aQ»_—_ CHAPTER 7 The Residue Theorem Evaluation formula for a1 given by eet Sale - afte 7 Tele -aVo (75) If k= 1 (simple pole), then the result is especially simple and is given by a, lime a) f@) (76) which is a special case of (7.5) with k = 1 if we define 0! EXAMPLE 7.1: If We have, using (7.6) and (7.5) with k 2M We-+ 1), then z Residue at Residue at es expansions. EXAMPLE 7.2: Let f(c) with w= —1/z, we find from which we see that the residue at z = 0 is the coefficient of 1/z and equals ~1. 7.3 The Residue Theorem Let /(2) be single-valued and analytic inside and on a simple closed curve C except at the singularities a,b, ¢,.. inside C, which have residues given by a1, b-1, C-1,.+. [see Fig. 7-1]. Then, the residue theorem states that fila de= arta +b Hea ) a) © ice., the integral of f(z) around Cis 277 times the sum of the residues of f(z) at the singularities enclosed by €. Note that (7.7) is a generalization of (7.3). Cauchy's theorem and integral formulas are special cases of this theorem (see Problem 7.75) Fig. 74,CHAPTER 7 The Residue Theorem Evaluation —ap zeoth 7.5. Find the residue of F(z) = SESE ay Solution ‘We have, as in Method 2 of Problem 7.4(b), and so the residue (coefficient of 1/2) is 7/45 Another Method. The result can also be obtained by finding 1 dt fs coszeosh a ae ls but this method is much more laborious than that given above. e 1 a e de around the circle C with equation |z| Solution ‘The integrand e" /{z"(z? + 2z + 2)} has a double pole at z 2-4 2e-42.= 0]. All hese poles ate inside C. Residue at and two simple poles at z= —1 + i [roots of 2yte*) — (€)22 +2) (2422427 Residue at eu Cre Residue at asta}a>——_ CHAPTER 7 The Residue Theorem Evaluation that is, on cost fe Definite Integrals of the Type |“, Fix) dx 7.7. Let |F(2)| = M/R* for z = Re where k > 1 and M are constants. Prove that Himg-sasfj-F(2)dz = 0 where Dis > the semi-circular are of radius R shown in Fig. 7-5. F Solution By Property (@) page 112, we have x x ® M =M yds) = 7 Fig. 75 [roe Sq Rae i since the length of are L= 7. Then lim =0 and so jim | F2)dz =0 aa an [ro de i Re, |f(2)| SM/R', k > Lit f@ = 1/@° + D. Solution Suppose 2 = Re". Then el l brary i ieee 1 where R is large enough (say R > 2, for example), so that M Note that we have made use of the inequality [zy + 2] > [ei] — with zy = Re" and 22 = 1 jae 79. Evaluate [ea Solution Consider f. d/(? +1), where Cis the closed contour of Fig. 7-5 consisting of the line from —R to Rand the semicircle T, traversed in the positive (counterclockwise) sense. Since 26 + 1 =0 when = €%, e895, e596, c726, 69/5, e!1/6, these are simple poles of 1/(26 + 1). Only the poles e*, e*"*, and e°* lie within C. Then, using L’Hospital’s rule, Residue at o Jim eS Residue at e°/5 Residue at e'CHAPTER 7 The Residue Theorem Evaluation ——ail> 73. 74. ‘The proof given here establishes the residue theorem for simply-connected regions containing a finite number of singularities of 2). I-can be extended to regions with infinitely many isolated singularities and ‘to multiply-connected regions (see Problems 7.96 and 7.97) Let f(2) be analytic inside and on a simple closed curve C except at a pole a of onder m inside C. Prove that the residue of f(z) at a is given by ap et im al det af} Solution Method 1. Suppose f(2) bas a pole a of order m, Then the Laurent series of fc) is tot aye Fag tay art a a", we have Then multiplying both sides by = A"f2) =n HO gle a) Ho ae ay! Hagle = ay" + 2 This represents the Taylor series about z =a of the analytic function on the let, Differentiating both sides ‘m— 1 times with respect to 2, we have el ro (= arf) = m— Dlg mln — 1) +-2ag(s = a) = Thus, on letting 2 —> a, (2 @)"fe2)} = m= I)! from which the requited result follows. Method 2. The required result also follows directly from Taylor's theorem on noting that the coefficient of (—a)""" in the expansion (2) is fea] im Method 3. See Problem 5.28, page 161 Find the residues of (a) f(@) plane. and (b) f(z) = e esc? z at all its poles in the finite G+Ve +4 Solution (a) fic) has a double pole at < 1 and simple poles at 2 = +21 Method 1. Reside at : = Lis 1d [egy 22% “4 inkl Resal 5 Residue ata» CHAPTER 7 The Residue Theorem Evaluation Residue at : = —2F is 444i FIP 2E +20) CAF HH) 25 tne Method 2. Residue a Lois Gia 4 25 using "Hospital's. In similar mane, orby placing /by—nthe esl we can obisin the residue a= tb) fz sh 2, ‘Method 1. Residue at <= mais ah 1 mar S| tim 1m sin 216 — mm sing = tea Sint) = ee sin’ Letting z— mm =u or 2 = w+ mz, this limit can be written =e" iim inem{ (i sin 2usinu — 217 cos a wf sinu + 2usinu — 2 cos ‘The limit in braces can be obtained using L’Hospital’s rule. However, it is easier to fist note that we wy Bray ENG) and thas write the Timitas cong Sees Bone a) _ pry Ha Dn = BF using L’Hospital’s rule several times. In evaluating this limit, we can instead use the series expansions. sinw bees cos = 12/21 + Method 2 (using Lauren's series) In this method, we expand f(z) = e* ese z ina Laurent series about z 1/(¢ — ma) a the required residue. To make the calculation easier, let be expanded in a Laurent series about u = 0 is &"** ese2(mz-+ u) ‘expansions for e and sin, we find using long division nrand obiain the coefficient of w+ mex. Thea, the function 10 mre! ese? u, Using the Maclaurin and so the residue isCauchy’s Integral Formulas and Related Theorems 5.1__Cauchy’s Integral Formulas Let f(¢) be analytic inside and on a simple closed curve C and let a be any point inside C [Fig. 5-1]. Then FO) de 6.) where C is traversed in the positive (counterclockwise) sense, Also, the nth derivative of fiz) at z = ais given by rea) Eth fe pe n=1,23, (52) ‘The result (5.1) can be considered a special case of (5.2) with n = 0 if we define 0! = 1 Fig. 51 The results (5.1) and (5.2) are called Cauchy's integral formulas and are quite remarkable because they show that if a function f(z) is known on the simple closed curve C, then the values of the function and all its derivatives. can be found at all points inside C. Thus, if a function of a complex variable has a first Gerivative, ie. is analytic, in a simply-connected region %, all its higher derivatives exist in R. This is not necessarily true for functions of real variablesCHAPTER 5 Cauchy’s Integral Formulas and Related Theorems © ————@U> 5.3. Prove that under the conditions of Problem 5.2, ra = ff 2ai J (a) z n=0, 1,23, Solution ‘The cases where n = 0 and | follow from Problems 5.1 and 5.2, respectively, provided we define f(a) = fla) and 0! = 1 To establish the case where Lath -fta) We use Problem 5.2 where « and a+/rlie in R to obtain aati earl af fo hb { 3(¢-a)~2h 2h 12 ay bf Heo 2h Pi Tea Imi Te me— a The result follows on taking the Himit as h —> 0 if we can show that the last term approaches zero. The proof is similar to that of Problem 5.2, for using the Fact that the integral around C equals the integral around I’, wwe have hf 3¢-a)-2% ai J (@-a-hyF Gay lhl MQne) _ alnine @—_cuaPTERS Cauchy's Integral Formulas and Related Theorems Fig. 512 If FL) has a pole of order m at z = ay, then Fa) = where fi(2) is analytic and f(a) #0 IF Flz) has a pole of order mz at z = az, then os Fa) = here f(z) is analytic and fi(az) #0 =o a fala) is analytic and fla) 4: ‘Then, by (1) and part (a), L t raft z mae 2) 1 fz) actif 2 =a aa |e ar ae -ay"Fe a. fing = Dida —1 + tim Ke — a2)" FO) Ifthe limits on the right are denoted by Ry and Ro, we can write fro © 2mi(Ry + Ro) where Ry and Rare called the residues of Fz) atthe poles 2 = a) and 2 = ay In genera, if F(2) has a number of poles inside C with residues Ry, Ro...» then fc F(2)de = 2x times the sum of the residues. This result is called the residue theorem. Applications of this theorem, together with generalization to singularities other than poles, are treated in Chapter 7. e 5.29. Evaluate 6 ———.; dz where C is the circle |z| = feta a z ‘Solution The pk 0 ty = a at = ine Can ath of ot 2 e ati 1d ma Residue at z= 77 is lim | — 7) Gwar a : le a ws wn nae amet my] aa ld oe Residue at =~ is tim, 3 {+ aeawl con 7 a: 2am iden = 2 (2! res faeComplex Differentiation and the Cauchy—-Riemann Equations 3.1_ Derivatives If fic) is single-valued in some region R of the ¢ plane, the derivative of f(z) is defined as 1 provided that the limit exists independent of the manner in which Az — 0. In such a ease, we say that f(z) is differentiable at z. In the definition (3.1), we sometimes use h instead of Az. Although differentiability implies continuity, the reverse is not true (see Problem 3.4), 3.2. Analytic Functions If the derivative fe) exists at all points: of a region R, then fiz) is said to be analytic in R-and is referred to as an analytic function in R or a function analytic in R. The terms regular and holomorphic are sometimes used as synonyms for analytic. ‘A funetion f(z) is said to be analytic at a point zy if there exists a neighborhood |z — zp| < Sat all points of which /'() exists 3.3 Cauchy-Riemann Equations A necessary condition that w = f(z) = u(x, ») + io(x, y) be analytic in a region R is that, in R, wand ¢ satisfy the Cauchy—Riemann equations aude xy 2) If the partial derivatives in (3.2) are continuous in R, then the Caucl conditions that f(z) be analytic in R. See Problem 3.5. The functions u(x, y) and v(x, y) are sometimes called conjugate functions. Given w having continuous first partials on a simply connected region PR (sce Section 4.6), we can find c (within an arbitrary additive constant) so that u + iv = f(2) is analytic (see Problems 3.7 and 3.8), Riemann equations are sufficientaf-— CHAPTER 3 Complex Differentiation 3.4 Harmonic Functions If the second partial derivatives of wand v with respect to x and y exist and are continuous in a region R, then we find from (3.2) that (see Problem 3.6) ar” ay (3) It follows that under these conditions, the real and imaginary parts of an analytic funetion satisfy Laplace's ‘equation denoted by ay ew =0 or VW=0 where Be Py Ba) The operator V? is often called the Laplacian. Fanetions such as u(x, ») and o(x, ») which satisfy Laplace’s equation in a region Rare called harmonic Junctions and are said to be harmonic in R. 3.5 Geometric Interpretation of the Derivative Let co [Fig. 3-1] be a point P in the z plane and let wi (Fig. 3-2] be its image P’ in the w plane under the transformation w = f(z). Since we suppose that f(2) is single-valued, the point zp maps into only one point = plane w plane Fig. 34 Fig. 32 If we give zp an increment Az, we obtain the point Q of Fig. 3-1. This point has image Q" in the w plane, ‘Thus, from Fig. 3-2, we see that PQ represents the complex number Aw = fico + Az) — flo). It follows, that the derivative at zp (if it exists) is given by S20 + Az) = feo) PQ ay ach Po G5) that is, the limit ofthe ratio PG to PQ as point Q approaches point P. The above interpretation clearly holds when zo is replaced by any point z.CHAPTER 3 Complex Differentiation <> ‘Similarly, by differentiating both sides of (1) with respect to y and (2) with respect tox, we find ee ae and » is harmonic. It will be shown later (Chapter 5) that iff(c) is analytic in al its derivatives exist and are continuous in RR. Hence, the above assumptions will not be necessary. 3.7. (a) Prove that u = e~*(xsiny — ycosy) is harmonic. (b) Find w such that f(2) = u + iv is analytic Solution 9) H _ e-yoainyy + (-e-DUasiny — yoosy) = esiny — xe" siny + ye cos eT MGIny) + e-tusiny ~ yeosy) = vt . — e-revcosy + ysiny —cosy) = xe" cosy +36 Mt eccosy yaa eony ty pelte cosy + ye siny — e* cosy) = —xe™'siny +26 siny + ye oosy 2 Adding (1) and (2) yields (u/s?) + (u/ay2) = 0 and wis barron. (b) From the Cauchy—Riemann equations, a rs ave & iny—xe™* siny + ye~*cosy @ cosy — re cosy = ye sin y “ Integrate (3) with respect to y, keeping x constant. Then cosy + xeeosy +e "ysiny + 6089) + FO) siny xe cosy + FO) o ‘where F(x) is an arbitrary real function of x: Substitute (5) into (4) and obtain ye siny ~ xe cosy + &* cosy + Flex) = =ye*siny — xe cosy — ye" siny or P(x) = 0 and Foy ‘constant. Then, from (5), mysiny +.xe0sy) +e For another method, see Problem 3.40, 3.8. Find flo Solution Method 1 We have—?__—_ CHAPTER 3 Complex Differentiation Putting y = 0 Fla) = lx, 0) + WC, 0). Replacing x by 2, f(z) = ule, 0) + inte, 0). ‘Then, from Problem 37, (2, 0) = 0, r(2, 0) arbitrary additive constant Method 2 ind 0 fc) = ule, 0) + in, 0) ice, apart from an Apart from an arbitrary additive constant, we have from the results of Problem 3.7, fla) =utiv = e"rsiny ~ ycosy) + ie siny +.xe0sy) hE) De hES HES) = Het eH = tee Method 3 We have x = (:+/2,y =e —2/2i. Then, substituting into u(x, )+ie(x,»), we find after much tedious labor that 2 disappears and we are left with the result ize In general, method 1 is preferable over methods 2 and 3 when both w and v are known, If only u (oF v) is known, another procedure is given in Problem 3.101 Differentials 3.9. Given w = 2:2, Find: (a) Aw, (b) div, (©) Aw— di. Solution (@ aw=se+A9-f)= [¢ (b) dw = principal pat of Aw = (3c? — 49.8 = Note that (2) = 32? = dz and dw = G2 4: (©) From (a) and (b), Aw — dw = Bz — 2M Az? + (2)? Note that ¢ > O.as Az —> 0, iL¢., (Aw ~ diw)/Az —> Oas Ac —» 0. It follows that Aw — div is an infinitesi- ‘mal of higher order than Az. Differentiation Rules. Derivatives of Elementary Functions 3.10. Prove the following assuming that f(2) and (2) are analytic in a region R d a a Y@)+8@) =F f+ 7s (a) & Fy. a d b) Fest) GSO + OF IO a if giz) #0 p) BESO = © 4 fie de (2) [gor