- 209 -Chapter VIResidues and PolesThe Cauchy integral theorem (Sec. 4.9) states that if a function is analytic at allpoints interior to and on a simple closed path C, then the value of the integral ofthe function around that path is zero.If, however, the function fails to be analytic ata finite number of points interior to C,there is, as we shall see in this chapter, aspecific number, called a residue,which each of those points contributes tothevalue of the integral. We develop here the theory of residues; and, in Chapter VI,weshallgivesomeapplicationsofthetheory$6.1.ResiduesRecall (Sec.2.13) thata point zois a singular point ofa function f if f failsto be analytic at zobut is analytic at some point in every neighborhood of =oDefinition 6.1.1. A singular point zo of a function f is said to beisolated if there is an >0 such that f is analytic for 0z-zk8z +1Example 1. The function f(=)=has the three isolated singular2(22 +1)points z=O and z=±iExample 2. The origin is a singularypoint of the principal branch (Sec.3.3)logz=lnr+io(z=rei0)of the logarithmic function.It is not,however,an isolated singular point sinceeverydeletedneighborhoodofitxcontains points on the negative real axis (seeFig. 6-1) and the branch is not even definedthereExample3.Thefunction1f(a)=Fig. 6-1sin(元 /)
- 209 - Chapter Ⅵ Residues and Poles The Cauchy integral theorem (Sec. 4.9) states that if a function is analytic at all points interior to and on a simple closed path , then the value of the integral of the function around that path is zero. If, however, the function fails to be analytic at a finite number of points interior to , there is, as we shall see in this chapter, a specific number, called a residue, which each of those points contributes to the value of the integral. We develop here the theory of residues; and, in Chapter Ⅶ, we shall give some applications of the theory. C C §6.1. Residues Recall (Sec. 2.13) that a point is a singular point of a function if fails to be analytic at but is analytic at some point in every neighborhood of . 0 z f f 0 z 0 z Definition 6.1.1. A singular point of a function is said to be isolated if there is an 0 z f ε > 0 such that f is analytic for < − ||0 < ε 0 zz . Example 1. The function )1( 1 )( 23 + + = zz z zf has the three isolated singular points z = 0 and ±= iz . Example 2. The origin is a singular point of the principal branch (Sec. 3.3) )(lnlog θ θ i =+= rezirz of the logarithmic function. It is not, however, an isolated singular point since every deleted ε neighborhood of it contains points on the negative real axis (see Fig. 6-1) and the branch is not even defined there. Example 3. The function )/sin( 1 )( z zf π = Fig. 6-1
- 210 -has the singular point z =0 and z =1/n(n =±1, ±2,...), all lying on thesegment of the real axis from z =-1 toz =1. Each singular point except z = 0is isolated. The singular point z = O is not isolated because every deleted neighborhood of the origin contains other singular points of the function. Moreprecisely,whena positivenumber is specified and m is any positiveintegersuch that m >1/ , the fact that 0<1/m< means that the point z =1/ mlies in the deleted neighborhood 0<=k(Fig.6-2)yy60x-m0xFig. 6-3Fig. 6-2When zo is an isolated singular point of a function f,thereis a positivenumber R, such that f is analytic for 0z-zkR,.Consequently, f(z)isrepresentedbyaLaurentseriesbhb,b,f(a)= a,(2-z0)" +(6.1.1)Z-20(2-20)(z--0)"n=0where O<z-zokR, and the coefficients an and b, have certain integralrepresentations (Sec. 5.4). In particular,f(=)dzb.=(n =1,2,..)2元/ Jc(2-20)-1+/where C is any positively oriented simple closed path around zo and lying inthe punctured disk 0<z-zo kR,(Fig. 6-3). When n =1, this expression forb,canbewritten
- 210 - has the singular point z = 0 and = nnz = ± ± K),2,1(/1 , all lying on the segment of the real axis from z = −1 to z = 1. Each singular point except is isolated. The singular point z = 0 z = 0 is not isolated because every deleted ε neighborhood of the origin contains other singular points of the function. More precisely, when a positive number ε is specified and is any positive integer such that m m > /1 ε , the fact that < /10 m < ε means that the point lies in the deleted = /1 mz ε neighborhood < z ||0 < ε (Fig. 6-2) Fig. 6-2 Fig. 6-3 When is an isolated singular point of a function , there is a positive number such that is analytic for 0 z f R2 f 20 < − ||0 < Rzz . Consequently, is represented by a Laurent series zf )( ∑ ∞ = + − + − + − +−= 0 0 2 0 2 0 1 0 )()( )()( n n n n n zz b zz b zz b zzazf L L (6.1.1) where and the coefficients and have certain integral representations (Sec. 5.4). In particular, 20 ||0 <−< Rzz an bn ∫ = − = +− C n n n zz dzzf i b ),2,1( )( )( 2 1 1 0 K π where is any positively oriented simple closed path around and lying in the punctured disk C 0 z 20 −< ||0 < Rzz (Fig. 6-3). When n = 1, this expression for can be written bn
211[(2)dz= 2元ib,(6.1.2)Definition 6.1.2. If zo is an isolated singular point of a function f, then the complexnumber b,, which is the coefficient of 1/(z-z)in expansion (6.1.1), is called the residue off at zo and denoted by Res f(2), or Res(f(z),z). Thus,Res f(z) =[cJ(z)dz and J. f(=)dz = 2mi·Res(f(2),z0))2mJEquation (6.1.2) provides a powerful method for evaluating certain integrals around simpleclosed paths.Example 4.Considerthe integraldz(6.1.3)(2-2)4where C is the positively oriented circle Iz-2 -1(Fig. 6-4). Since the integrand is analyticeverywhere in the finite plane except at the points z =O and z =2, it has a Laurent seriesrepresentation that is valid in the punctured disk 0 z-2k2, also shown in Fig. 6-4.yx0Fig. 6-4Thus, according to equation (6.1.2), the value of integral (6.1.3) is 2i times the residue of itsintegrand at z=2.To determine that residue, werecall (Sec.5.3)theMaclaurin series expansion1.=="(Izk1)1-zne0and use it to write-z(z - 2)4(z - 2)42 +(z - 2)112(z - 2)41-(-2-2(-1)"(z-2)"=4(0 </z-2k2)2n+IIn this Laurent series, the coefficient of 1/(=-2) is the desired residue, namely -1/16.Consequently,dz1元i(6.1.4)=2元／Jc z(2 - 2)4816)
211 . (6.1.2) ∫ = C ibdzzf 2)( π 1 Definition 6.1.2. If is an isolated singular point of a function , then the complex number , which is the coefficient of 0 z f 1 b − zz 0 )/(1 in expansion (6.1.1), is called the residue of f at and denoted by , or . Thus, 0 z )(Res0 zf = zz )),((Res 0 zzf )(Res0 zf = zz ∫ = C dzzf i )( 2 1 π and . ∫ ⋅= C zzfidzzf )),((Res2)( π 0 Equation (6.1.2) provides a powerful method for evaluating certain integrals around simple closed paths. Example 4. Consider the integral ∫C zz − dz 4 )2( , (6.1.3) where C is the positively oriented circle z − = 1|2| (Fig. 6-4). Since the integrand is analytic everywhere in the finite plane except at the points z = 0 and z = 2 , it has a Laurent series representation that is valid in the punctured disk < z − < 2|2|0 , also shown in Fig. 6-4. Fig. 6-4 Thus, according to equation (6.1.2), the value of integral (6.1.3) is 2πi times the residue of its integrand at z = 2 . To determine that residue, we recall (Sec. 5.3) the Maclaurin series expansion )1|(| 1 1 0 = < − ∑ ∞ = zz z n n and use it to write ( ) ( ).2|2|0)2( 2 1 2 2 1 1 )2(2 1 )2(2 1 )2( 1 )2( 1 0 4 1 4 4 4 ∑ ∞ = − + − <−< − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − −− ⋅ − = −+ ⋅ − = − n n n n z z z z zzz z In this Laurent series, the coefficient of z − )2/(1 is the desired residue, namely . Consequently, − 16/1 ∫ ⎟ −=⎠ ⎞ ⎜ ⎝ ⎛ −= C − i i zz dz 816 1 2 )2( 4 π π . (6.1.4)
212Example 5. Let us show thatz=0, where Cis the unit circle|==1.dzexpZSince 1/ z2 is analytic everywhere except at the origin, so is the integrand. The isolated singularpoint z = 0 is interior to C; and, with the aid of the Maclaurin series (Sec. 5.3)心e"=1+-+..(1=<8),3!1!2!onecanwritetheLaurent series expansion(1)111111[=1+=(0△=k8)expl.++1222!24ZTherefore, the residue of the integrand at its isolated singular point z = O is zero, and the valueof the integral is established.We are reminded in this example that, although the analyticity of a function within and on asimple closed path C isa sufficient conditionforthevalue ofthe integral around C tobe zero,it is not a necessary condition
212 Example 5. Let us show that ∫ ⎟ = ⎠ ⎞ ⎜ ⎝ ⎛ C dz z 0 1 exp 2 , where is the unit circle . Since is analytic everywhere except at the origin, so is the integrand. The isolated singular point is interior to ; and, with the aid of the Maclaurin series (Sec. 5.3) C z = 1|| 2 /1 z z = 0 C )|(| !3!2!1 1 32 z ∞<++++= zzz ez L , one can write the Laurent series expansion )||0( 1 !3 11 !2 11 !1 1 1 1 exp 2 ⎟ 2 4 6 ∞<<+⋅+⋅+⋅+=⎠ ⎞ ⎜ ⎝ ⎛ z z zzz L . Therefore, the residue of the integrand at its isolated singular point z = 0 is zero, and the value of the integral is established. We are reminded in this example that, although the analyticity of a function within and on a simple closed path is a sufficient condition for the value of the integral around to be zero, it is not a necessary condition. C C
s6.2.Cauchy'sResidueTheoremIf, except for a finite number of singular points, a function f is analytic inside a simple closedpath C, those singular points must be isolated (Sec. 6.1). The following theorem, which is knownas Cauchy's residue theorem,is a precise statement of the fact that iffis also analytic on Cand if C is positively oriented, then the value of the integral of f around C is 2πi timesthe sum of the residues of f at the singular points inside CTheorem 6.2.1(Cauchy).Let C be a simple closed path,described in the positive sense.Ifa functionfis analytic inside and on C exceptforafinite numberof singularpointsz,(k =,2,..,n) inside C, then[. J(=)dz = 2m之 Res f(a),(6.2.1)二lProof. To prove the theorem, let the point z,(k =1,2,..,n) be centers of positivelyoriented circles Ck which are interior to C and are so small that no two ofthem have points incommon (Fig. 6-5). The circles Ck,together with the simple closed path C, form the boundaryof a closed region throughout which f is analytic and whose interior is a multiply connecteddomain.Hence,according to the Theorem4.11.2,Sec.4.11,we haveJ J()d-], (=)= 0.k=This reduces to equation (6.2.1) because (Sec. 6.1)Jc (=)dz = 2mResf() (k =1,2.,n),and the proof is complete.y2,.22CxoFig. 6-5Example1.Use the Cauchy's residue theorem to evaluate the integral[5z-2dJcz(z-1)whenCisthecircle==2,describedcounterclockwise.Solution. The integrand has the two isolated singularities z = O and z =1, both of whichare interior to C. We can find the residues B, at z = O and B, at z=1 with the aid oftheMaclaurinseries11+=+=2+...(zk1)1- zWe observe first that when 0 </ =k 1(Fig. 6-6)
§6.2. Cauchy’s Residue Theorem If, except for a finite number of singular points, a function is analytic inside a simple closed path , those singular points must be isolated (Sec. 6.1). The following theorem, which is known as Cauchy’s residue theorem, is a precise statement of the fact that if is also analytic on and if is positively oriented, then the value of the integral of around is f C f C C f C 2π i times the sum of the residues of f at the singular points inside C . Theorem 6.2.1(Cauchy). Let be a simple closed path, described in the positive sense. If a function is analytic inside and on C except for a finite number of singular points inside , then C f nkz ),2,1( k = K C ∑ . (6.2.1) ∫ = = = n k C zz zfidzzf k 1 π )(Res2)( Proof. To prove the theorem, let the point nkz ),2,1( k = K be centers of positively oriented circles which are interior to and are so small that no two of them have points in common (Fig. 6-5). The circles , together with the simple closed path , form the boundary of a closed region throughout which is analytic and whose interior is a multiply connected domain. Hence, according to the Theorem 4.11.2, Sec. 4.11, we have Ck C Ck C f ∫ ∑∫ = − = C n k Ck dzzfdzzf 1 )( 0)( . This reduces to equation (6.2.1) because (Sec. 6.1) ∫ = = k = k C zz π kzfidzzf K n),2,1()(Res2)( , and the proof is complete. Example 1. Use the Cauchy’s residue theorem to evaluate the integral Fig. 6-5 ∫ − − C dz zz z )1( 25 when C is the circle z = 2|| , described counterclockwise. Solution. The integrand has the two isolated singularities z = 0 and , both of which are interior to . We can find the residues at z = 1 C B1 z = 0 and at B2 z = 1 with the aid of the Maclaurin series 1 )1|(| 1 1 2 <+++= − zz z z L . We observe first that when z << 1||0 (Fig. 6-6)