Chapter 10: Rational Functions and the Riemann Sphere. By a rational function we mean a function f which can be expressed in the form

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1 Chapter 10: Rational Functions and the Riemann Sphere By a rational function we mean a function f which can be expressed in the form f(z) = p(z) q(z) = a nz n +a n 1 z n 1 + +a 1 z +a 0 b m z m +b m 1 z m 1 + +b 1 z +b 0 (10.1) where p(z) = a n z n +a n 1 z n 1 + +a 1 z +a 0, and q(z) = b m z m +b m 1 z m 1 + +b 1 z +b 0 are polynomials (with complex coefficients) with a n, b m 0. If α 1, α 2,..., α n are roots of p(z) and β 1, β 2,..., β m are roots of q(z), then we may write f(z) = C (z α 1)(z α 2 ) (z a n ) (z β 1 )(z β 2 ) (z β m ), (10.2) where C issomenonzeroconstant. Weassumethat p(z) and q(z) donothaveacommon root. Otherwise, say α k = β j, then the right hand side of (10.2) can be simplified by canceling the factors (z α k ) and (z β j ). The complex numbers β 1, β 2,..., β m are 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) k r(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) k g(z) = 1. Thus we have f(z) = p(z) q(z) = p(z)(r(z)h(z)+(z β)k g(z)) (z β) k = p(z)h(z) r(z) (z β) k + p(z)g(z). r(z) Now we use the Taylor expansion of p(z)h(z) at β, say p(z)h(z) = N 1 j= 0 c j(z β) j.

2 Let f 0 (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 f(z) = N j= 0 c j(z β) j k +f 0 (z) N c j(z β) j k + k 1 j= k j= 0(z β) k j +f 0(z), with c 0 0, which is a sum of continued fractions or partial fractions in z β and a rational function f 0 (z) such that β is not a pole of f 0 (z). Continuing with the same argument to f 0 (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) isarationalfunctionhavingpoles β 1, β 2,..., β r withmultiplicities m 1, m 2,..., m r respectively, then we can write f(z) = r k= 1 where g(z) is some polynomial and c (k) j mk j= 0 c j c (k) j +g(z), (10.3) (z β k ) j are some constants with c (k) m k 0. It is easy 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 the coefficients c (k) j, as shown in the following example. Example To find the continued fractions expansion of we begin by putting f(z) = z3 +4z 2 4z +1 z 2 (z 1) 2, z 3 +4z 2 4z +1 z 2 (z 1) 2 = a z 2 + b z + c z 1 + d (z 1) 2. Then a(z 1) 2 +bz(z 1) 2 +cz 2 (z 1)+dz 2 = (z 1) 2 2z(z 1) 2 +3z 2 (z 1)+2z 2. Simplifying the right hand side, we have a(z 1) 2 +bz(z 1) 2 +cz 2 (z 1)+dz 2 = z 3 +4z 2 4z +1. (10.4) 2

3 Differentiate both sides, we obtain 2a(z 1)+b(3z 2 4z +1)+c(3z 2 2z)+2dz = 3z 2 +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 z 3 +4z 2 4z +1 z 2 (z 1) 2 = 1 z 2 2 z + 3 z (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) = 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) = and g(z) = cz +d for some constants a, b, c, d, giving us f(z) = p(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 cz +d = b 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 0 = cz +d b d ()d b(cz +d) d(cz +d) 3 (ad bc)z d(cz +d).

4 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 = a b 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 f(z) = µ A (z) = cz +d and we call it the fractional linear transform or the Móbius transform determined by A. Now suppose we have another matrix with its Möbius transform µ A (z) given by A a b = c d, µ A (z) = a z +b c z +d Then the composite µ A µ A (z), which is defined to be µ A (µ A (z)), is given by µ A µ A (z) = aa z+ b c z+ d +b c a z+ b c z+ d +d = a(a z +b )+b(c z +d ) c(a z +b )+d(c z +d ) = (aa +bc )z +ab +bd (ca +dc )z +bd +dd = µ B(z), where B = aa +bc ab +bd 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 µ A µ A = µ AA. We have seen that if µ A is invertible (that is, µ 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 µ A µ B (z) = µ B µ A (z) = µ I (z) = z, showing that µ A is invertible with its inverse given by µ B. We have shown the following fact µ A (z) = cz +d is invertible if and only if det(a) ad bc 0, Notice that if λ is any nonzero constant, then λa and A gives rise to the same Möbius transform. Indeed, λaz +λb µ λa (z) = = λcz +λd cz +d = µ A(z) 4

5 and hence we may replace A by λa in T A (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)λ 2 = 1. Since det(a) 0, det(a)λ 2 = 1 treated as an algebraic equation in λ always has a solution. From now on, whenever we write µ 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 det(a 1 ) = det(a) 1 = 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 Mö( C) the set of all Möbius transforms: Mö( C) = {µ A : A SL(2,C)}. The identities µ A µ B = µ AB and µ 1 A = µ A 1 tell us that Mö( C) is a group under the usual composition of transforms. They also tells us that the mapping µ: SL(2) Mö( C) given by µ(a) = µ A is a group homomorphism from SL(2) onto Mö( C). Bythefirstisomorphismtheoremingrouptheory, weknowthat Mö( C) isisomorphic to the quotient group SL(2)/kerµ, where kerµ {A SL(2): µ A = I} is the kernel of the homomorphism µ. What is kerµ? Suppose that A = Then µ A (z) cz +d = z a b kerµ. c d for all z, or az + b = cz 2 + dz for all z. Hence we must have b = 0, c = 0 and a = d. Consequently A = ai. Now det(a) = 1 gives a 2 = 1 and hence a = 1 or a = 1. So kerµ = {I, I}. We conclude Mö( C) = SL(2,C)/{I, I}. 5

6 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 µ : SL(2) Mö( C) we know that conjugate matrices A and B in SL(2) gives rise to conjugate Möbius transforms µ A and µ B. Recall that A and B are conjugate if there exists S in SL(2) such that S 1 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 = a b c d with det(a) = ad bc = 1. Notice that when A = I or A = I, we have µ A (z) z, that is, µ A is the identity transform. So from now on we assume A ±I. The eigenvalues λ 1, λ 2 of A are the roots of characteristic polynomial λ a b p A (λ) = det(λi A) = = λ 2 (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 λ 1 and λ 2 are roots of p A (λ), we have p A (λ) = (λ λ 1 )(λ λ 2 ), or p A (λ) = λ 2 (λ 1 +λ 2 )λ+λ 1 λ 2 ). Thus we have λ 1 +λ 2 = a+d = tr(a) and λ 1 λ 2 = 1. We consider two cases: λ 1 = λ 2 and λ 1 λ 2. In view of λ 1 λ 2 = 1, in the first case we have λ 1 = λ 2 = ±1, and A is conjugate to ±1 1 P = with µ 0 ±1 P (z) = z ±1. Notice that is the unique fixed point of P. In this case the Möbius transform µ A also has a unique fixed point. In the is we say that the transform µ A is parabolic. In the second case A is conjugate to the diagonal matrix λ1 0 D = with µ 0 λ D (z) = cz. 2 In this case m D has precisely two fixed points, namely 0 and. In this case, 6

7 1. when c = 1 but c 1, we say that the Möbius transform µ A is elliptic; 2. when c is real and positive but c 1, we say that the Möbius transform µ A is hyperbolic; 3. when c is neither real nor satisfying c = 1, we say that the Möbius transform µ 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 Móbius 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 R 3 will be designated as x = (x 1,x 2,x 3 ), or its like kind. The unit sphere in R 3, denoted by S 2, is the set consisting of points x = (x 1,x 2,x 3 ) satisfying x 2 x 2 1 +x2 2 +x2 3 = 1: S 2 = {x = (x 1,x 2,x 3 ) R 3 : x 2 1 +x2 2 +x2 3 = 1}. The North pole of S 2 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 x 1 x 2 plane. We may identify a point z = x+iy in the complex plane with the point (x 1,x 2,0) in R 3, where x 1 = x and x 2 = y. Take any point x = (x 1,x 2,x 3 ) 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 1, x 2, x 3 )+(1 t)(0, 0, 1) (tx 1, tx 2, 1+t(x 3 1)). 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 R 3. To find this point of intersection z? Well, choose t in such a way that (tx 1, tx 2, 1+t(x 3 1)) = (x, y, 0). 7

8 Thus, here t must be the one satisfying 1+t(x 3 1) = 0, or t = 1 1 x 3. Notice that, since x = (x 1, x 2, x 3 ) is any point on the unit sphere other than the North Pole N = (0, 0, 1), the third coordinate x 3 must be different from 1 and hence the expression legitimate. Thus x = tx 1 and y = tx 2, or 1 1 x 3 x = x 1 1 x 3, y = x 2 1 x 3 and hence z = x 1 +ix 2 1 x 3. (11.1) Theseidentitiestell ushowto findtheimage z = x+iy ofany givenpoint x = (x 1,x 2,x 3 ) on the sphere other than the North Pole, under the SL(2)stereographic projection. However, it is more interesting to consider the reverse problem of asking for the point x = (x 1, x 2, x 3 ) 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 x 1, x 2, x 3 in terms of x, y. Recall that we have 1+t(x 3 1) = 0. Rewrite this as 1+tx 3 t = 0, or tx 3 = t 1, and finally x 3 = t 1 (t 1). We also have x = tx 1 and y = tx 2, which give x 1 = t 1 x and x 2 = t 1 y. Substituting x 1 = t 1 x, x 2 = t 1 y and x 3 = t 1 (t 1) (11.2) into x 2 1 +x2 2 +x2 3 = 1, then multiplying both sides by t2, we have x 2 +y 2 +(1 t) 2 = t 2, or x 2 +y t+t 2 = t 2. Canceling t 2, we can isolate t and express it in terms of x and y: t = 1 2 (x2 +y 2 +1). Substituting this t back to (11.2), we arrive at x 1 = 2x x 2 +y 2 +1, x 2 = 2y x 2 +y 2 +1, x 3 = x2 +y 2 1 x 2 +y (11.3) Letuswrite σ(z) = (x 1,x 2,x 3 ), whee x 1, x 2, x 3 aregivenby(11.3)abovewith x+iy = z. Exercise 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 R 3. Verify that κ(z,w) = z w 1+ z 2 1+ w 2. is Also verify the following identities κ(z,w) = κ(z,w), κ(z 1,w 1 ) = κ(z,w). 8

9 (Note: The chordal distance is used in a number of places, for example, in the study of robustness of control systems.) Exercise 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 = (x 1, x 2, x 3 ), is given by u = x 1 1+x 3, v = x 2 1+x 3, and furthermore, if σ(z) = (x 1,x 2,x 3 ), then w = 1/z. Finally, show that x 1 = 2u u 2 +v 2 +1, x 2 = 2v u 2 +v 2 +1, x 3 = x2 +y 2 +1 x 2 +y 2 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 a 2 +b 2 = c 2, meaning that they represent three sides of right triangles, such as = 5 2, = 13 2, = 17 2, etc. We call such triples as Pythagoras triples. Rewrite a 2 +b 2 = c 2 as (a/c) (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 u = a/c 1+b/c = a b+c, and v = 0 which are positive rational numbers. Write u = m/n, where m and n are positive integers. Then a/c = 2u/(u 2 + 1) = 2mn/(m 2 + n 2 ) and b/c = (u 2 1)/(u 2 + 1) = (m 2 n 2 )/(m 2 +n 2 ). If we let a = 2mn, b = m 2 n 2, and c = m 2 +n 2, then a 2 +b 2 = c 2, in other words, (a, b, c). For example, taking m = 3 and n = 2, we get a = 2mn = 12, b = m 2 n 2 = 5 and c = m 2 +n 2 = 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 S 2 gives by σ(x+iy) = (x 1, x 2, x 3 ) according to (11.3). Notice that, as z 2 = x 2 + y 2, the point (x 1, x 2, x 3 ) approaches to (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 S 2, is a bijection. 9

10 The natural transformation group for the sphere S 2 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 x 3 axis can be represented by a matrix of the form So if we write y = Rx, then cosα sinα 0 R = sinα cosα y 1 = (cosα)x 1 +(sinα)x 2, y 2 = ( sinα)x 1 +(cosα)x 2, y 3 = x 3. Suppose that σ(z) = x and σ(w) = y, with z = x+iy and w = u+iv. Then y 2 w = u+iv = y 1 +i = (cosα)x 1 +(sinα)x 2 +i ( sinα)x 1 +(cosα)x 2 1 y 3 1 y 3 1 x 3 1 x 3 = (cosα)x+(sinα)y +i(( sinα)x+(cosα)y) = (cosα isinα)(x+iy) = e iα z. Thus we have w = µ A (z), where A = e iα 0 SU(2). 0 1 Next we study the 90 o rotation with the x 2 -axis as the axis of rotation, represented by the matrix S = Thus, y = Sx means y 1 = x 3, y 2 = x 2, y 3 = x 1. Suppose that σ(z) = x and σ(w) = y, with z = x+iy and w = u+iv. Then w = u+iv = y 1 + iy 2 = (x 3 +ix 2 )(x 2 +y 2 +1) 1 y 3 1 y 3 (1+x 1 )(x 2 +y 2 +1) = x2 +y 2 1 2yi x 2 +y x = zz 1+z z zz +1+z +z = (z 1)(z +1) (z +1)(z +1) = z 1 z +1. Thus we have w = µ B (z), where B = SU(2)

z, w = z 1 w 1 + z 2 w 2 z, w 2 z 2 w 2. d([z], [w]) = 2 φ : P(C 2 ) \ [1 : 0] C ; [z 1 : z 2 ] z 1 z 2 ψ : P(C 2 ) \ [0 : 1] C ; [z 1 : z 2 ] z 2 z 1

$z, w = z 1 w 1 + z 2 w 2 z, w 2 z 2 w 2. d([z], [w]) = 2 φ : P(C 2 ) \ [1 : 0] C ; [z 1 : z 2 ] z 1 z 2 ψ : P(C 2 ) \ [0 : 1] C ; [z 1 : z 2 ] z 2 z 1$ 3 3 THE RIEMANN SPHERE 31 Models for the Riemann Sphere One dimensional projective complex space P(C ) is the set of all one-dimensional subspaces of C If z = (z 1, z ) C \ 0 then we will denote by [z]