Complex Analysis: Converting Integrals to Path Integrals & Power Series, Essays (university) of Applied Mathematics

Download Complex Analysis: Converting Integrals to Path Integrals & Power Series and more Essays (university) Applied Mathematics in PDF only on Docsity! Math 115 (2006-2007) Yum-Tong Siu 1 Power and Laurent Series Expansion, Classification of Isolated Sin- gularities, and Computation of Residues. We are going to take the next step into complex analysis from the practical viewpoint of the computation of definite integrals. We start out with the problem of computation of definite integrals of the form ∫ 2π θ=0 R (cos θ, sin θ) dθ, where R (X,Y ) is a rational function of two indeterminates X and Y over the complex number field. Recall that we used the following four steps. (i) Use the parametrization z = eiθ, we get dθ = dz iz , cos θ = 1 2 ( z + 1 z ) , sin θ = 1 2i ( z − 1 z ) to convert the given definite integral to a path integral over the unit circle of the form ∫ |z|=1 S (z) dz, where S(z) is a rational function of the complex variable z. (ii) Use partial fraction decomposition S(z) = ℓ ∑ j=1   kj ∑ ν=1 Aj,ν (z − aj)ν   , where Aj,ν ∈ C, k1, · · · , kℓ ∈ N, and a1, · · · , aℓ are distinct complex numbers. (iii) Use special path integrals ∮ |z−a|=r (z − a)n dz = { 0 for n 6= −1 2πi for n = −1 for a ∈ C and r > 0 and n ∈ Z by explicitly using the parametrization θ 7→ z = a + reiθ for 0 ≤ θ ≤ 2π. Math 115 (2006-2007) Yum-Tong Siu 2 (iv) Finally use the Cauchy theorem for the smooth case and use small circles around those aj’s which are inside {|z| < 1} to compute the given definite integral as ∑ j∈J 2πi (kj − 1)! ( dk1−1 dzkj−1 ( (z − aj)kj iz R ( 1 2 ( z + 1 z ) , 1 2i ( z − 1 z )) )) z=aj , where J consists of all 1 ≤ j ≤ ℓ with |aj| < 1. This procedure converts the computation of a definite integral to the com- putation of derivatives. For our next step into complex analysis from the practical viewpoint of the computation of definite integrals, we would like to be able to compute definite integrals for a more general class of integrands, also by converting the computation of definite integrals to the computations of derivatives. The key is to make this procedure work is to get the second step (ii) for functions more general than rational functions. Of course, these functions must be holomorphic (at least outside a finite set of points) so that we can apply the theorem of Cauchy-Goursat. The analog of the second step (ii) for general holomorphic functions is the following statement on power series expansion and its generalization to Laurent series expansion so that to each term in the convergent power or Laurent series we can apply the computation of special path integrals in (iii). Power Series Expansion of a Holomorphic Function on a Disk. Let 0 < R ≤ ∞ and a ∈ C and f(z) be a holomorphic function on {|z − a| < R}. Then there exist cn ∈ C for n ∈ N ∪ {0} such that f(z) = ∞ ∑ n=0 cn (z − a)n where the infinite series is absolute and uniform convergent on any compact subset of {|z − a| < R}. Moreover, cn ∈ C is unique and is given by cn = 1 n! f (n)(a) = 1 2πi ∫ |z|=r f(z)dz (z − a)n+1 for any 0 < r < R. Math 115 (2006-2007) Yum-Tong Siu 5 (known as the Laurent series of f on the R1 < |z − c| < R2) with cν = 1 2πi ∫ C f(ζ) (ζ − c)ν+1dζ for any circle C with radius r between R1 ′ and R2 ′ and center c. This Laurent series expansion (*) is convergent uniformly and absolutely on any compact subset of the annulus R1 < |z− c| < R2. (Note that for the special case when f(z) is holomorphic on |z − c| < R2 the theorem of Cauchy-Goursat yields cn = 0 for n ≤ −1 and the Laurent series becomes a power series so that the statement on power series is a special case for the statement on Laurent series. Moreover, for this special case, by term-by-term differentiation and then evaluating at z = a we obtain cn = 1 n! f (n)(a) for n ≥ 0.) Suppose we have an expansion (*) which converges uniformly and abso- lutely on any compact subset of the annulus R1 < |z − c| < R2. Then by applying the operator ∫ C 1 (ζ − c)n+1 (·)dζ to both sides of (*) and using the fact that ∫ Cr,c (z − c)ndz = 2πi or 0 according as n is −1 or not, we conclude that cn in the expansion (*) must be equal to 1 2πi ∫ C f(ζ) (ζ − c)n+1dζ, giving us the uniqueness of the coefficients cn in a Laurent series. Q.E.D. Isolated singularities. We now consider the special case R1 = 0. In other words, a is an isolated singularity of the holomorphic function f(z). The series −1 ∑ ν=−∞ cν(z − a)ν is known as the principal part of f(z) at a. When the principal part −1 ∑ ν=−∞ cν(z − a)ν Math 115 (2006-2007) Yum-Tong Siu 6 is zero, we say that we have a removable singularity at c. In the case of a removable singularity at a the function f near a is a convergent power series in z − a and is therefore holomorphic at c. When the principal part −1 ∑ ν=−∞ cν(z − a)ν is nonzero and consists only of a finite number of nonzero terms, we say that we have a pole at a. The number k such that c−k 6= 0 but c−ℓ = 0 for ℓ > k is called the order of the pole. When the principal part −1 ∑ ν=−∞ cν(z − a)ν contains an infinite number of nonzero terms, we say that we have an essential singularity at a. Characterization of Removable Singularities. A removable singularity is char- acterized by the fact that f is L2 near c. Suppose f is L2 near a. We would like to verify that f is holomorphic at a. For r > 0, we have ∫ 2π θ=0 |f(reiθ)|2dθ = ∞ ∑ µ,ν=−∞ ∫ 2π θ=0 cµcνr µ+νei(µ−ν)θdθ = ∞ ∑ ν=−∞ 2π cνcνr 2ν ≥ 2π cνcνr2ν . Suppose cν 6= 0 for some ν ≤ −1. Then the finiteness of ∫ ε r=0 rdr ∫ 2π θ=0 |f(reiθ)|2dθ for some ε > 0 would imply the finiteness of ∫ ε r=0 2π cν c̄ − ν r 2ν+1dr which is not possible for any ν ≤ −1. Characterization of Poles. Suppose a is a pole of order k. Then we can write f(z) = (z − a)−kg(z), where g(z) is a convergent power series in z − a near a and g(a) 6= 0. Thus |f(z)| goes to infinity of order k as z → a. Conversely, if |f(z)| goes to infinity of order k as z → c, then (z − a)kf(z) has a removable singularity at a and (z − a)kf(z) is equal to a convergent power series g(z) in z−a near a with g(a) 6= 0. Hence f(z) = (z−a)−kg(z) is a Laurent series with finite principal part and a is a pole of order k. Math 115 (2006-2007) Yum-Tong Siu 7 We call a function meromorphic if it is holomorphic except for isolated singularities which are poles. Characterization of Essential Singularities. Suppose a is an essential singu- larity. Then the image under f(z) of any deleted disk neighborhood of a is dense in C. Suppose not. Then there exists some a such that 1 f−a is holo- morphic on 0 < |z − a| < ε and is uniformly bounded there. Hence 1 f−a is holomorphic on |z − a| < ε. Let 1 f − a = ∞ ∑ ν=k bν(z − a)ν be its power series expansion on |z − a| < ε with bk 6= 0. Then f(z) = a + 1 (z − a)k 1 ∑∞ ν=0 bν+k(z − a)ν and the power series expansion of 1 ∑∞ ν=0 bν+k(z − a)ν would make a a pole of order k for f(z) which is a contradiction. Conversely when the image under f(z) of any deleted disk neighborhood of a is dense in C, a is an essential singularity of f(z), because the other two cases of removable singularity and pole do not have this density property. Residues. The residue of f at a is defined as c−1 which is by definition the integral 1 2πi ∫ C f(ζ)dζ for any sufficiently small circle C centered at a. When a is a pole of order k, an alternative description of c−1 is 1 (k − 1)! ( dk−1 dzk−1 (z − a)kf(z) ) z=c , because of the power series expansion (z − a)kf(z) = ∞ ∑ ν=0 cν−k(z − a)k. Math 115 (2006-2007) Yum-Tong Siu 10 Proof. For (a) we assume that (∗∗) R < lim inf n→∞ |cn| |cn+1| and show that it leads to a contradiction. From (∗∗) we can choose z0 ∈ C with R < |z0 − a| < lim inf n→∞ |cn| |cn+1| and we are going to show the power series ∑∞ n=0 cn (z0 − a) n converges, which would contradict R < |z0 − a|. Take r > 0 with |z0 − a| < r < lim inf n→∞ |cn| |cn+1| . Then for some N ∈ N with N ≥ N0 we have r ≤ |cn||cn+1| for n ≥ N. Since cn 6= 0 for n ≥ N , by induction on n ≥ N we conclude that |cnrn| ≤ |cN | rN for n ≥ N and, as a consequence, ∞ ∑ n=N |cn (z0 − a)n| = ∞ ∑ n=N |cnrn| ( |z0 − a| r )n ≤ |cN | rN ∞ ∑ n=N ( |z0 − a| r )n = |cN | rN 1 − |z0−a| r < ∞. For the verification of (b) we assume that (††) lim inf n→∞ |cn| |cn+1| < R and show that it leads to a contradiction. From (††) we can choose z0 ∈ C with lim inf n→∞ |cn| |cn+1| < |z0 − a| < R Math 115 (2006-2007) Yum-Tong Siu 11 and we are going to show the power series ∑∞ n=0 cn (z0 − a) n does not con- verge, which would contradict |z0 − a| < R. Take r > 0 with lim sup n→∞ |cn| |cn+1| < r < |z0 − a| . Then there exists a subsequence {nν}ν∈N of {n}n∈N such that |cnν | ∣ ∣cnν+1 ∣ ∣ ≥ r for n ≥ N. Then for some N ∈ N with N ≥ N0 we have |cn| |cn+1| ≥ r for n ≥ N. Since cn 6= 0 for n ≥ N , by induction on n ≥ N we conclude that |cnrn| ≥ |cN | rN for n ≥ N and |cn (z0 − a)n| ≥ |cN | rN > 0 for n ≥ N. In particular, the sequence cn (z0 − a)n cannot approach 0 as n → ∞. As a consequence, the power series ∑∞ n=0 cn (z0 − a) n does not converge, which would contradict |z0 − a| < R. Q.E.D.