the dictionary.

Points are numbers. The complex number \(z = x + iy\) is the point \((x, y)\). The modulus \(|z| = \sqrt{x^2 + y^2}\) is the distance from \(0\) to \(z\), so the distance between two points is

\[ |z - w| . \]

The argument \(\arg z\) is the angle from the positive real axis to \(z\), and every nonzero \(z\) can be written in polar form \(z = r e^{i\theta}\) with \(r = |z|\) and \(\theta = \arg z\). The conjugate \(\bar z = x - iy\) is the mirror image of \(z\) in the real axis, and \(z \bar z = |z|^2\).

Moves are formulas. Each basic motion of the plane is a short formula.

Motion	Formula	Why
translation by \(b\)	\(z \mapsto z + b\)	adding \(b\) adds its coordinates
rotation by \(\theta\) about \(0\)	\(z \mapsto e^{i\theta} z\)	multiplying moduli and adding arguments
scaling by \(r > 0\) about \(0\)	\(z \mapsto r z\)	multiplies every distance from \(0\) by \(r\)
reflection in the real axis	\(z \mapsto \bar z\)	changes the sign of \(y\)

The rotation rule is the heart of it: for \(a = re^{i\alpha}\), the product \(az = r|z| e^{i(\alpha + \arg z)}\) has modulus multiplied by \(r\) and angle increased by \(\alpha\). So multiplying by \(a\) turns the whole plane through \(\arg a\) about \(0\) and stretches it by \(|a|\), a spiral similarity.

Example. Multiply the triangle with corners \(1\), \(3\), \(2 + i\) by \(a = 1 + i\). Since \(|1 + i| = \sqrt2\) and \(\arg(1 + i) = 45^\circ\), the image is the same triangle turned \(45^\circ\) about \(0\) and enlarged by \(\sqrt 2\). The new corners are \((1+i)\cdot 1 = 1 + i\), \((1+i) \cdot 3 = 3 + 3i\) and \((1+i)(2+i) = 2 + i + 2i + i^2 = 1 + 3i\). Check: the side from \(1\) to \(3\) had length \(2\); the image side from \(1+i\) to \(3+3i\) has length \(|2 + 2i| = 2\sqrt2\).

Theorem 13.1. Every similarity of the plane that keeps orientation (a direct similarity) has the form \(z \mapsto az + b\) with \(a \neq 0\), and every one that reverses orientation has the form \(z \mapsto a\bar z + b\).

Proof. A direct similarity \(S\) with ratio \(k\) is determined by where it sends two points, say \(0 \mapsto b\) and \(1 \mapsto c\). The map \(T(z) = (c - b)z + b\) is a direct similarity with the same effect on \(0\) and \(1\). Then \(T^{-1} S\) is a direct similarity fixing \(0\) and \(1\), with ratio \(1\) (it keeps the distance from \(0\) to \(1\)), so it is an orientation-keeping isometry with two fixed points. By Chapter 10 such an isometry is a rotation about each fixed point, and a rotation fixing a second point is the identity. So \(S = T\). For an orientation-reversing \(S\), apply this to \(z \mapsto S(\bar z)\). \(\square\)

Angles and lines. For three points \(a, b, c\) with \(a \neq b\), the quotient

\[ \frac{c - a}{b - a} \]

has modulus \(|c - a|/|b - a|\) and argument equal to the angle turned from the direction \(a \to b\) to the direction \(a \to c\). So:

\(a, b, c\) are collinear exactly when \(\dfrac{c - a}{b - a}\) is real;
\(ac \perp ab\) exactly when \(\dfrac{c - a}{b - a}\) is purely imaginary;
triangles \(abc\) and \(a'b'c'\) are similar with the same orientation exactly when \(\dfrac{c - a}{b - a} = \dfrac{c' - a'}{b' - a'}\) (the two triangles have the same angle at \(a\) and the same ratio of the two sides there, which is SAS similarity from Chapter 5).

Equilateral triangles. Let \(\omega = e^{2\pi i/3} = -\tfrac12 + \tfrac{\sqrt3}{2} i\), a cube root of \(1\). It satisfies \(1 + \omega + \omega^2 = 0\) (because \(\omega^3 - 1 = (\omega - 1)(\omega^2 + \omega + 1) = 0\) and \(\omega \neq 1\)).

Lemma 13.2. Three points \(a, b, c\) form an equilateral triangle in counterclockwise order if and only if

\[ a + \omega b + \omega^2 c = 0 . \]

Proof. The triangle is equilateral and counterclockwise exactly when \(c\) is \(b\) turned \(60^\circ\) counterclockwise about \(a\): \(c - a = e^{i\pi/3}(b - a)\). Now \(e^{i\pi/3} = -\omega^2\) (both have modulus \(1\), and \(\arg(-\omega^2) = 180^\circ + 240^\circ = 420^\circ = 60^\circ\)). So the condition is \(c = a - \omega^2 b + \omega^2 a = (1 + \omega^2) a - \omega^2 b = -\omega a - \omega^2 b\), that is \(\omega a + \omega^2 b + c = 0\). Multiply by \(\omega^2\) and use \(\omega^3 = 1\): \(a + \omega b + \omega^2 c = 0\). Every step can be reversed. \(\square\)

Theorem 13.3 (Napoleon). The centres of the equilateral triangles built outward on the sides of any triangle form an equilateral triangle.

Proof. Label the triangle \(a, b, c\) counterclockwise. The outward equilateral triangle on side \(ab\) lies to the right of the direction \(a \to b\), so its third corner is \(b\) turned \(60^\circ\) clockwise about \(a\): \(p = a + e^{-i\pi/3}(b - a)\). Since \(e^{-i\pi/3} = -\omega\), this is \(p = a - \omega b + \omega a = (1 + \omega)a - \omega b = -\omega^2 a - \omega b\). The centre is the average of the corners:

\[ 3 g_1 = a + b + p = (1 - \omega^2) a + (1 - \omega) b . \]

In the same way \(3g_2 = (1 - \omega^2) b + (1 - \omega) c\) and \(3g_3 = (1 - \omega^2) c + (1 - \omega) a\). Now collect \(3(g_1 + \omega g_2 + \omega^2 g_3)\) by the letters:

\[ \begin{aligned} a&: \ (1 - \omega^2) + \omega^2(1 - \omega) = 1 - \omega^3 = 0, \\ b&: \ (1 - \omega) + \omega(1 - \omega^2) = 1 - \omega^3 = 0, \\ c&: \ \omega(1 - \omega) + \omega^2 (1 - \omega^2) = \omega - \omega^4 = \omega - \omega = 0 . \end{aligned} \]

So \(g_1 + \omega g_2 + \omega^2 g_3 = 0\), and by Lemma 13.2 the centres form an equilateral triangle. \(\square\)

Check. Take \(a = 0\), \(b = 4\), \(c = 1 + 3i\). The three centres come out at distance \(3.7764\ldots\) from each other, all three the same, as a short computation (below) confirms.

Try it in Python

import cmath, math
a, b, c = 0, 4, 1 + 3j                      # counterclockwise
rot = cmath.exp(-1j * math.pi / 3)          # turn 60 degrees clockwise
centres = [(p + q + p + rot * (q - p)) / 3 for p, q in [(a, b), (b, c), (c, a)]]
print([abs(centres[i] - centres[(i + 1) % 3]) for i in range(3)])

The method is always the same: name the points, write each condition as an equation in complex numbers, and let the algebra of \(\omega\), conjugates and quotients do the work. Its weakness is that formulas can grow large; its strength is that it never needs a clever extra line.

13.2 Inversion in a circle¶

Level: advanced. Reflection in a line swaps the two sides of the line and fixes the line itself. Is there a "reflection in a circle" that swaps inside and outside and fixes the circle? There is, and it is one of the most useful maps in geometry.

Definition. Let \(\Gamma\) be a circle with centre \(O\) and radius \(r\). Inversion in \(\Gamma\) sends each point \(P \neq O\) to the point \(P'\) on the ray from \(O\) through \(P\) with

\[ OP \cdot OP' = r^2 . \]

\(O\) is the centre of inversion and \(r\) its radius. Points on \(\Gamma\) stay put (\(OP = r\) gives \(OP' = r\)). A point inside, close to the circle, goes to a point outside, close to the circle; a point near \(O\) goes very far away. Doing the inversion twice gives back \(P\), since \(OP'' = r^2/OP' = OP\): inversion is its own inverse.

Constructing \(P'\). For \(P\) inside \(\Gamma\), draw the perpendicular to \(OP\) at \(P\); it meets \(\Gamma\) at \(T\). The tangent to \(\Gamma\) at \(T\) meets the ray \(OP\) at \(P'\). Why: the tangent is perpendicular to the radius \(OT\), so \(OTP'\) is a right triangle with altitude \(TP\). By the altitude theorem of Chapter 5 (triangles \(OPT\) and \(OTP'\) are similar, sharing the angle at \(O\), each with a right angle), \(OP/OT = OT/OP'\), so \(OP \cdot OP' = OT^2 = r^2\).

In complex numbers. Put \(O\) at \(0\). The point \(P'\) has the same argument as \(z\) and modulus \(r^2/|z|\), so

\[ z' = \frac{r^2}{|z|^2} z = \frac{r^2 z}{z \bar z} = \frac{r^2}{\bar z} . \]

Example. Invert \(z = 1 + i\) in the unit circle: \(z' = 1/\overline{(1+i)} = 1/(1 - i) = (1+i)/2\). Check: \(|z| \cdot |z'| = \sqrt2 \cdot \tfrac{\sqrt2}{2} = 1\), and both lie on the line \(y = x\).

Inversion is not defined at \(O\), and nothing goes to \(O\). It is natural to add one point at infinity, written \(\infty\), and to say that \(O\) and \(\infty\) are swapped. We make this precise in 13.6.

How inversion changes distances. Let \(P, Q\) go to \(P', Q'\) (with \(O\), \(P\), \(Q\) not on one line). From \(OP \cdot OP' = OQ \cdot OQ' = r^2\) we get \(OP/OQ = OQ'/OP'\), and the triangles \(OPQ\) and \(OQ'P'\) share the angle at \(O\). By SAS similarity they are similar, with \(P'Q'/PQ = OQ'/OP = r^2/(OP \cdot OQ)\). So

\[ P'Q' = \frac{r^2 \cdot PQ}{OP \cdot OQ} . \]

(When \(O, P, Q\) are collinear the same formula follows from \(OP' = r^2/OP\) by subtraction.) Note also from the similarity that \(\angle OPQ = \angle OQ'P'\): inversion swaps angles in this mirror-like way. We shall use the distance formula for Ptolemy's theorem in 13.5.

Lines and circles¶

The main theorem says that inversion sends lines and circles to lines and circles. Precisely:

Theorem 13.4. Under inversion with centre \(O\):

a line through \(O\) goes to itself;
a line not through \(O\) goes to a circle through \(O\);
a circle through \(O\) goes to a line not through \(O\);
a circle not through \(O\) goes to a circle not through \(O\).

(In each case, "goes to" means the image of the figure with \(O\) removed, plus or minus the point \(O\) itself.)

First proof (geometric), for case 2. Let \(\ell\) be a line not through \(O\), and let \(A\) be the foot of the perpendicular from \(O\) to \(\ell\). Take any other point \(B\) of \(\ell\), with images \(A'\) and \(B'\). As above, triangles \(OAB\) and \(OB'A'\) are similar, so \(\angle OB'A' = \angle OAB = 90^\circ\). So \(B'\) sees the fixed segment \(OA'\) at a right angle, and by Thales' theorem (Chapter 6) \(B'\) lies on the circle with diameter \(OA'\). Conversely every point of that circle except \(O\) comes from a point of \(\ell\), by the same argument run backwards. Case 3 is case 2 read in reverse, since inversion is its own inverse. Case 1 is clear from the definition.

For case 4, let the circle \(\gamma\) not pass through \(O\), and let a line through \(O\) meet it at \(P\) and \(Q\). By the power of a point (Chapter 6), \(OP \cdot OQ = p\), a constant (the power of \(O\) with respect to \(\gamma\), taken positive, if \(O\) is outside \(\gamma\)). Then \(OP' = r^2/OP = (r^2/p)\, OQ\). So \(P'\) is the image of \(Q\) under the scaling about \(O\) by the factor \(r^2/p\); as \(Q\) runs over \(\gamma\), \(P'\) runs over the scaled copy of \(\gamma\), which is a circle. (If \(O\) is inside \(\gamma\), the same works with \(P\) and \(Q\) on opposite sides of \(O\), using the scaling by \(-r^2/p\).) \(\square\)

Second proof (algebraic), all cases at once. Every circle and every line can be written as

\[ \alpha z \bar z + \beta \bar z + \bar\beta z + \gamma = 0, \qquad \alpha, \gamma \text{ real}, \ \beta \text{ complex}. \]

For \(\alpha \neq 0\) this is a circle: divide by \(\alpha\) and complete the square to get \(|z + \beta/\alpha|^2 = |\beta|^2/\alpha^2 - \gamma/\alpha\) (a real circle when the right side is positive). For \(\alpha = 0\) it is \(2\,\mathrm{Re}(\bar\beta z) + \gamma = 0\), a line. The curve passes through \(0\) exactly when \(\gamma = 0\). Now invert in the unit circle: \(w = 1/\bar z\), so \(z = 1/\bar w\), \(\bar z = 1/w\) and \(z \bar z = 1/(w \bar w)\). Substituting and multiplying by \(w \bar w\):

\[ \alpha + \beta w + \bar\beta \bar w + \gamma w \bar w = 0 . \]

This is the same kind of equation with \(\alpha\) and \(\gamma\) swapped. So circles and lines go to circles and lines, and "passing through \(0\)" (\(\gamma = 0\)) is swapped with "being a line" (\(\alpha = 0\)), which is exactly the list above. \(\square\)

The algebraic proof is worth a moment's thought. Lines and circles are one family, the generalised circles, and a line is simply a circle through the point \(\infty\).

Example. Invert the circle \(|z - 2| = 1\) in the unit circle. It is symmetric in the real axis and crosses it at \(1\) and \(3\), so its image is a circle symmetric in the real axis crossing it at \(1/1 = 1\) and \(1/3\). Hence the image has diameter from \(\tfrac13\) to \(1\): centre \(\tfrac23\), radius \(\tfrac13\). Notice that the centre \(2\) did not go to the centre: \(2\) goes to \(\tfrac12\), not \(\tfrac23\). In general, inverting the circle with centre \(c\) and radius \(s\) (not through \(O = 0\)) in the circle of radius \(r\) gives the circle with

\[ \text{centre } \frac{r^2 c}{|c|^2 - s^2}, \qquad \text{radius } \frac{r^2 s}{\big||c|^2 - s^2\big|} . \]

Check with \(c = 2\), \(s = 1\), \(r = 1\): centre \(2/3\), radius \(1/3\).

Angles are kept¶

Theorem 13.5. Inversion preserves the size of the angle between two curves at a point (other than \(O\)), and reverses its sense.

Proof for lines and circles. Let two lines \(\ell_1, \ell_2\) meet at \(P \neq O\) at angle \(\theta\). If they do not pass through \(O\), their images are circles through \(O\) and \(P'\). The image circle of \(\ell_1\) has its centre on the perpendicular from \(O\) to \(\ell_1\) (by the first proof above), so its tangent at \(O\) is parallel to \(\ell_1\). The same holds for \(\ell_2\). So the two image circles cross at \(O\) at angle \(\theta\). Two circles meeting at two points meet at the same angle at both (the figure is symmetric in the line through the two centres), so they also cross at \(P'\) at angle \(\theta\). If a line passes through \(O\) it is its own image and a similar, easier argument applies. For curves in general, the angle between curves is defined as the angle between their tangent lines; inversion is smooth away from \(O\), so it carries tangent lines to tangent circles of the images, and the result for lines carries over. The reversal of sense comes from the mirror symmetry: inversion is the identity on \(\Gamma\) but swaps its two sides, just as a reflection does. \(\square\)

A map that keeps angles is called conformal. Inversions are conformal but reverse orientation, like reflections. In 1850 Joseph Liouville proved a striking converse in three dimensions: every conformal map of a region of space is a composition of inversions in spheres, reflections and similarities. In the plane there are far more conformal maps (every complex differentiable function with nonzero derivative is one), which is why complex analysis is so rich.

Tangency is kept. Two curves that touch (meet at angle \(0\)) still touch after inversion. That is the key to the next section.

A little history. Jakob Steiner used inversion in unpublished work in the 1820s, and several others found it independently in the following decades; William Thomson (later Lord Kelvin) used it in the 1840s to solve problems of electric charge on conductors, his "method of images". The name and the systematic theory grew through the 19th century.

13.3 Apollonius' problem and Steiner chains¶

Level: advanced. Apollonius of Perga (about 200 BC) wrote a book, Tangencies, now lost, that solved this problem: given three circles, draw a circle touching all three. We know of the book from a summary by Pappus of Alexandria. The ten special cases where some of the "circles" are points or lines are included. Inversion turns the hardest case into the easiest.

How many answers? A circle touching a given circle can touch it from outside or contain it (or be contained by it). With three circles there are \(2^3 = 8\) choices of "inside or outside" for each, and for three circles that are far apart and lie outside each other, all eight choices give exactly one solution each: eight solutions in general. For special positions there can be fewer (or, in degenerate cases, infinitely many).

The method. Suppose the three circles \(C_1, C_2, C_3\) have centres \(O_1, O_2, O_3\) and radii \(r_1 \le r_2, r_3\), and lie outside one another. Look for a circle \(K\) with centre \(X\) and radius \(\rho\) that touches all three from outside, so \(XO_i = \rho + r_i\).

Shrink. Reduce all three radii by \(r_1\) and enlarge \(\rho\) by \(r_1\). The distances \(XO_i = (\rho + r_1) + (r_i - r_1)\) are unchanged, so the circle with centre \(X\) and radius \(\rho + r_1\) touches the two shrunken circles (radii \(r_2 - r_1\), \(r_3 - r_1\)) and passes through the point \(O_1\). We now need a circle through a point touching two circles.
Invert about \(O_1\). The wanted circle passes through \(O_1\), so it becomes a line. The two other circles become circles (they do not pass through \(O_1\)). Tangency is kept, so the line touches both image circles: it is a common tangent of two circles, and two circles outside each other have four of them, easy to construct.
Invert back each common tangent to get a circle through \(O_1\), and then undo the shrinking.

The other sign choices (touching some circles from inside) are handled the same way, enlarging some radii instead of shrinking them. This approach, reducing a circle to a point, goes back to François Viète, who published a solution of Apollonius' problem in 1600; the use of inversion came later. Since each step uses only a straightedge and compass, every solution of Apollonius' problem is constructible, the question taken up in Chapter 14.

Descartes' circle theorem. A special case has a beautiful formula. If four circles each touch the other three, at six distinct points, write their curvatures \(k_i = 1/r_i\), with a minus sign for a circle that encloses the others. Then

\[ (k_1 + k_2 + k_3 + k_4)^2 = 2(k_1^2 + k_2^2 + k_3^2 + k_4^2) , \]

so that, given three, \(k_4 = k_1 + k_2 + k_3 \pm 2\sqrt{k_1 k_2 + k_2 k_3 + k_3 k_1}\). René Descartes stated it in a 1643 letter to Princess Elisabeth of Bohemia; Frederick Soddy rediscovered it in 1936 and published it as a poem, "The Kiss Precise", in Nature. We will not prove it here. One proof inverts about a point of tangency, so that two of the circles become parallel lines and the other two become equal circles between them; the article by Lagarias, Mallows and Wilks in the further reading gives proofs and far-reaching generalisations.

Example. Three unit circles touch each other. Then \(k_1 = k_2 = k_3 = 1\) and \(k_4 = 3 \pm 2\sqrt3\). The plus sign gives the small circle in the middle, of radius \(1/(3 + 2\sqrt3) = (2\sqrt3 - 3)/3 \approx 0.1547\). Check by plain geometry: the centres form an equilateral triangle of side \(2\), whose centre is at distance \(2/\sqrt3 \approx 1.1547\) from each corner, so the middle circle has radius \(1.1547 - 1 = 0.1547\). The minus sign gives \(k_4 = 3 - 2\sqrt3 < 0\), the big enclosing circle, of radius \(1/(2\sqrt3 - 3) = 1 + 2/\sqrt3 \approx 2.1547\).

Steiner's porism¶

Take two circles, one inside the other, not touching. Draw a circle in the ring between them touching both; then another touching both and the first; then another, and so on around the ring. Usually the chain does not close up: the last circle overlaps the first. But sometimes it closes exactly, with the last circle touching the first.

Theorem 13.6 (Steiner's porism). If a chain closes for one choice of first circle, it closes, with the same number of circles, for every choice of first circle.

(A porism is an old word for a statement of this kind: either the construction never works, or it works for every starting point.)

The proof needs one lemma.

Lemma 13.7. Two circles, one inside the other and not touching, can be inverted into two concentric circles.

Proof. Put the line through their centres along the real axis, with circles \(|z - c_1| = s_1\) and \(|z - c_2| = s_2\). We look for two points \(p, q\) on the real axis that are inverse to each other in both circles, that is \((p - c_i)(q - c_i) = s_i^2\) for \(i = 1, 2\). Subtracting the two equations gives a linear relation between \(p + q\) and \(c_1, c_2\); substituting back gives a quadratic for \(p\) whose discriminant is positive exactly when the circles are nested or separate (not crossing). These two points are called the limiting points of the pair. Now invert about \(p\). Every circle through \(p\) and \(q\) is orthogonal to both given circles (a circle through two points inverse in a circle \(\gamma\) is mapped to itself by inversion in \(\gamma\), so it meets \(\gamma\) at right angles). After inverting at \(p\), the circles through \(p\) and \(q\) become lines through \(q'\), the image of \(q\). The images of the two given circles are orthogonal to all lines through \(q'\), so both are circles centred at \(q'\): concentric. \(\square\)

Proof of Theorem 13.6. Invert the two circles into concentric ones, with radii \(a < b\). Tangency is kept, so a Steiner chain becomes a Steiner chain between concentric circles. There, each chain circle touches both, so it has diameter \(b - a\) and its centre lies on the circle of radius \((a + b)/2\). A chain is just a set of equal circles around a ring, and rotating it about the common centre gives another chain. The chain closes with \(n\) circles exactly when the angle each circle takes up, seen from the centre, is \(2\pi/n\), that is when

\[ \sin \frac{\pi}{n} = \frac{(b - a)/2}{(a + b)/2} = \frac{b - a}{b + a}, \]

a condition on \(a\) and \(b\) alone, not on where the chain starts. Invert back. \(\square\)

Example. For \(n = 6\) we need \((b - a)/(b + a) = \sin 30^\circ = \tfrac12\), so \(b = 3a\): six circles of radius \(1\) fit exactly in the ring between radii \(1\) and \(3\), the left half of the figure. The right half is its image under an inversion whose centre lies inside the small circle; every tangency survives.

13.4 Möbius transformations and the cross-ratio¶

Level: advanced to expert. Combine the maps we have: translations \(z \mapsto z + b\), rotations and scalings \(z \mapsto az\), and the map \(z \mapsto 1/z\). (This last one is inversion in the unit circle, \(z \mapsto 1/\bar z\), followed by reflection in the real axis.) What do all their compositions look like?

Definition. A Möbius transformation is a map

\[ f(z) = \frac{az + b}{cz + d}, \qquad a, b, c, d \in \mathbb{C}, \quad ad - bc \neq 0 . \]

It acts on the extended plane \(\hat{\mathbb{C}} = \mathbb{C} \cup \{\infty\}\), with the conventions \(f(-d/c) = \infty\) and \(f(\infty) = a/c\) when \(c \neq 0\), and \(f(\infty) = \infty\) when \(c = 0\). The condition \(ad - bc \neq 0\) makes sure \(f\) is not constant.

Theorem 13.8. Every Möbius transformation is a composition of translations, rotation-scalings and \(z \mapsto 1/z\). Hence it is a bijection of \(\hat{\mathbb{C}}\), it sends generalised circles to generalised circles, and it preserves angles and their sense.

Proof. If \(c = 0\), \(f(z) = (a/d) z + b/d\), a rotation-scaling followed by a translation. If \(c \neq 0\), divide:

\[ \frac{az + b}{cz + d} = \frac{a}{c} - \frac{ad - bc}{c} \cdot \frac{1}{cz + d} , \]

which is: \(z \mapsto cz + d\), then \(w \mapsto 1/w\), then multiply by \(-(ad - bc)/c\), then add \(a/c\). Each step sends generalised circles to generalised circles (Theorem 13.4 and the reflection) and is conformal; \(1/z\) is two orientation-reversing maps in a row, so it keeps orientation. \(\square\)

Matrices. Write \(f\) as the matrix \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\). A direct calculation shows that composing two Möbius maps corresponds to multiplying their matrices, and that the matrices \(M\) and \(\lambda M\) (\(\lambda \neq 0\)) give the same map. So the Möbius transformations form a group, isomorphic to \(\mathrm{PGL}(2, \mathbb{C})\), the invertible \(2 \times 2\) complex matrices modulo scalars. Scaling so that \(ad - bc = 1\) identifies it with \(\mathrm{PSL}(2, \mathbb{C})\). (Groups and quotients are in the Algebra book, Chapter 13 and Chapter 15.)

Three points decide everything¶

Theorem 13.9. Given three distinct points \(z_1, z_2, z_3\) of \(\hat{\mathbb{C}}\), there is exactly one Möbius transformation sending them to \(0, 1, \infty\) in that order. For finite \(z_i\) it is

\[ f(z) = \frac{(z - z_1)(z_2 - z_3)}{(z - z_3)(z_2 - z_1)} . \]

Proof. Existence: \(f(z_1) = 0\) because of the factor \(z - z_1\); \(f(z_3) = \infty\) because of \(z - z_3\) in the denominator; and \(f(z_2) = 1\) because the factors cancel. (If one \(z_i\) is \(\infty\), delete the two factors containing it.) Uniqueness: if \(g\) also works, then \(h = g \circ f^{-1}\) fixes \(0\), \(1\) and \(\infty\). Write \(h(z) = (az + b)/(cz + d)\). Fixing \(\infty\) forces \(c = 0\); fixing \(0\) forces \(b = 0\); so \(h(z) = (a/d) z\), and fixing \(1\) forces \(a = d\). So \(h\) is the identity and \(g = f\). \(\square\)

Corollary. For any two triples of distinct points there is exactly one Möbius transformation sending the first to the second (send the first to \(0, 1, \infty\), then follow the inverse of the map that sends the second there). And since a generalised circle is fixed by any three of its points, any generalised circle can be moved to any other.

Definition. The cross-ratio of four points is the value at \(z\) of the map of Theorem 13.9:

\[ (z, z_1; z_2, z_3) = \frac{(z - z_1)(z_2 - z_3)}{(z - z_3)(z_2 - z_1)} . \]

Be warned that books order the four points in several different ways, so their formulas differ by a permutation; the facts below hold in every convention. The projective cross-ratio of Chapter 18 is the same idea on a line.

Theorem 13.10. Möbius transformations preserve the cross-ratio: \((Tz, Tz_1; Tz_2, Tz_3) = (z, z_1; z_2, z_3)\).

Proof. Let \(f\) send \(z_1, z_2, z_3\) to \(0, 1, \infty\). Then \(f \circ T^{-1}\) sends \(Tz_1, Tz_2, Tz_3\) to \(0, 1, \infty\), so by uniqueness it is the map that defines the cross-ratio of \(Tz\) with \(Tz_1, Tz_2, Tz_3\). Its value at \(Tz\) is \(f(T^{-1}Tz) = f(z)\). \(\square\)

Theorem 13.11. Four distinct points lie on one generalised circle if and only if their cross-ratio is real.

Proof. Let \(f\) send \(z_1, z_2, z_3\) to \(0, 1, \infty\). It sends the generalised circle through \(z_1, z_2, z_3\) to the generalised circle through \(0, 1, \infty\), which is the real line (with \(\infty\)). So \(z\) lies on the first exactly when \(f(z)\) is real, and \(f(z)\) is the cross-ratio. \(\square\)

A worked example: from the disc to the half-plane¶

Find the Möbius map sending \(1, i, -1\) to \(0, 1, \infty\). By the formula,

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

It sends the unit circle (through \(1, i, -1\)) to the real line. Which side goes where? Compute \(f(0)\):

\[ f(0) = \frac{-(1 + i)}{i - 1} = \frac{-(1 + i)(-1 - i)}{(i - 1)(-1 - i)} = \frac{(1 + i)^2}{2} = \frac{2i}{2} = i . \]

So the inside of the disc goes to the upper half-plane \(\{\mathrm{Im}\, w > 0\}\). Circles about \(0\) become circles around \(i\) (not centred at \(i\)), and diameters become arcs through \(i\) that meet the real line at right angles, because angles are kept and diameters meet the unit circle at right angles. Check: \(f(-i) = -1\), real, as it must be for a point of the unit circle; so the cross-ratio \((-i, 1; i, -1) = -1\) and the four points \(1, i, -1, -i\) are concyclic. This map will carry the Poincaré disc to the half-plane in Chapter 17.

Fixed points and the four kinds¶

Level: expert. A Möbius map other than the identity has one or two fixed points in \(\hat{\mathbb{C}}\), the roots of \(cz^2 + (d - a)z - b = 0\) (with \(\infty\) fixed when \(c = 0\)). Normalise so that \(ad - bc = 1\), and let \(t = a + d\) be the trace (defined up to sign). Then:

parabolic if \(t = \pm 2\): one fixed point; conjugate to a translation \(z \mapsto z + 1\);
elliptic if \(t\) is real with \(|t| < 2\): two fixed points; conjugate to a rotation \(z \mapsto e^{i\theta} z\);
hyperbolic if \(t\) is real with \(|t| > 2\): conjugate to a scaling \(z \mapsto kz\), \(k > 0\);
loxodromic otherwise: conjugate to a spiral \(z \mapsto kz\) with \(k\) not real and not of modulus \(1\).

Why: if there are two fixed points, move them to \(0\) and \(\infty\); a map fixing both is \(z \mapsto kz\), with matrix conjugate to \(\mathrm{diag}(\sqrt k, 1/\sqrt k)\) and trace \(\sqrt k + 1/\sqrt k\). That trace is real with modulus below \(2\) exactly when \(|k| = 1\), and real with modulus above \(2\) exactly when \(k\) is real and positive (not \(1\)). With one fixed point, move it to \(\infty\); the map is \(z \mapsto z + b\), conjugate to \(z \mapsto z + 1\), with trace \(\pm 2\).

Example. \(f(z) = 1/(2 - z)\), matrix \(\begin{pmatrix} 0 & 1 \\ -1 & 2 \end{pmatrix}\), determinant \(1\), trace \(2\): parabolic. Fixed points: \(z(2 - z) = 1\), so \(z^2 - 2z + 1 = 0\), a double root at \(z = 1\). Check: \(f(1) = 1\).

The images of circles under groups of Möbius maps make the astonishing fractal pictures in Indra's Pearls by Mumford, Series and Wright; that is where this subject meets current research on Kleinian groups.

13.5 Ptolemy again, via inversion¶

Level: advanced. In Chapter 12 you met Ptolemy's theorem: for a quadrilateral \(ABCD\) inscribed in a circle,

\[ AB \cdot CD + AD \cdot BC = AC \cdot BD . \]

Claudius Ptolemy (2nd century AD) used it in the Almagest to compute his table of chords, the ancestor of the sine table. Here are two short proofs, which also give more: an inequality for every four points.

Theorem 13.12 (Ptolemy's inequality). For any four points \(A, B, C, D\) in the plane,

\[ AB \cdot CD + AD \cdot BC \ \ge \ AC \cdot BD , \]

with equality exactly when \(A, B, C, D\) lie in this order on a circle (or on a line, with \(C\) between \(B\) and \(D\) seen from \(A\); more on this below).

Proof by inversion. Invert about \(A\) with radius \(1\). By the distance formula of 13.2,

\[ B'C' = \frac{BC}{AB \cdot AC}, \qquad C'D' = \frac{CD}{AC \cdot AD}, \qquad B'D' = \frac{BD}{AB \cdot AD} . \]

The triangle inequality for \(B', C', D'\) says \(B'C' + C'D' \ge B'D'\):

\[ \frac{BC}{AB \cdot AC} + \frac{CD}{AC \cdot AD} \ \ge \ \frac{BD}{AB \cdot AD} . \]

Multiply both sides by \(AB \cdot AC \cdot AD\): \(BC \cdot AD + CD \cdot AB \ge BD \cdot AC\), which is the inequality. Equality holds exactly when \(C'\) lies on the segment \(B'D'\). By Theorem 13.4, \(B', C', D'\) lie on a line not through \(A\) exactly when \(B, C, D\) lie on a circle through \(A\), and \(C'\) is between \(B'\) and \(D'\) exactly when \(C\) is on the arc \(BD\) that does not contain \(A\), that is, when \(A, B, C, D\) are in this order around the circle. \(\square\)

Proof by complex numbers. For any four complex numbers there is an identity, which you can check by expanding both sides:

\[ (a - b)(c - d) + (a - d)(b - c) = (a - c)(b - d) . \]

Take moduli and use the triangle inequality \(|u + v| \le |u| + |v|\): \(|a - c||b - d| \le |a - b||c - d| + |a - d||b - c|\), which is Ptolemy's inequality again. Equality in \(|u + v| \le |u| + |v|\) holds when \(u\) and \(v\) point the same way, that is when

\[ \frac{(a - b)(c - d)}{(a - d)(b - c)} \]

is a positive real number. This quotient is a cross-ratio of \(a, b, c, d\) (up to the ordering convention), so by Theorem 13.11 it is real exactly when the four points are concyclic; the sign sorts out the order. \(\square\)

Examples. For a square of side \(1\): \(1 \cdot 1 + 1 \cdot 1 = \sqrt2 \cdot \sqrt2\). For a \(3 \times 4\) rectangle, whose diagonals are \(5\): \(3 \cdot 3 + 4 \cdot 4 = 5 \cdot 5\), which is Pythagoras' theorem. For a regular pentagon with side \(1\) and diagonal \(d\), take four of its vertices: three sides are \(1\), the fourth side and both diagonals are \(d\), and Ptolemy gives \(d \cdot d = 1 \cdot d + 1 \cdot 1\), so \(d^2 = d + 1\) and \(d = (1 + \sqrt5)/2\), the golden ratio of Chapter 5.

13.6 The Riemann sphere¶

Level: advanced to expert. We have been adding a point \(\infty\) to the plane by decree. Is there a place where it really lives, as an ordinary point? Yes: wrap the plane onto a sphere.

Stereographic projection. Let \(S\) be the unit sphere \(X^2 + Y^2 + Z^2 = 1\) and \(N = (0, 0, 1)\) its north pole. Identify the plane \(Z = 0\) with \(\mathbb{C}\). For each point \(P \neq N\) of the sphere, the line from \(N\) through \(P\) meets the plane at one point \(z\); this is stereographic projection from \(N\). Points of the southern hemisphere go inside the unit circle, the equator stays where it is, points of the northern hemisphere go outside, and as \(P\) approaches \(N\), \(z\) runs off to infinity. So we send \(N\) to \(\infty\), and the sphere becomes a picture of \(\hat{\mathbb{C}}\): the Riemann sphere.

Formulas. The line from \(N\) through \(P = (X, Y, Z)\) is \(N + t(P - N) = (tX, tY, 1 + t(Z - 1))\). It reaches \(Z = 0\) when \(t = 1/(1 - Z)\), so

\[ z = \frac{X + iY}{1 - Z} . \]

Going back, for \(z = x + iy\):

\[ (X, Y, Z) = \left( \frac{2x}{|z|^2 + 1}, \ \frac{2y}{|z|^2 + 1}, \ \frac{|z|^2 - 1}{|z|^2 + 1} \right) . \]

Check: the sum of squares is \((4|z|^2 + (|z|^2 - 1)^2)/(|z|^2 + 1)^2 = (|z|^2 + 1)^2/(|z|^2 + 1)^2 = 1\), so the point lies on the sphere; and \(1 - Z = 2/(|z|^2 + 1)\), so \((X + iY)/(1 - Z) = x + iy\).

Theorem 13.13. Stereographic projection sends circles on the sphere to generalised circles in the plane: to lines if the circle passes through \(N\), to circles otherwise. It also preserves angles.

Proof of the first part. A circle on the sphere is its intersection with a plane \(aX + bY + cZ = d\). Put in the formulas for \(X, Y, Z\) and multiply by \(|z|^2 + 1\):

\[ 2ax + 2by + c(|z|^2 - 1) = d(|z|^2 + 1), \quad\text{that is}\quad (c - d)|z|^2 + 2ax + 2by - (c + d) = 0 . \]

If \(c \neq d\) this is a circle (or empty, or a point, but a real circle on the sphere gives a real image). If \(c = d\) it is a line, and \(c = d\) says exactly that the plane passes through \(N = (0, 0, 1)\). \(\square\)

The angle part is true too; one proof shows that stereographic projection is the restriction of an inversion in space, in the sphere with centre \(N\) and radius \(\sqrt2\), and inversions in space are conformal just as in the plane.

Rotations are Möbius maps. Turn the sphere through half a turn about the \(X\)-axis: \((X, Y, Z) \mapsto (X, -Y, -Z)\). What does this do in the plane? The point \(z = (X + iY)/(1 - Z)\) goes to \((X - iY)/(1 + Z)\). Compare \(1/z = (1 - Z)/(X + iY) = (1 - Z)(X - iY)/(X^2 + Y^2) = (1 - Z)(X - iY)/(1 - Z^2) = (X - iY)/(1 + Z)\). The same. So the mysterious map \(z \mapsto 1/z\), which swaps \(0\) and \(\infty\), is simply a half turn of the Riemann sphere swapping the poles. In general (we state this without proof) every rotation of the sphere becomes a Möbius map of the form

\[ z \mapsto \frac{az + b}{-\bar b z + \bar a}, \qquad |a|^2 + |b|^2 = 1, \]

and this is the link between the rotation group \(\mathrm{SO}(3)\) and the group \(\mathrm{SU}(2)\) of \(2 \times 2\) unitary matrices that physicists use to describe the spin of an electron.

Maps of planets. Because it keeps angles and sends circles to circles, stereographic projection is a favourite for maps of polar regions: polar maps of the Earth, and of Mars's polar ice caps, are often drawn in the polar stereographic projection. Chapter 16 explains why no flat map can be perfect, and what each projection gives up.

History. Stereographic projection is ancient: it is the geometry of the astrolabe, and Ptolemy wrote about it. That it keeps angles was proved much later: Thomas Harriot found it in unpublished notes around 1590, and Edmond Halley published a proof in the 1690s. August Ferdinand Möbius studied the transformations that bear his name in the 1850s, and Bernhard Riemann made the sphere the natural home of complex functions in the same decade.

Check yourself¶

1. Invert the point \(3 + 4i\) in the unit circle. Check your answer.

The image is \(1/\bar z = 1/(3 - 4i) = (3 + 4i)/25 = 0.12 + 0.16i\). Check: it lies on the same ray (a positive multiple of \(3 + 4i\)), and \(|3 + 4i| \cdot |0.12 + 0.16i| = 5 \cdot 0.2 = 1\).

2. The points \(0\) and \(2\) are two corners of an equilateral triangle, labelled counterclockwise. Find the third corner, and check Lemma 13.2.

Turn \(2\) by \(60^\circ\) counterclockwise about \(0\): \(c = 2e^{i\pi/3} = 1 + \sqrt3\, i\). Check: \(0 + \omega \cdot 2 + \omega^2(1 + \sqrt3 i)\) with \(\omega = -\tfrac12 + \tfrac{\sqrt3}{2} i\) and \(\omega^2 = -\tfrac12 - \tfrac{\sqrt3}{2} i\). Then \(2\omega = -1 + \sqrt3 i\) and \(\omega^2 (1 + \sqrt3 i) = -\tfrac12 - \tfrac{\sqrt3}{2} i - \tfrac{\sqrt3}{2} i - \tfrac{3}{2} i^2 = 1 - \sqrt3 i\). The sum is \(0\).

3. What does inversion in the unit circle do to the line \(y = 1\)?

The line does not pass through \(0\), so it becomes a circle through \(0\). Its closest point to \(0\) is \(i\), which is on the unit circle and stays fixed; it goes to the far end of a diameter, so the image is the circle with diameter from \(0\) to \(i\): centre \(i/2\), radius \(1/2\), equation \(x^2 + y^2 = y\). Check with the formula: if \(w = u + iv\) is the image of \(z\), then \(z = w/|w|^2\), and \(\mathrm{Im}\, z = 1\) means \(v/(u^2 + v^2) = 1\), that is \(u^2 + v^2 = v\).

4. Invert the circle \(|z - 3| = 1\) in the unit circle. Give the centre and radius of the image.

The circle crosses the real axis at \(2\) and \(4\), which go to \(\tfrac12\) and \(\tfrac14\). The image is symmetric in the real axis, so these are the ends of a diameter: centre \(\tfrac38\), radius \(\tfrac18\). The formula agrees: centre \(c/(|c|^2 - s^2) = 3/8\), radius \(s/(|c|^2 - s^2) = 1/8\).

5. Show that \(z \mapsto 1/z\) sends the unit circle to itself. Where does it send the upper half-plane?

If \(|z| = 1\) then \(|1/z| = 1\). Also \(1/z = \bar z/|z|^2\), so \(\mathrm{Im}(1/z) = -\mathrm{Im}(z)/|z|^2\), which has the opposite sign. So the upper half-plane goes to the lower half-plane. (On the Riemann sphere this is the half turn about the \(X\)-axis from 13.6, which swaps the two halves of the sphere cut by the plane \(Y = 0\).)

6. Find the Möbius transformation sending \(0, 1, \infty\) to \(1, i, -1\).

Write \(f(z) = (az + b)/(cz + d)\). From \(f(\infty) = a/c = -1\), \(a = -c\). From \(f(0) = b/d = 1\), take \(b = d = 1\). From \(f(1) = (a + 1)/(c + 1) = i\) with \(a = -c\): \(1 - c = i(c + 1)\), so \(c = (1 - i)/(1 + i) = -i\) and \(a = i\). So \(f(z) = (iz + 1)/(-iz + 1)\). Check: \(f(1) = (1 + i)/(1 - i) = (1 + i)^2/2 = i\). This is the inverse of the map in the worked example of 13.4.

7. Use the cross-ratio to decide whether \(0\), \(1\), \(1 + i\) and \(i\) lie on one circle, and also \(0\), \(1\), \(2 + i\), \(i\).

Compute \((z, z_1; z_2, z_3)\) with \(z = i\), \(z_1 = 0\), \(z_2 = 1\), \(z_3 = 1 + i\): \(\dfrac{(i - 0)(1 - 1 - i)}{(i - 1 - i)(1 - 0)} = \dfrac{i \cdot (-i)}{-1} = -1\), real: concyclic (they are the corners of a square). With \(z_3 = 2 + i\): \(\dfrac{i(1 - 2 - i)}{(i - 2 - i) \cdot 1} = \dfrac{i(-1 - i)}{-2} = \dfrac{1 - i}{-2}\), not real: not concyclic.

8. A regular pentagon has side \(1\). Use Ptolemy's theorem to find its diagonal.

Take four vertices \(A, B, C, D\) in order. Then \(AB = BC = CD = 1\), while \(AD\), \(AC\) and \(BD\) are diagonals, all equal to \(d\). Ptolemy: \(AB \cdot CD + AD \cdot BC = AC \cdot BD\) gives \(1 + d = d^2\), so \(d = (1 + \sqrt5)/2 \approx 1.618\).

9. Three circles of radii \(1\), \(2\) and \(3\) touch each other from outside. Find the radii of the two circles touching all three.

Curvatures \(1, \tfrac12, \tfrac13\), with sum \(\tfrac{11}{6}\) and \(k_1k_2 + k_2k_3 + k_3k_1 = \tfrac12 + \tfrac16 + \tfrac13 = 1\). So \(k_4 = \tfrac{11}{6} \pm 2\). The plus sign gives \(\tfrac{23}{6}\): a small circle of radius \(\tfrac{6}{23}\) in the gap. The minus sign gives \(-\tfrac16\): a circle of radius \(6\) enclosing all three. Check the second: put the centres at \(A = (0, 0)\), \(B = (3, 0)\), \(C = (0, 4)\) (the distances are \(1 + 2 = 3\), \(1 + 3 = 4\), \(2 + 3 = 5\)). The point \(X = (3, 4)\) is at distances \(5, 4, 3\) from them, which are \(6 - 1\), \(6 - 2\), \(6 - 3\): the circle of radius \(6\) about \(X\) touches all three from outside them.

10. For which ratio \(b/a\) of the radii of two concentric circles does a Steiner chain of \(4\) circles close?

We need \((b - a)/(b + a) = \sin(\pi/4) = \tfrac{\sqrt2}{2}\). Solving, \(b/a = (1 + s)/(1 - s)\) with \(s = \tfrac{\sqrt2}{2}\), which simplifies to \(3 + 2\sqrt2 \approx 5.83\).

11. Show that inversion in the circle of radius \(r\) about \(0\), followed by inversion in the circle of radius \(s\) about \(0\), is a scaling. By what factor?

The first sends \(z\) to \(r^2/\bar z\). The second sends \(w\) to \(s^2/\bar w\), and \(\bar w = r^2/z\), so the result is \(s^2 z/r^2\). It is the scaling about \(0\) by \((s/r)^2\). (Two reflections in parallel lines make a translation; two inversions in concentric circles make a scaling.)

12. Which point of the unit sphere does stereographic projection send to \(z = 1 + i\)? Which point goes to \(0\), and which to the unit circle?

\(|z|^2 = 2\), so \((X, Y, Z) = (2/3, 2/3, 1/3)\). Check: \(4/9 + 4/9 + 1/9 = 1\), and \((X + iY)/(1 - Z) = (2/3)(1 + i)/(2/3) = 1 + i\). The south pole \((0, 0, -1)\) goes to \(0\), and the points with \(Z = 0\) (the equator) go to the unit circle, since there \(z = X + iY\) with \(X^2 + Y^2 = 1\).

What you learned¶

Complex numbers turn geometry into algebra: distance is \(|z - w|\), multiplying by \(a\) turns by \(\arg a\) and scales by \(|a|\), and the quotient \((c - a)/(b - a)\) records the angle at \(a\); Napoleon's theorem follows in a few lines.
Inversion \(z \mapsto r^2/\bar z\) swaps inside and outside of a circle, sends generalised circles (lines and circles) to generalised circles, keeps angles and reverses their sense.
Inversion turns hard tangency problems into easy ones: Apollonius' problem becomes a common tangent problem, and Steiner chains become rings of equal circles, which proves Steiner's porism.
Möbius maps \((az + b)/(cz + d)\) form a group; three points fix one; they preserve the cross-ratio, and four points are concyclic exactly when their cross-ratio is real.
Ptolemy's inequality \(AB \cdot CD + AD \cdot BC \ge AC \cdot BD\) holds for any four points, with equality for a cyclic quadrilateral; inversion or a one-line complex identity proves it.
Stereographic projection makes the extended plane into the Riemann sphere, sends circles to circles, and turns rotations of the sphere into Möbius maps.

Next: Chapter 14 asks which points a straightedge and compass can reach, and proves that some of the oldest problems in geometry have no solution.