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Chapter 10. Rational Functions and the Riemann Sphere

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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) = µ (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 µA′ (z) given by a′ b′ a′z + b′ A′ = , µA′ (z) = c′ d′ c z + d ′ ′ Then the composite µAµA′ (z), which is defined to be µA(µA′ (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′ µAµA′ (z) = a z+ b = = = µB(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 µAµA′ = µAA′ . We have seen that if µA is invertible (that is, µA 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 µAµB(z) = µBµA(z) = µI (z) = z, showing that µA is invertible with its inverse given by µB. We have shown the following fact az + b µA(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.
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