n } is convergent, lim n in+1

Transcription

1 hapter 3 Series y residuos redit: This notes are 00% from chapter 6 of the book entitled A First ourse in omplex Analysis with Applications of Dennis G. Zill and Patrick D. Shanahan (2003) [2]. auchy s integral formula for derivatives indicates that if a function f is analytic at a point z 0, then it possesses derivatives of all orders at that point. As a consequence of this result we shall see that f can always be expanded in a power series centered at that point. On the other hand, if f fails to be analytic at z 0, we may still be able to expand it in a different kind of series known as a Laurent series. The notion of Laurent series leads to the concept of a residue, and this, in turn, leads to yet another way of evaluating complex and, in some instances, real integrals. 3. Sequences and Series Sequences A sequence {z n } is a function whose domain is the set of positive integers and whose range is a subset of the complex numbers. If lim n z n L, we say the sequence {z n } is convergent. In other words, {z n } converges to the number L if for each positive real number ε an N can be found such that z n L < ε whenever n > N. A sequence that is not convergent is said to be divergent. Example: The sequence { in+ n } is convergent, lim n in+ n 0. Theorem (6.): riterion for onvergence A sequence {z n } converges to a complex number L a+ib if and only if R(z n ) converges to R(L) a and I(z n ) converges to I(L) b. 209

2 Example onsider the sequence { 3+in n+i2n } Solution: it converges since, z n 3+in n+i2n 2n2 +3n 5n 2 +i n2 6n 5n 2 (3.) R(z n ) 2 5 I(z n ) 5 (3.2) (3.3) as n. Series An infinite series or series of complex numbers z k z +z 2 + +z n + (3.4) k is convergent if the sequence of partial sums {S n }, where S n z +z 2 + +z n (3.5) converges. If S n L as n, we say that the series converges to L or that the sum of the series is L. Geometric Series A geometric series is any series of the form az k a+az +az 2 + +az n + (3.6) k the nth term of the sequence of partial sums is S n a+az +az 2 + +az n (3.7) When an infinite series is a geometric series, it is always possible to find a formula for S n : zs n az +az 2 + +az n (3.8) S n zs n (az +az 2 + +az n ) (a+az +az 2 + +az n )(3.9) S n ( z) a az n a( z n ) S n a zn z for n, z n 0 whenever z <, then S n a/( z), then (3.0) a+az +az 2 + +az n + a z (3.) A geometric series diverges when z. Special Geometric Series. (i) For a z +z +z2 +z 3 + (3.2) 20

3 (ii) For a and z z (iii) For a (iv) By writing zn z z + zn z Example The infinite series +z z +z2 z 3 + (3.3) z n z +z +z2 +z 3 + +z n (3.4) we get z +z +z2 +z 3 + +z n + zn z k (3.5) (+2i) k 5 k +2i + (+2i) (3.6) is a geometric series with a 5 (+2i) and z 5 (+2i). Since z 5/5 <, the series is convergent k (+2i) k 5 k +i2 5 +i2 5 i 2 (3.7) Theorem (6.2): A Necessary ondition for onvergence If k z k converges, then lim n z n 0. Theorem (6.3): The nth Term Test for Divergence If lim n z n 0, then k z k diverges. Absolute and onditional onvergence An infinite series k z k is said to be absolutely convergent if k z k converges. An infinite series k z k is said to be conditionally convergent if it converges but k z k diverges. Absolute convergence implies convergence. Example Theseries k ik k isabsolutelyconvergentsincetheseries 2 is the same as the real convergent p-series k k. 2 k 2 k ik Tests for onvergence Two of the most frequently used tests for convergence of infinite series are given in the next theorems. Theorem (6.4): Ratio Test Suppose k z k is a seriesofnonzerocomplex terms such that lim z n+ n z n L (3.8) (i) If L <, then the series converges absolutely. (ii) If L > or L, then the series diverges. (iii) If L, the test is inconclusive. 2

4 Theorem (6.5): Root Test Suppose k z k is a series of nonzero complex terms such that n lim zn L (3.9) n (i) If L <, then the series converges absolutely. (ii) If L > or L, then the series diverges. (iii) If L, the test is inconclusive. Power Series The notion of a power series is important in the study of analytic functions. An infinite series of the form a k (z z 0 ) k a 0 +a (z z 0 )+a 2 (z z 0 ) 2 + (3.20) k wherethecoefficientsa k arecomplexconstants,iscalledapowerseriesinz z?0. It is said to be centered at z 0, called the center of the series. ircle of onvergence Every complex power series k0 a k(z z 0 ) k has a radius of convergence and a circle of convergence, which is the circle centered at z 0 of largest radius R > 0 for which the series converges at every point within the circle z z 0 R. A power series converges absolutely at all points z within its circle of convergence, that is, for all z satisfying z z 0 < R, and diverges at all points z exterior to the circle, that is, for all z satisfying z z 0 > R. The radius of convergence can be: (i) R 0 (in which case the serie converges only at its center z z 0 ), (ii) R a finite positive number (in which case the serie converges at all interior points of the circle z z 0 R), or (iii) R (in which case the serie converges for all z). A power series may converge at some, all, or at none of the points on the actual circle of convergence. Example onsider the power series k lim n z n+2 n+ z n+ n lim n z k+ k. By the ratio test, n z z (3.2) n+ Thus the series converges absolutely for z <. The circle of convergence is z and the radius of convergence is R. On the circle of convergence z, the series does not converge absolutely since k k is the well-known divergent harmonic series. For a power series k a k(z z 0 ) k, the limit depends only on the coefficients a k. Thus, if a (i) lim n+ n a n L 0, the radius of convergence is R /L. Esto es así porque para que la serie converja el cociente lim an+ z z0 n+ a n z z 0 n LR <. a (ii) lim n+ n a n 0, the radius of convergence is R (iii) lim n a n+ a n, the radius of convergence is R 0 Similar conclusions can be made for the root test. lim an+ a n R 22

5 ( ) k+ k! (z i) k. With the iden- Example onsider the power series k tification a n ( ) n+ /n! we have ( ) n+2 lim (n+)! n lim ( ) n+ n! n n+ 0 (3.22) Hence the radius of convergence is ; the power series with center z 0 +i converges absolutely for all z, that is, for z i <. Example onsider the power series ( n, 6n+ 2n+5) the root test gives lim n k ( 6k+ 2k+5) k(z 2i) k. With a n n 6n+ an lim n 2n+5 3 (3.23) Then, the radius of convergence of the series is R /3. The circle of convergence is z 2i /3; the power series converges absolutely for z 2i < /3. The Arithmetic of Power Series Some facts A power series k0 a k(z z 0 ) k can be multiplied by a nonzero complex constant c without affecting its convergence or divergence. A power series k0 a k(z z 0 ) k converges absolutely within its circle of convergence. As a consequence, within the circle of convergence the terms ofthe series can be rearrangedand the rearrangedseries has the same sum L as the original series. A power series k0 a k(z z 0 ) k and k0 b k(z z 0 ) k can be added and subtracted by adding or subtracting like terms. In symbols: a k (z z 0 ) k ± b k (z z 0 ) k (a k ±b k )(z z 0 ) k (3.24) k0 k0 If both series have the same nonzero radius R of convergence, the radius of convergence of k0 (a k ±b k )(z z 0 ) k is R. k0 Two power series can (with care) be multiplied and divided. 3.2 Taylor Series Throughout the discussion in this section we will assume that a power series has either a positive or an infinite radius R of convergence. Differentiation and Integration of Power Series The three theorems that follow indicate a function f that is defined by a power series is continuous, differentiable, and integrable within its circle of convergence. 23

6 Theorem (6.6): ontinuity A power series k0 a k(z z 0 ) k represents a continuous function f within its circle of convergence z z 0 R. Theorem (6.7): Term-by-Term Differentiation Apowerseries k0 a k(z z 0 ) k can be differentiated term by term within its circle of convergence z z 0 R. It follows as a corollary to Theorem 6.7 that a power series defines an infinitely differentiable function within its circle of convergence and each differentiated series has the same radius of convergence R as the original power series. Theorem (6.8): Term-by-Term Integration A power series k0 a k(z z 0 ) k can be integratedterm-by-termwithin its circleofconvergence z z 0 R, for every contour lying entirely within the circle of convergence. The theorem states that a k (z z 0 ) k dz a k (z z 0 ) k dz (3.25) k0 whenever lies in the interior of z z 0 R. Indefinite integration can also be carried out term by term: a k (z z 0 ) k dz a k (z z 0 ) k a k dz k + (z z 0) k+ +constant k0 k0 k0 k0 The ratio test can be used to be prove that both (3.26) a k (z z 0 ) k (3.27) k0 and k0 a k k + have the same circle of convergence z z 0 R. (3.28) Taylor Series Suppose a powerseriesrepresentsafunction f within z z 0 R, that is, f(z) a k (z z 0 ) k a 0 +a (z z 0 )+a 2 (z z 0 ) 2 +a 3 (z z 0 ) 3 + (3.29) k0 It follows from Theorem 6.7 that the derivatives of f are the series f (z) f (z) a k k(z z 0 ) k a +2a 2 (z z 0 )+3a 3 (z z 0 ) 2 + k a k k(k )(z z 0 ) k 2 2 a a 3 (z z 0 )+ k2. 24

7 Since the power series f(z) k0 a k(z z 0 ) k represents a differentiable function f within its circle of convergence z z 0 R, where R is either a positive number or infinity, we conclude that a power series represents an analytic function within its circle of convergence. By evaluating the derivatives at z z 0 we get, In general, with n 0. Then f(z) f(z 0 ) a 0 (3.30) f (z 0 )!a (3.3) f (z 0 ) 2!a 2 (3.32) f (z 0 ) 3!a 3 (3.33) a n f(n) (z 0 ) n! k0. (3.34) (3.35) f (k) (z 0 ) (z z 0 ) k (3.36) k! This series is called the Taylor series for f centered at z 0. Maclaurin series It is the Taylor series for z 0 0, f(z) k0 f (k) (0) z k (3.37) k! Theorem (6.9): Taylor s Theorem Let f be analytic within a domain D and let z 0 be a point in D. Then f has the series representation f(z) k0 f (k) (z 0 ) (z z 0 ) k (3.38) k! valid for the largest circle with center at z 0 and radius R that lies entirely within D. Proof: See pag. 36 of Ref. [2]. We can find the radius of convergence of a Taylor series as the distance from the center z 0 of the series to the nearest isolated singularity of f. Where, an isolated singularity is a point at which f fails to be analytic but is, nonetheless, analytic at all other points throughout some neighborhood of the point. For example, z 5i is an isolated singularity of f(z) /(z 5i). If the function f is entire, then the radius of convergence R. 25

8 Some Important Maclaurin Series e z + z! + z2 2! + z k k! k0 sinz z z3 3! + z5 5! ( ) k z 2k+ (2k +)! k0 cosz z2 2! + z4 4! k0 ( ) k z2k (2k)! (3.39) (3.40) (3.4) Example Suppose the function f(z) (3 i)/( i + z) is expanded in a Taylor series with center z 0 4 2i. What is its radius of convergence R? Solution: Observe that the function is analytic at every point except at z +i, which is an isolated singularity of f. The distance from z +i to z 0 4 2i is z z 0 ( 4) 2 +( ( 2)) 2 34 R (3.42) The power series expansion of a function, with center z 0, is unique. On a practical level this means that a power series expansion of an analytic function f centered at z 0, irrespective of the method used to obtain it, is the Taylor series expansion of the function. Example For example, we can obtain cosz z2 2! + z4 4! k0 ( ) k z2k (2k)! (3.43) by simply differentiating sinz z z3 3! + z5 5! ( ) k z 2k+ (2k +)! k0 (3.44) term by term. Example For example, the Maclaurin series for e z2 can be obtained by replacing the symbol z in e z + z! + z2 2! + z k k! k0 (3.45) by z 2,i.e. e z2 + (z2 )! + (z2 ) 2 2! + + z2! + z4 2! + k0 (z 2 ) k k0 z 2k k! k! (3.46) (3.47) 26

9 Example Find the Maclaurin expansion of f(z) ( z) 2. Solution: From, z +z +z2 +z 3 + (3.48) valid for z <, we differentiate both sides with respect to z to get ( z) 2 +2z +3z2 + kz k (3.49) Since we are using Theorem 6.7, the radius of convergence of the last power series is the same as the original series, R. Example We can often build on results such as the above one. For example, if we want the Maclaurin expansion of f(z) z3 ( z) 2, we simply multiply the above equation by z 3 : z 3 ( z) 2 z3 +2z 4 +3z 5 + Its radius of convergence is still R. k kz k+2 (3.50) Example Expand f(z) /( z) in a Taylor series with center z 0 2i. Solution: By using the geometric series we have k z z +2i 2i 2i (z 2i) 2i z 2i 2i (3.5) (3.52) Next, we use the power series by replacing z z 2i 2i, z 2i [ 2i z 2i 2i + z 2i 2i + ( ) 2 z 2i + z 2i 2i 2i + (3.53) ( ) 3 z 2i + ] (3.54) 2i 2i + z 2i (z 2i)2 + + (3.55) ( 2i) 2 ( 2i) 3 Because the distance from the center z 0 2i to the nearest singularity z is 5, we conclude that the circle of convergence is z 2i 5. Remark As a consequence of Theorem 5., we know that an analytic function f is infinitely differentiable. As a consequence of Theorem 6.9, we know that an analytic function f can always be expanded in a power series with a nonzero radius R of convergence. In real analysis, a function f can be infinitely differentiable, but it may be impossible to represent it by a power series. 27

10 3.3 Laurent Series If a complex function f fails to be analytic at a point z z 0, then this point is said to be a singularity or singular point of the function. For example, the complex numbers z ±2i are singularities of the function f(z) z/(z 2 +4); the nonpositive x-axis and the branch point z 0 are singular points of Lnz. In this section we will be concerned with a new kind of power series expansion of f about an isolated singularity z 0. Isolated Singularities Suppose that z z 0 is a singularity of a complex functionf. Thepointz z 0 issaidtobeanisolatedsingularityofthefunctionf ifthere existssomedeleted neighborhood, orpunctured open disk, 0 < z z 0 < R of z 0 throughout which f is analytic. For example, z ±2i are isolated singularities of f(z) z/(z 2 +4). On the other hand, the branch point z 0 is not an isolated singularity of Lnz. We say that a singular point z z 0 of a function f is nonisolated if every neighborhood of z 0 contains at least one singularity of f other than z 0. For example, the branch point z 0 is a nonisolated singularity of Lnz. Series with negative powers If z z 0 is a singularity of a function f, then certainly f cannot be expanded in a power series with z 0 as its center. However, about an isolated singularity z z 0, it is possible to represent f by a series involving both negative and nonnegative integer powers of z z 0, f(z) + a 2 (z z 0 ) 2 + a +a 0 +a (z z 0 )+a 2 (z z 0 ) 2 z z 0 a k (z z 0 ) k + a k (z z 0 ) k (3.56) k k0 The series with negative powers is called principal part and will converge for z z 0 < r, i.e. z z 0 > /r r. The part of nonnegative powers is called the analytic part and will converge for z z 0 < R. Then, the sum converges when z is a point in an annular domains defined by r < z z 0 < R. Example The function f(z) sinz/z 4 is not analytic at the isolated singularity z 0 and hence cannot be expanded in a Maclaurin series. Since sinz is an entire function with Maclaurin series given by, for z <, we have sinz z z3 3! + z5 5! z7 + (3.57) 7! f(z) sinz z 4 z 3 3!z + z 5! z3 + (3.58) 7! The analytic part of this series converges for z <. The principal part is valid for z > 0. Thus, it converges for all z except at z 0, i.e. 0 < z <. 28

11 Figure 3.: (From Ref. [2]) Theorem (6.0): Laurent s Theorem Let f be analytic within the annular domain D defined by r < z z 0 < R. Then f has the series representation f(z) k a k (z z 0 ) k (3.59) valid for r < z z 0 < R. The coefficients a k are given by a k f(s) 2πi (s z 0 ) k+ds (3.60) with k 0,±,, where is a simple closed curve that lies entirely within D and has z 0 in its interior. See Figure 3. Proof: See pag. 327 in Ref. [2]. Regardlesshow a Laurentexpansion ofafunction f is obtained in a specified annular domain it is the Laurent series; that is, the series we obtain is unique. Inthecasewhena k 0fork,2,3,, theprincipalpartofthelaurent series is zero and it reduces to a Taylor series. Thus, a Laurent expansion can be considered as a generalization of a Taylor series. Example Expand f(z) z(z ) in a Laurent series valid for the following annular domains: (a) 0 < z < (b) < z (c) 0 < z < (d) < z (a) We expand the geometric series /( z) valid for z <, f(z) z which converges for 0 < z <. z (3.6) z [+z +z2 +z 3 + ] (3.62) z z z2 (3.63) 29

12 (b) To obtain a series that converges for < z, we start by constructing a series that converges for /z < < z, f(z) z 2 z [ z 2 + z + + ] z 2 (3.64) (3.65) (c) We rewrite f(z) and proceed like in (a), f(z) z 2 + z (3.66) z4 ( +z)(z ) (3.67) z +(z ) (3.68) z [ (z )+(z )2 (z ) 3 + ] (3.69) z +(z ) (z )2 + (3.70) The requirement that z is equivalent to 0 < z, and the geometric series in brackets converges for z <. Thus the last series converges for z satisfying 0 < z and z <, that is, for 0 < z <. (d) Proceeding as in part (b), we write f(z) (z )(+(z )) (z ) 2 + (z ) 2 z [ z + (z ) 2 (z ) 3 + ] (3.7) (3.72) (3.73) (z ) 2 (z ) 3 + (3.74) (z ) 4 Because the series within the brackets converges for /(z ) <, the final series converges for < z. Example Expand f(z) (z ) 2 (z 3) in a Laurent series valid for: (a) 0 < z < 2 and (b) 0 < z 3 < 2. Solution: 220

13 (a) We need to express z 3 in terms of z f(z) (z ) 2 (z 3) [ ] (z ) 2 2+(z ) [ ] 2(z ) 2 2(z ) 2 [ z 2 + z 2 (3.75) (3.76) (3.77) ( ) 2 ( ) 3 z z ] (3.78) 2 2 2(z ) 2 4(z ) 8 (z ) (3.79) 6 (b) To obtain powers of z 3, we write z 2+(z 3) and f(z) (z ) 2 (3.80) (z 3) z 3 [2+(z 3)] 2 (3.8) [ + z 3 ] 2 (3.82) 4(z 3) 2 [ + ( 2) ( ) z 3 + ( 2)( 3) ( ) 2 z 3 + ] 4(z 3)! 2 2! 2 4(z 3) (z 3) 8 (z 3)2 + (3.83) where we have used the binomial series (( + z) α + αz + α(α ) 2! z 2 + α(α )(α 2) 3! z 3 + valid for z < ). The binomial expansion is valid for (z 3)/2 <, i.e. z 3 < 2. Example Expand f(z) 8z+ z( z) in a Laurent series valid for 0 < z <. Solution: by partial fractions we write 8z + f(z) z( z) z + 9 z (3.84) (3.85) z +9+9z +9z2 + (3.86) the geometric series convergesfor z <, but due to the term /z, the resulting Laurent series is valid for 0 < z <. Example Expand f(z) e 3/z in a Laurent series valid for 0 < z <. Solution: For all finite z <, is valid the expansion e z +z + z2 + (3.87) 2! 22

14 z z 0 Laurent series for 0 < z z 0 < R Removable singularity a 0 +a (z z 0 )+a 2 (z z 0 ) 2 + Pole of order n a n (z z 0) n + + a z z 0 +a 0 +a (z z 0 )+ Simple pole a z z 0 +a 0 +a (z z 0 )+ Essential singularity + a 2 (z z 0) 2 +a 0 +a (z z 0 )+ We obtain the Laurent series f by simply replacing z 3/z, for z 0, valid for 0 < z <. e 3/z + 3 z (3.88) 2!z2 3.4 Zeros and Poles We will assign different names to the isolated singularity z z 0 according to the number of terms in the principal part of the Laurent series. lassification of Isolated Singular Points An isolated singular point z z 0 of a complex function f is given a classification depending on whether the principal part of its Laurent expansion contains zero, a finite number, or an infinite number of terms. (i) Removable singularity: If the principal part is zero, that is, all the coefficients a k are zero, then z z 0 is called a removable singularity. (ii) Pole: If the principal part contains a finite number of nonzero terms, then z z 0 is called a pole. If, in this case, the last nonzero coefficient is a n, n, then we say that z z 0 is a pole of order n. If z z 0 is pole of order, then the principal part contains exactly one term with coefficient a. A pole of order is commonly called a simple pole. (iii) Essential singularity: If the principal part contains an infinitely many nonzero terms, then z z 0 is called an essential singularity. See table 3.4 Removable Singularity In the series, sinz z z2 3! + z4 (3.89) 5! that allthe coefficientsinthe principalpartofthelaurentseriesarezero. Hence z 0 is a removable singularity of the function f(z) (sinz)/z. If a function f has a removable singularity at the point z z 0, then we can always supply an appropriate definition for the value of f(z 0 ) so that f 222

15 becomes analytic at z z 0. For instance, since the right-hand side of the series expansion of (sinz)/z is when we set z 0, it makes sense to define f(0). Hence the function f(z) (sinz)/z, as given by sinz z z2 3! + z4 (3.90) 5! is now defined and continuous at every complex number z. Indeed, f is also analytic at z 0 because it is representedby the Taylorseries z 2 /3!+z 4 /5! centered at 0. Example. (i) sinz z 2 z z 3! + z3 (3.9) 5! sinz z 4 z 3 3!z + z (3.92) 5! for 0 < z <, the z 0 is a simple pole of the function f(z) (sinz)/z 2 and a pole of order 3 of the function g(z) (sinz)/z 4. (ii) The expansion of f of the Example 3 of Section 6.3 valid for 0 < z < 2 was given by the equation f(z) (z ) 2 (z 3) 2(z ) 2 4(z ) 8 z (3.93) 6 Then, z is a pole of order 2. (iii) The value z 0 is an essential singularity of f(z) e 3/z. Zeros A number z 0 is zero of a function f if f(z 0 ) 0. We say that an analytic function f has a zero of order n or a zero of multiplicity n at z z 0 if but A zero of order is called a simple zero. f(z 0 ) 0 (3.94) f (z 0 ) 0 (3.95) f (z 0 ) 0 (3.96). (3.97) f (n ) (z 0 ) 0 (3.98) f (n) (z 0 ) 0 (3.99) Example For f(z) (z 5) 3 we see that f(5) 0, f (5) 0, f (5) 0, but f (5) 6 0. Thus f has a zero of order (or multiplicity) 3 at z 0 5. Theorem (6.): Zero of Order n A function f that is analytic in some disk z z 0 < R has a zero of order n at z z 0 if and only if f can be written where φ is analytic at z z 0 and φ(z 0 ) 0 f(z) (z z 0 ) n φ(z) (3.00) 223

16 Example The analytic function f(z) zsinz 2 has a zero at z 0. If we replace z by z 2 in the series expansion of sinz we get f(z) zsinz z 3 z7 3! + z (3.0) 5! z 3 [ z4 3! + z8 ] 5! (3.02) z 3 φ(z) (3.03) with φ(0) 0, then z 0 is a zero of f of order 3. Theorem (6.2): Pole of Order n A function f analytic in a punctured disk 0 < z z 0 < R has a pole of order n at z z 0 if and only if f can be written f(z) φ(z) (z z 0 ) n (3.04) where φ is analytic at z z 0 and φ(z 0 ) 0. More about zeros A zero z z 0 of an analytic function f is isolated in the sense that there exists some neighborhood of z 0 for which f(z) 0 at every point z in that neighborhood except at z z 0. As a consequence, if z 0 is a zero of a nontrivial analytic function f, then the function /f(z) has an isolated singularity at the point z z 0. Theorem (6.3): Pole of Order n If the functions g and h are analytic at z z 0 and h has a zero of order n at z z 0 and g(z 0 ) 0, then the function f(z) g(z)/h(z) has a pole of order n at z z 0. Examples. (i) The rational function f(z) 2z +5 (z )(z +5)(z 2) 4 (3.05) shows that the denominator has zeros of order at z and z 5, and a zero of order 4 at z 2. Since the numerator is not zero at any of these points, it follows that f has simple poles at z and z 5, and a pole of order 4 at z 2. (ii) The value z 0 is a zero of order 3 of zsinz 2. Then, we conclude that the reciprocal function f(z) /(zsinz 2 ) has a pole of order 3 at z 0. Remarks. (i) From the preceding discussion, it should be intuitively clear that if a function f has a pole at z z 0, then f(z) as z z 0 from any direction and we can write lim z z0 f(z). (ii ) A function f is meromorphic if it is analytic throughout a domain D, except possibly for poles in D. It can be proved that a meromorphic function can have at most a finite number of poles in D. For example, the rational function f(z) /(z 2 +) is meromorphic in the complex plane. 224

17 3.5 Residues and Residue Theorem Residue The coefficient a of /(z z 0 ) in the Laurent series is called the residueofthefunctionf attheisolatedsingularityz 0, notedasa Res(f(z),z 0 ). Examples. (i) z is a pole of order two of the function f(z) (z ) 2 (z 3). From the Laurent series obtained above valid for the deleted neighborhood of z defined by 0 < z < 2, f(z) /2 (z ) 2 + /4 z 8 z... (3.06) 6 we have Res(f(z),) /4. (ii) z 0 is an essential singularity of f(z) e 3/z. From its Laurent series e 3/z + 3 z (3.07) 2!z2 valid for 0 < z <, we get Res(f(z),0) 3. The following theorem gives a way to obtain the residues of a function f without the necessity of expanding f in a Laurent series. Theorem (6.4): Residue at a Simple Pole If f has a simple pole at z z 0, then Res(f(z),z 0 ) lim (z z 0 )f(z) (3.08) z z 0 Prof: Since f has a simple pole at z z 0, its Laurent expansion convergent on a punctured disk 0 < z z 0 < R has the form f(z) a z z 0 +a 0 +a (z z 0 )+ (3.09) where a 0. By multiplying both sides of this series by z z 0 and then taking the limit as z z 0 we obtain the above relation. Theorem (6.5): Residue at a Pole of Order n If f has a pole of order n at z z 0, then Res(f(z),z 0 ) Proof: See pag. 344 of Ref. [2]. d n lim (n )! z z 0 dz n (z z 0) n f(z) (3.0) Example The function f(z) (z ) 2 (z 3) has a simple pole at z 3 and a pole of order 2 at z. Find the residues. Solution: Since z 3 is a simple pole, we have: Res(f(z),3) lim(z 3)f(z) lim z 3 z 3 (z ) 2 4 (3.) 225

18 For the pole of order 2, we have Res(f(z),) (2 )! lim d 2 z dz 2 (z )2 f(z) (3.2) d lim z dz (z )2 (z ) 2 (z 3) (3.3) d lim z dz (z 3) (3.4) lim z (z 3) 2 4 (3.5) When f is not arationalfunction, calculatingresiduesbymeans ofthe above limits can sometimes be tedious. An alternative residue formula can be obtain if the function f can be written as a quotient f(z) g(z)/h(z), where g and h are analytic at z z 0. If g(z 0 ) 0 and if the function h has a zero of order at z 0, then f has a simple pole at z z 0 and Proof: Let us write the derivative of h Res(f(z),z 0 ) g(z 0) h (z 0 ) (3.6) h h(z) h(z 0 ) h(z) (z) lim lim (3.7) z z 0 z z 0 z z 0 z z 0 by using the definition of residues and h(z 0 ) 0, we get Res(f(z),z 0 ) lim z z 0 (z z 0 )f(z) (3.8) lim (z z 0 ) g(z) (3.9) z z 0 h(z) g(z) lim (3.20) z z 0 h(z) z z 0 lim z z 0 g(z 0) h (z 0 ) g(z) (3.2) h(z) h(z 0) z z 0 (3.22) Example The polynomial z 4 + can be factored as (z z )(z z 2 )(z z 3 )(z z 4 ), where z e iπ/4, z 2 e 3iπ/4, z 3 e 5iπ/4, and z 4 e 7iπ/4 are its four distinct roots. Then, the function f(z) /(z 4 +) has four simple poles. By using Res(f(z),z 0 ) g(z0) h (z 0) we get Res(f(z),z ) Res(f(z),z 2 ) Res(f(z),z 3 ) Res(f(z),z 4 ) 4z 3 4z 3 2 4z 3 3 4z e 3iπ/4 4 2 i e 9iπ/4 4 2 i e 5iπ/ i e 2iπ/ i (3.23) (3.24) (3.25) (3.26)

19 Figure 3.2: n singular points within contour (From Ref. [2]) Alternatively we can use the expression lim z z0 (z z 0 )f(z) for each, pole Res(f(z),z i ) lim (z z i ) z zi (z z )(z z 2 )(z z 3 )(z z 4 ) (3.27) for example, Res(f(z),z ) lim (z z ) z z (z z )(z z 2 )(z z 3 )(z z 4 ) (z z 2 )(z z 3 )(z z 4 ) (e iπ/4 e 3iπ/4 )(e iπ/4 e 5iπ/4 )(e iπ/4 e 7iπ/4 ) and then, work out the above expression to reduce it to 4 2 i 4 2. Theorem (6.6): auchy s Residue Theorem Let D be a simply connected domain and a simple closed contour lying entirely within D. If a function f is analytic on and within, except at a finite number of isolated singular points z,z 2,,z n within, then f(z)dz 2πi n Res(f(z),z k ) (3.28) k Proof: Suppose, 2,, n are circles centered at z,z 2,,z n, respectively. Suppose further that each circle k has a radius r k small enough so that, 2,, n are mutually disjoint and are interior to the simple closed curve, Fig Weknownfromearlierdevelopthat k f(z)dz 2πiRes(f(z),z k ), and so by Theorem 5.5 we have f(z)dz n k k f(z)dz 2πi n Res(f(z),z k ) (3.29) k 227

20 Example Evaluate (z ) 2 (z 3) dz for the following two contours: (i) a rectangle defined by x 0, x 4, y, y, (ii) the circle z 2. Solution: (i) Since both z and z 3 are poles within the rectangle we have (z ) 2 dz 2πi[Res(f(z),)+Res(f(z),3)] (3.30) (z 3) [ 2πi 4 + ] 0 (3.3) 4 (3.32) (ii) Since only the pole z lies within the circle z 2, we have (z ) 2 dz 2πiRes(f(z),) (3.33) (z 3) ( ) 2πi i π (3.34) 4 2 Example Evaluate 2z+6 z 2 +4dz where the contour is the circle z i 2: Solution: By factoring the denominator as z 2 +4 (z 2i)(z+2i) we see that the integrand has simple poles at 2i and 2i. Because only 2i lies within the contour, we get 2z +6 z 2 dz 2πiRes(f(z),2i) (3.35) +4 ( ) 3+2i 2πi π(3+i2) (3.36) 2i Example Evaluate e z z 4 +5z dz where the contour is the circle z 2: 3 Solution: By factoring the denominator as z 4 +5z 3 z 3 (z+5) we see that the integrand has a pole of order 3 at z 0 and a single pole at z 5. But only the first one is inside, then e z z 4 +5z3dz 2πiRes(f(z),0) (3.37) 2πi 2! lim z 0 d 2 πi lim z 0 dz 2z3 d 2 dz 2z3 e z z 4 +5z 3 (3.38) e z z 3 (z +5) d 2 e z πi lim z 0 dz 2 (z +5) (3.39) (3.40) (z 2 +8z +7)e z πi lim z 0 (z +5) 3 (3.4) i 7π 25 (3.42) 228

21 Example Evaluate tanzdz, where the contour is the circle z 2. c Solution: The integrand f(z) tanz sinz/cosz has simple poles at the points where cosz 0, i.e. z (2n+)π/2, n 0,±,. Since only π/2 and π/2 are within the circle z 2, we have tanzdz 2πi[Res(f(z), π/2)+res(f(z),π/2)] (3.43) With the identifications g(z) sinz, h(z) cosz, and h (z) sinz, we get [ sin( π/2) tan zdz 2πi sin( π/2) + sin(π/2) ] sin(π/2) (3.44) 2πi[ +( )] 4πi (3.45) Example Evaluate c e3/z dz, where the contour is the circle z. Solution: z 0 is an essential singularity of the integrand f(z) e 3/z and so neither of the two above procedure are applicable to find the residue of f at that point. Nevertheless, we saw demonstrate above that i.e. Res(f(z),0) a 3. From, c e 3/z + 3 z (3.46) 2!z2 f(z)dz 2πi n Res(f(z),z k ) (3.47) k where z k are the isolate singularities of f, we get c e 3/z dz 2πi n Res(f(z),z k ) 2πiRes(f(z),0) 6πi (3.48) k 3.6 Some onsequences of the Residue Theorem 3.6. Evaluation of Real Trigonometric Integrals Integrals of the Form 2π 0 F(cosθ,sinθ)dθ The basic idea here is to convert a real trigonometric integral into a complex integral, where the contour is the unit circle z centered at the origin. We begin by parametrizing the contour by z e iθ, 0 θ 2π, then dz ie iθ dθ (3.49) cosθ eiθ +e iθ 2 sinθ eiθ e iθ 2i (3.50) (3.5) 229

22 or dθ dz iz (3.52) cosθ 2 (z +z ) (3.53) sinθ 2i (z z ) (3.54) then 2π F(cosθ,sinθ)dθ 0 where is the unit circle z. F( 2 (z +z ), 2i (z z )) dz iz (3.55) Example Evaluate 2π 0 (2+cosθ) dθ 2 Solution: using the above substitutions we get dz [2+ 2 (z +z )] 2 iz dz (3.56) (2+ z2 + 2z )2 iz 4 z i (z 2 +4z +) 2dz (3.57) 4 z i [(z z )(z z 2 )] 2dz (3.58) with z 2 3 and z Because only z 2 is inside the unit circle, we have dz [2+ 2 (z +z )] 2 iz 4 i 4 i z [(z z )(z z 2 )] 2dz (3.59) z (z z ) 2 (z z 2 ) 2dz (3.60) 4 i 2πiRes(f(z),z 2) (3.6) where then d Res(f(z),z 2 ) lim z z 2 dz (z z 2) 2 f(z) (3.62) d lim z z 2 dz (z z 2) 2 z (z z ) 2 (z z 2 ) 2 (3.63) d z lim z z 2 dz (z z ) 2 (3.64) z z lim z z 2 (z z ) (3.65) dz [2+ 2 (z +z )] 2 iz 4 i 2πiRes(f(z),z 2) (3.66) 4 i 2πi 6 3 4π 3 3 (3.67) (3.68) 230

23 and, finally, 2π 0 4π (2+cosθ) 2dθ 3 3 (3.69) Evaluation of Real Improper Integrals Integrals of the Form f(x)dx Suppose y f(x) is a real function that is defined and continuous on the interval [0, ) defined as with f(x)dx I I f(x)dx+ 0 f(x)dx lim R f(x)dx I +I 2 (3.70) R f(x)dx lim R 0 0 f(x)dx (3.7) R f(x)dx (3.72) providedboth integralsi andi 2 areconvergent. If either one, I ori 2, isdivergent, then f(x)dx is divergent. It is important to remember that the above R definition for the improper integral is not the same as lim R R f(x)dx. For the integral f(x)dx to be convergent, the limits lim R R 0 f(x)dx 0 and lim R Rf(x)dx must exist independently of one another. But, in the event that we know (a priori) that an improper integral f(x)dx converges, we can then evaluate it by R f(x)dx lim f(x)dx (3.73) R R R On the other hand, the symmetric limit lim R f(x)dx may exist even R though the improper integral f(x)dx is divergent. The limit in lim R R R f(x)dx (3.74) if it exists, is called the auchy principal value (P.V.) of the integral and is written R P.V. f(x)dx lim f(x)dx (3.75) R R auchy Principal Value When an integral of form f(x)dx converges, its auchy principal value is the same as the value of the integral. If the integral diverges, it may still possess a auchy principal value. About even functions Suppose f(x) is continuous on (, ) and is an even function, that is, f( x) f(x). If the auchy principal value exists, 0 f(x)dx 2 P.V. f(x)dx (3.76) f(x)dx P.V. 23 f(x)dx (3.77)

24 Figure 3.3: (From Ref. [2]) About evaluation of the improper integral To evaluate an integral f(x)dx, where the rational function f(x) p(x)/q(x) is continuous on (, ), byresiduetheorywereplacexbythecomplexvariablez andintegrate the complex function f over a closed contour that consists of the interval [ R,R] on the real axis and a semicircle R of radius large enough to enclose all the poles of f(z) p(z)/q(z) in the upper half-plane Im(z) > 0, Fig By Theorem 6.6 we have R f(z)dz f(z)dz + f(x)dx (3.78) R R n 2πi Res(f(z),z k ) (3.79) k where z k, k,2,,n denotes poles in the upper half-plane. If we can show that the integral R f(z)dz 0 as R, then we have P.V. k R f(x)dx lim f(x)dx (3.80) R R n 2πi Res(f(z),z k ) (3.8) Example Evaluate the auchy principal value of Solution: let us write f(z) (x 2 +)(x 2 +9) dx. (x 2 +)(x 2 +9) (z i)(z +i)(z 3i)(z +3i) (3.82) we take be the closed contour consisting of the interval [ R,R] on the x-axis and the semicircle R of radius R > 3. R f(z)dz f(z)dz + f(z)dz I +I 2 (3.83) R R ( 2πi[Res(f(z),i)+Res(f(z),3i)] 2πi 6i ) 48i π (3.84) 2 We now want to let R. Before doing this, we use the following inequality valid in the contour R (z 2 +)(z 2 +9) z 2 z 2 9 (R 2 )(R 2 9) (3.85) 232

25 Since the length L of the semicircle is πr, it follows from the ML-inequality, Theorem 5.3, that I 2 (z 2 +)(z 2 +9) πr (R 2 )(R 2 (3.86) 9) R Then, I 2 0 as R, and so and then, lim I 2 0 (3.87) R (3.88) Finally, lim R I π 2 (3.89) R lim R R (x 2 +)(x 2 dx P.V. +9) (x 2 +)(x 2 +9) dx π 2 (3.90) Because the integrand f(z) is an even function, the existence of the auchy principal value implies that the original integral converges to π/2, i.e. (x 2 +)(x 2 +9) dx π 2 (3.9) Sufficient conditions under which the contour integral along R approaches zero as R is always true are summarized in the next theorem. Theorem (6.7): Behavior of Integral as R Suppose f(z) p(z)/q(z) is a rational function, where the degree of p(z) is n and the degree of q(z) is m n+2. If R is a semicircular contour z Re iθ, 0 θ π, then R f(z)dz 0 as R. Example Evaluate the auchy principal value of x 4 + dx. Solution: The conditions given in Theorem 6.7 are satisfied. f(z) /(z 4 +) has simple poles in the upper half-plane at z e πi/4 and z 2 e 3πi/4, with residues then PV Res(f(z),z ) 4 2 i Res(f(z),z 2 ) i 4 2 (3.92) (3.93) x 4 + dx 2πi[Res(f(z),z )+Res(f(z),z 2 )] π 2 (3.94) Since the integrand f(z) is an even function, the original integral converges to π/ 2, i.e. x 4 + dx π (3.95) 2 233

26 Fourier Integrals: f(x)cosαxdx and f(x)sinαxdx Fourier integralsappearastherealandimaginarypartsintheimproperintegral f(x)eiαx dx. We can write f(x)e iαx dx f(x)cosαxdx+i f(x) sin αxdx (3.96) whenever both integrals on the right-hand side of converge. Suppose f(x) p(x)/q(x) is a rational function that is continuous on (, ). Then both Fourier integrals in can be evaluated at the same time by considering the complex integral f(z)eiαz dz, where α > 0, and the contour consists of the interval [ R,R] on the real axis and a semicircular contour R with radius large enough to enclose the poles of f(z) in the upper-half plane. The next theorem gives sufficient conditions under which the contour integral along R approaches zero as R. Theorem (6.8): Behavior of Integral as R Suppose f(z) p(z)/q(z) is a rational function, where the degree of p(z) is n and the degree of q(z) is m n+2. If R is a semicircular contour z Re iθ, 0 θ π, and α > 0, then R f(z)e iαz dz 0 as R. Example Evaluate the auchy principal value of 0 Solution: First we rewrite the integral 0 xsinx x 2 +9 dx 2 xsinx x 2 +9 dx. xsinx x 2 dx (3.97) +9 With α we build z z 2 +9 eiz dz (3.98) with a semicircle in the upper complex plane. Using theorem 6.6 where R z z 2 +9 eiz dz + R R x x 2 +9 eix dx 2πiRes(f(z)e iz,3i) (3.99) Res(f(z)e iz z,3i) Res( z 2 +9 eiz,3i) z z3i 2z eiz e 3 2 (3.200) The integral in the contour R goes to zero, then ( ) x e 3 PV x 2 +9 eix dx 2πi i π 2 e 3 (3.20) Then, x xcosx x 2 +9 eix dx x 2 +9 dx+i xsinx x 2 +9 dx i π e 3 (3.202) Equating real and imaginary parts we get PV PV xcosx x 2 dx 0 (3.203) +9 xsinx x 2 +9 dx π e 3 (3.204) 234

27 Figure 3.4: (From Ref. [2]) Finally, in view of the fact that the integrand is an even function, we obtain the value of the required integral, 0 xsinx x 2 +9 dx 2 xsinx x 2 +9 dx π 2e 3 (3.205) Indented ontours In the situation where f has poles on the real axis, we must use an indented(mellado) contour as illustrated in Figure 3.4. The symbol r denotes a semicircular contour centered at z c and oriented in the positive direction. The next theorem is important to this discussion. Theorem (6.9): Integral of functions with pole on the real axis Suppose f has a simple pole z c on the real axis. If r is the contour defined by z c+re iθ, 0 θ π, then lim r 0 Proof: See pag. 359 in Ref. [2] r f(z)dz πires(f(z),c) (3.206) Example Evaluate the auchy principal value of Solution: Let us consider the contour integral sinx x(x 2 2x+2) dx. e iz z(z 2 dx (3.207) 2z +2) e The function f(z) iz z(z 2 2z+2) has a pole at z 0 and at z +i in the upper half-plane. The contour, shown in Figure 3.5, is indented at the origin, then r + R R R + + 2πiRes(f(z)e iz,+i) (3.208) r r By taking the limits R and r 0, it follows from Theorems 6.8 and 6.9 that PV e ix x(x 2 2x+2) dx πires(f(z)eiz,0) 2πiRes(f(z)e iz,+i) 235

28 Figure 3.5: (From Ref. [2]) where Res(f(z)e iz,0) 2 (3.209) then PV Res(f(z)e iz,+i) e +i (+i) (3.20) 4 e ix ) ( x(x 2 2x+2) dx πi 2 +2πi e +i (+i) (3.2) 4 Using e +i e (cos+isin) and equating real and imaginary parts, we get PV PV cosx x(x 2 2x+2) dx π 2 e (sin+cos) (3.22) sinx x(x 2 2x+2) dx π 2 [+e (sin cos)] (3.23) Integration along a Branch ut Branch Point at z 0 Here we examine integrals of the form 0 f(x)dx, where the integrand f(x) is algebraic but when it is converted to a complex function, the resulting integrand f(z) has, in addition to poles, a nonisolated singularity at z 0. Example: Integration along a Branch ut Evaluate 0 x(x+) dx. Solution: The above real integral is improper for two reasons: (i) an infinite discontinuity at x 0 and (ii) the infinite limit of integration. Moreover, it can be argued from the facts that the integrand behaves like x /2 near the origin and like x 3/2 as x, that the integral converges. We form the integral dz (3.24) z /2 (z +) where is the closed contour shown in Figure 3.6 consisting of four components. The integrand f(z) of the contour integral is single valued and analytic on and within, except for the simple pole at z e iπ. Hence we can 236

29 Figure 3.6: (From Ref. [2]) write dz 2πiRes(f(z), ) (3.25) z /2 (z +) + 2πiRes(f(z), ) (3.26) R + + ED r AB (3.27) with f(z). The segment AB coincides with the upper side of the z /2 (z+) positive real axis for which θ 0, z xe 0i ; while, the segment ED coincides with the lower side of the positive real axis for which θ 2π, z xe 0+2π i, then ED r (xe 2πi) /2 R r R R r xe 2πi + (e2πi dx) (3.28) x /2 dx (3.29) x+ x /2 dx (3.220) x+ AB R r R r (xe 0i) /2 xe 0i + (e0i dx) (3.22) x /2 dx (3.222) x+ Now with z re iθ and z Re iθ on r and R, respectively, it can be shown that r 0 (3.223) R 0 (3.224) 237

30 as r 0 and R, respectively. Then [ ] + 2πiRes(f(z), ) (3.225) lim r 0,R R + ED r + AB is the same as with then 2 dx 2πiRes(f(z), ) (3.226) 0 x(x+) Res(f(z), ) z /2 ze e πi/2 i (3.227) iπ 0 x(x+) dx π (3.228) The Argument Principle and Rouché s Theorem Argument Principle Unlike the foregoing discussion in which the focus was on the evaluation of real integrals, we next apply residue theory to the location of zeros of an analytic function. Theorem (6.20): Argument Principle Let be a simple closed contour lying entirely within a domain D. Suppose f is analytic in D except at a finite number of poles inside, and that f(z) 0 on. Then f (z) 2πi f(z) dz N 0 N p (3.229) where N 0 is the total number of zeros of f inside and N p is the total number of poles of f inside. In determining N 0 and N p, zeros and poles are counted according to their order or multiplicities. Proof: See pag. 363 in Ref. [2]. Theorem (6.2): Rouché s Theorem Let be a simple closed contour lying entirely within a domain D. Suppose f and g are analytic in D. If the strict inequality f(z) g(z) < f(z) holds for all z on, then f and g have the same number of zeros (counted according to their order or multiplicities) inside. Proof: See pag. 365 in Ref. [2]. The Rouché s Theorem is helpful in determining the number of zeros of an analytic function (in a given region). 238

31 Bibliography [] David Poole, Linear Algebra. A Modern Introduction. Second Edition. Thomson. Australia [2] Dennis G. Zill and Patrick D. Shanahan. A First ourse in omplex Analysis with Applications. Jones and Bartlett Publishers