Chapter 10. Rational Functions and the Riemann Sphere

Chapter 10: Rational Functions and the Riemann Sphere

By a rational function we mean a function f which can be expressed in the form

n n 1 p(z) anz + an 1z − + + a1z + a0 − f(z) = = m m 1 (10.1) q(z) bmz + bm 1z − + + b1z + b0 − where n n 1 p(z) = anz + an 1z − + + a1z + a0, and − m m 1 q(z) = bmz + bm 1z − + + b1z + b0 − are polynomials (with complex coeﬃcients) with an, bm = 0. If α , α , . . . , αn are roots 1 2 of p(z) and β1, β2, . . . , βm are roots of q(z), then we may write

(z α )(z α ) (z an) f(z) = C − 1 − 2 − , (10.2) (z β )(z β ) (z βm) − 1 − 2 − where C is some nonzero constant. We assume that p(z) and q(z) do not have a common

root. Otherwise, say αk = βj, then the right hand side of (10.2) can be simplified by canceling the factors (z αk) and (z βj). The complex numbers β , β , . . . , βm are − − 1 2 points where f(z) is undefined. They are called the poles of f(z). The multiplicity of a pole β of f(z) is defined to be the multiplicity of β as a root of the denominator polynomial q(z). The roots of p(z) are zeros of f(z) – they are exactly the places where f(z) vanishes. The multiplicity of a zero α of f(z) is exactly the multiplicity of α as a root of p(z).

Now we single out a pole of f(z), say β. Then β is a root of q(z) and let k be the multiplicity of this root. Write q(z) = (z β)kr(z), where r(z) is a polynomial with − r(β) = 0, that is, β is not a root of r(z). Now the polynomials (z β)k and r(z) − are relatively prime and hence from the factorization theory of polynomials we know that there exist polynomials g(z) and h(z) such that r(z)h(z) + (z β)kg(z) = 1. Thus we − have p(z) p(z)(r(z)h(z) + (z β)kg(z)) p(z)h(z) p(z)g(z) f(z) = = − = + . q(z) (z β)kr(z) (z β)k r(z) − − Now we use the Taylor expansion of p(z)h(z) at β, say

N j p(z)h(z) = cj(z β) . j= 0 − 1 Let f0(z) = p(z)g(z)/r(z), which is a rational function considered to be simpler than f(z) by noticing that the pole β of f(z) is deleted. Now

N j k f(z) = cj(z β) − + f0(z) j= 0 − N k 1 j k − cj cj(z β) − + k j + f0(z), ≡ j= k − j= 0 (z β) − − with c = 0, which is a sum of “continued fractions” or “partial fractions” in z β 0 − and a rational function f0(z) such that β is not a pole of f0(z). Continuing with the same argument to f0(z) until all poles of f(z) are exhausted, we arrive at the continued fractions expansion of f(z). Here let us make the statement precise: SL(2)if

f(z) is a rational function having poles β1, β2, . . . , βr with multiplicities m1, m2, . . . , mr respectively, then we can write

(k) r mk cj f(z) = j + g(z), (10.3) k= 1 j= 0 (z βk) − (k) (k) where g(z) is some polynomial and c are some constants with cm = 0. It is easy j k to see that, if the degree of the numerator p(z) of f(z) p(z)/q(z) is less than that ≡ of the denominator q(z), then the polynomial g(z) in the above expansion does not appear, that is, g(z) 0. If this is not the case, we can use the long division to reduce the ≡ degree of p(z), until the numerator has a lower degree than the denominator. Normally, we do not follow the above argument to obtain the partial fractions of a rational function. Instead, we make use of the above assertion to do the “reverse engineering” in search for (k) the coeﬃcients cj , as shown in the following example.

Example 10.1. To ﬁnd the continued fractions expansion of

z3 + 4z2 4z + 1 f(z) = − , z2(z 1)2 − we begin by putting

z3 + 4z2 4z + 1 a b c d − = + + + . z2(z 1)2 z2 z z 1 (z 1)2 − − − Then a(z 1)2 + bz(z 1)2 + cz2(z 1) + dz2 = (z 1)2 2z(z 1)2 + 3z2(z 1) + 2z2. − − − − − − − Simplifying the right hand side, we have

a(z 1)2 + bz(z 1)2 + cz2(z 1) + dz2 = z3 + 4z2 4z + 1. (10.4) − − − − 2 Diﬀerentiate both sides, we obtain

2a(z 1) + b(3z2 4z + 1) + c(3z2 2z) + 2dz = 3z2 + 8z 4. (10.5) − − − − Letting z = 0, and then letting z = 1, in both (10.4) and (10,5), we obtain a = 1, 2a + b = 4, d = 2 and b + c + 2d = 7. Thus a = 1, b = 2, c = 3 and d = 2. We − − − have arrived at z3 + 4z2 4z + 1 1 2 3 2 − = + + . z2(z 1)2 z2 − z z 1 (z 1)2 − − − Certainly, in practice, we can use some math software to get an answer quickly for more complicated rational functions.

If z = β is a pole of a rational function f(z), naturally we write f(β) = . This ∞ suggests adding a point denoted by to the complex plane C to form the so called ∞ extended complex plane, denoted by C˜ . Thus C˜ = C . Now let us consider the ∪ {∞} following question: which rational function maps bijectively from C˜ onto itself? Here let us recall that f : C˜ C˜ is a bijection means that, for each w C˜ , the equation → ∈ f(z) = w has a unique solution in z. An example of such a function is a linear conformal mapping, having the form f(z) = az + b for z C and f( ) = , where a, b are ∈ ∞ ∞ constants with a = 0. Another example is the reciprocal mapping f(z) = 1/z with f(0) = and f( ) = 0. ∞ ∞ Assume that f(z) is a rational function representing a bijection of C˜ onto itself. Let us write f(z) = p(z)/q(z), where p(z) and q(z) are polynomials without common polynomial factors. Since the the solution of f(z) = 0 must be unique, the degree of p(z) cannot exceed 1; otherwise p(z) would have more than one root and each root would give rise to one solution of f(z) = 0 (counting multiplicity). Similarly, each root of g(z) contributes to a solution of f(z) = and hence the degree of g(z) cannot exceed 1. Thus we can write ∞ p(z) = az + b and g(z) = cz + d for some constants a, b, c, d, giving us p(z) az + b f(z) = = . g(z) cz + d Notice that a and b cannot be zero simultaneously; otherwise we would have f(z) b/d, ≡ a constant function, which is certainly not bijective. Now consider the equation in z az + b b = , cz + d d assuming d = 0. Clearly z = 0 is a solution of this equation. Since f is assumed to be bijective, z = 0 is the only solution. Let us rewrite this equation as az + b b (az + b)d b(cz + d) (ad bc)z 0 = − − . cz + d − d ≡ d(cz + d) ≡ d(cz + d)

3 This tells us that ad bc is nonzero; otherwise f(z) would be the constant function b/d. − Now ad bc is the determinant of the matrix − a b A = c d and det(A) ad bc = 0 means that A is an invertible matrix. Clearly f(z) is completely ≡ − determined by matrix A. So let us put

az + b f(z) = (z) = A cz + d and we call it the fractional linear transform or the M´obius transform determined

by A. Now suppose we have another matrix with its M¨obius transform A′ (z) given by

a′ b′ a′z + b′ A′ = , A′ (z) = c′ d′ c z + d ′ ′

Then the composite AA′ (z), which is deﬁned to be A(A′ (z)), is given by

a a′ z+ b′ + b c′ z+ d′ a(a′z + b′) + b(c′z + d′) (aa′ + bc′)z + ab′ + bd′ AA′ (z) = a z+ b = = = B(z), c ′ ′ + d c(a′z + b′) + d(c′z + d′) (ca′ + dc′)z + bd′ + dd′ c′ z+ d′ where aa + bc ab + bd B = ′ ′ ′ ′ . ca′ + dc′ bd′ + dd′ Now a rather amazing thing happens: matrix B is just the product AA′ of the matrices A

and A′. So we conclude

AA′ = AA′ .

We have seen that if A is invertible (that is, A is a bijection of C˜ onto itself), then det(A) = 0. Now we can easily see that the converse is also true. Indeed, suppose that det(A) = 0. Then A is invertible and let us denote its inverse by B: AB = BA = I. Hence AB(z) = BA(z) = I (z) = z, showing that A is invertible with its inverse given by B. We have shown the following fact

az + b A(z) = is invertible if and only if det(A) ad bc = 0, cz + d ≡ −

Notice that if λ is any nonzero constant, then λA and A gives rise to the same M¨obius transform. Indeed, λaz + λb az + b (z) = = = (z) λA λcz + λd cz + d A

4 and hence we may replace A by λA in TA(z) for any judicially chosen λ. We can always choose λ in such a way that det(λA) = 1. Indeed, the last identity can be rewritten as det(A)λ2 = 1. Since det(A) = 0, det(A)λ2 = 1 treated as an algebraic equation in λ always has a solution. From now on, whenever we write A(z), we tacitly assume det(A) = 1. Denote by SL(2, C), or simply by SL(2), the set of all 2 2 complex matrices with × det(A) = 1. We claim that SL(2) forms a group under the usual matrix multiplication. Indeed, if A, B SL(2), then det(A) = det(B) = 1 gives ∈ det(AB) = det(A) det(B) = 1 and hence AB SL(2). Furthermore, if A SL(2), then det(A)=1 tells us that A ∈ ∈ is invertible with 1 1 det(A− ) = det(A)− = 1. The group SL(2, C) is called the (complex) special linear group (over C). This group is relevant to many branches of mathematics and is one of the key object of studies from various points of view. In fact, there is a book with SL(2, C) as its title. Denote by M¨o(C˜ ) the set of all M¨obius transforms:

M¨o(C˜ ) = A : A SL(2, C) . { ∈ } 1 1 M¨o C˜ The identities AB = AB and A− = A− tell us that ( ) is a group under the usual composition of transforms. They also tells us that the mapping

: SL(2) M¨o(C˜ ) given by (A) = A −→ is a group homomorphism from SL(2) onto M¨o(C˜ ).

By the ﬁrst isomorphism theorem in group theory, we know that M¨o(C˜ ) is isomorphic to the quotient group SL(2)/ ker , where

ker A SL(2): A = I ≡ { ∈ } a b is the kernel of the homomorphism . What is ker ? Suppose that A = ker . c d ∈ Then az + b A(z) = z ≡ cz + d for all z, or az + b = cz2 + dz for all z. Hence we must have b = 0, c = 0 and a = d. Consequently A = aI. Now det(A) = 1 gives a2 = 1 and hence a = 1 or a = 1. So − ker = I, I . We conclude { − } Mö(C˜ ) = SL(2, C)/ I, I . ∼ { − } 5 Using the terminology in the theory of covering spaces, the above isomorphism an be described by saying that the special linear group SL(2, C) is a double cover of the Möbius group Mö(C˜ ). The set C(z) of all rational functions can be consider as a (transcendental) field extension of the complex field C. The Galois group of automorphisms of the field C(z) over C can be identified with the Möbius group Mö(C˜ ). To study the Möbius geometry, we would like to classify Möbius transforms up to conjugacy. From the homomorphism : SL(2) Mö(C˜ ) we know that conjugate → matrices A and B in SL(2) gives rise to conjugate Möbius transforms A and B. Recall 1 that A and B are conjugate if there exists S in SL(2) such that S− AS = B. As we know, the problem of classifying matrices according to conjugacy is boiled down to the eigenvalues and eigenvectors problem. Let A be a matrix in SL(2), say

a b A = with det(A) = ad bc = 1. c d −

Notice that when A = I or A = I, we have A(z) z, that is, A is the identity − ≡ transform. So from now on we assume A = I. The eigenvalues λ , λ of A are the ± 1 2 roots of characteristic polynomial

λ a b 2 pA(λ) = det(λI A) = − − = λ (a + d)λ + 1. − c λ d − − − (Notice that we have used the fact that ad bc = 1.) The sum of the diagonal entries of − A, namely a + d, is called the trace of A and is denoted by tr(A). Since λ1 and λ2 2 are roots of pA(λ), we have pA(λ) = (λ λ )(λ λ ), or pA(λ) = λ (λ +λ )λ+λ λ ). − 1 − 2 − 1 2 1 2 Thus we have λ1 + λ2 = a + d = tr(A) and λ1λ2 = 1. We consider two cases: λ1 = λ2 and λ = λ . In view of λ λ = 1, in the first case we have λ = λ = 1, and A is 1 2 1 2 1 2 ± conjugate to 1 1 P = ± with P (z) = z 1. 0 1 ± ± Notice that is the unique fixed point of P . In this case the Möbius transform A also ∞ has a unique fixed point. In the is we say that the transform A is parabolic. In the second case A is conjugate to the diagonal matrix

λ 0 D = 1 with (z) = cz. 0 λ D 2

In this case mD has precisely two ﬁxed points, namely 0 and . In this case, ∞ 6 1. when c = 1 but c = 1, we say that the M¨obius transform A is elliptic; | |

2. when c is real and positive but c = 1, we say that the M¨obius transform A is hyperbolic;

3. when c is neither real nor satisfying c = 1, we say that the M¨obius transform | | A is loxodromic.

There are many interesting subgroups of SL(2, C), such as SL(2, R), which will be introduced for studying the hyperbolic plane in the last chapter, as well as the “arithmetic subgroup” SL(2, Z).

We have seen that the appropriate domain for rational functions is the extended complex plane C˜ = C and the special linear group SL(2) acts on tc as M´obius ∪ {∞} transforms, giving rise to automorphisms of the ﬁeld of rational functions. In the present chapter we represent the extended complex plane C˜ geometrically as a sphere, using a device related to map drawing called the stereographic projection. This pretty device has applications in both physics (such as Penrose’s twister theory) and engineering (such as guidance systems, robotics).

In the present chapter, a general point in the Euclidean space R3 will be designated 3 2 as x = (x1, x2, x3), or its like kind. The unit sphere in R , denoted by S , is the set consisting of points x = (x , x , x ) satisfying x 2 x2 + x2 + x2 = 1: 1 2 3 ≡ 1 2 3 S2 = x = (x , x , x ) R3 : x2 + x2 + x2 = 1 . { 1 2 3 ∈ 1 2 3 } The North pole of S2 is the point N = (0, 0, 1). The horizontal plane through the origin O, is taken to be the complex plane, which overlaps with the x1x2–plane. We 3 may identify a point z = x + iy in the complex plane with the point (x1, x2, 0) in R , where x1 = x and x2 = y.

Take any point x = (x1, x2, x3) on the unit sphere distinct fro the North pole N. Draw a line ℓ passing through x and N. Using the parametric equation for this line, we know that a general point on this line is of the form

t(x , x , x ) + (1 t)(0, 0, 1) (tx , tx , 1 + t(x 1)). 1 2 3 − ≡ 1 2 3 − Here t can be considered as a parameter for this line. The line ℓ will hit a point on the complex plane at a point, say z = x + iy, which is identiﬁed with the point (x, y, 0) in R3. To ﬁnd this point of intersection z? Well, choose t in such a way that

(tx , tx , 1 + t(x 1)) = (x, y, 0). 1 2 3 − 7 1 Thus, here t must be the one satisfying 1+ t(x3 1) = 0, or t = 1 x3 . Notice that, since − − x = (x1, x2, x3) is any point on the unit sphere other than the North Pole N = (0, 0, 1), 1 the third coordinate x3 must be diﬀerent from 1 and hence the expression 1 x3 is − legitimate. Thus x = tx1 and y = tx2, or

x x x + ix x = 1 , y = 2 and hence z = 1 2 . (11.1) 1 x 1 x 1 x − 3 − 3 − 3

These identities tell us how to ﬁnd the image z = x+iy of any given point x = (x1, x2, x3) on the sphere other than the North Pole, under the SL(2)stereographic projection. How- ever, it is more interesting to consider the reverse problem of asking for the point x =

(x1, x2, x3) on the unit sphere when the point z = x + iy on the complex plane is given.

In other words, we have to ﬁnd an expression of x1, x2, x3 in terms of x, y.

Recall that we have 1+t(x3 1) = 0. Rewrite this as 1+tx3 t = 0, or tx3 = t 1, 1 − − −1 and ﬁnally x3 = t− (t 1). We also have x = tx1 and y = tx2, which give x1 = t− x 1 − and x2 = t− y. Substituting

1 1 1 x = t− x, x = t− y and x = t− (t 1) (11.2) 1 2 3 −

2 2 2 2 into x1 + x2 + x3 = 1, then multiplying both sides by t , we have

x2 + y2 + (1 t)2 = t2, − or x2 + y2 + 1 2t + t2 = t2. Canceling t2, we can isolate t and express it in terms of 1− 2 2 x and y: t = 2 (x + y + 1). Substituting this t back to (11.2), we arrive at

2x 2y x2 + y2 1 x = , x = , x = − . (11.3) 1 x2 + y2 + 1 2 x2 + y2 + 1 3 x2 + y2 + 1

Let us write σ(z) = (x1, x2, x3), whee x1, x2, x3 are given by (11.3) above with x+iy = z. Exercise 11.1. Given two complex numbers z and w with x = σ(z) and y = σ(w), the chordal distance κ(z, w) between z and w is deﬁned to be x y , the Euclidean − distance between the points x and y in R3. Verify that

z w κ(z, w) = | − | . 1 + z 2 1 + w 2 | | | | Also verify the following identities

1 1 κ(z, w) = κ(z, w), κ(z− , w− ) = κ(z, w).

8 (Note: The chordal distance is used in a number of places, for example, in the study of robustness of control systems.) Exercise 11.2. Prove that, if we use the South pole S = (0, 0, 1) in our stereographic − projection, then the point w = u + iv on the complex plane, which is the image of x = (x1, x2, x3), is given by x x u = 1 , v = 2 , 1 + x3 1 + x3

and furthermore, if σ(z) = (x1, x2, x3), then w = 1/z. Finally, show that

2u 2v x2 + y2 + 1 x = , x = , x = . 1 u2 + v2 + 1 2 u2 + v2 + 1 3 x2 + y2 1 − Notice that the right hand sides in both (11.1) and (11.2) are rational functions. Hence the stereographic projection gives a correspondence between rational points on the sphere and rational points on the plane. Here, by rational points we mean points in a Euclidean space with rational numbers as their coordinates. This will help us to solve some special Diophantine equations describe as follows. Ancient Greek mathematicians found many triples (a, b, c) of positive integers satisfying a2 + b2 = c2, meaning that they represent three sides of right triangles, such as 32 + 42 = 52, 52 + 122 = 132, 82 + 152 = 172, etc. We call such triples as Pythagoras triples. Rewrite a2 +b2 = c2 as (a/c)2 +02 +(b/c)2 = 1, showing that x = (a/c, 0, b/c) is a rational point on the sphere. Let z = u + iv be a point on the complex plane corresponding to this point under the stereographic projection. Then

a/c a u = = , and v = 0 1 + b/c b + c which are positive rational numbers. Write u = m/n, where m and n are positive integers. Then a/c = 2u/(u2 + 1) = 2mn/(m2 + n2) and b/c = (u2 1)/(u2 + 1) = − (m2 n2)/(m2 +n2). If we let a = 2mn, b = m2 n2, and c = m2 +n2, then a2 +b2 = c2, − − in other words, (a, b, c). For example, taking m = 3 and n = 2, we get a = 2mn = 12, b = m2 n2 = 5 and c = m2 + n2 = 13. From our above discussion it is not hard to see − that all Pythagoras triples can be obtained in this way.

Let us return to the mapping σ : C S2 gives by σ(x+iy) = (x , x , x ) according → 1 2 3 to (11.3). Notice that, as z 2 = x2 + y2 , the point (x , x , x ) approaches to | | → ∞ 1 2 3 (0, 0, 1) = N, the North Pole. So it is natural to put σ( ) = N. In this way σ is | ∞ extended to a map, still denoted by σ from the extended complex plane C˜ to the sphere. This mapping, namely σ : C˜ S2, is a bijection. → 9 The natural transformation group for the sphere S2 is the rotation group SO(3). By deﬁnition, SO(3) consists of all real matrices in SU(3). Thus, a 3 3 matrix × A belongs to SO(3) if it is real and it satisﬁes the conditions A⊤A = AA⊤ = I and det(A) = 1.

A rotation about the x3–axis can be represented by a matrix of the form

cos α sin α 0 R = sin α cos α 0 .  − 0 0 1    So if we write y = Rx, then

y = (cos α)x + (sin α)x , y = ( sin α)x + (cos α)x , y = x . 1 1 2 2 − 1 2 3 3 Suppose that σ(z) = x and σ(w) = y, with z = x + iy and w = u + iv. Then

y y (cos α)x + (sin α)x ( sin α)x + (cos α)x w = u + iv = 1 + i 2 = 1 2 + i − 1 2 1 y3 1 y3 1 x3 1 x3 − − − − iα = (cos α)x + (sin α)y + i(( sin α)x + (cos α)y) = (cos α i sin α)(x + iy) = e− z. − − Thus we have iα e− 0 w = A(z), where A = SU(2). 0 1 ∈ o Next we study the 90 rotation with the x2-axis as the axis of rotation, represented by the matrix 0 0 1 S = 0 1 0 .  1 0 0  −   Thus, y = Sx means y = x , y = x , y = x . 1 3 2 2 3 − 1 Suppose that σ(z) = x and σ(w) = y, with z = x + iy and w = u + iv. Then

y iy (x + ix )(x2 + y2 + 1) w = u + iv = 1 + 2 = 3 2 1 y 1 y (1 + x )(x2 + y2 + 1) − 3 − 3 1 x2 + y2 1 2yi zz 1 + z z (z 1)(z + 1) z 1 = − − = − − = − = − . x2 + y2 + 1 + 2x zz + 1 + z + z (z + 1)(z + 1) z + 1

Thus we have 1 1 1 w = B(z), where B = − SU(2). √2 1 1 ∈