5z -2 = 5z -2.-1=/5-2(-1-2-22 -..);2(2-1)"21-2-(Z)and, by identifying the coefficient of 1/z in the product on the right here, we find that B, =2Also, since5z -225(z-1)+3135+-[1-(2-1)+(2-1) -.],z-1z(z -1)1+(z - 1)z-1when 0<z-1k1, it is clear that B, = 3. Thus5z-2dz=2元i(B,+B,)=10元iJc z(=- 1)yx2Fig. 6-6
1( ) 2 5 1 125 )1( 25 ⎟ 2 −−−− L ⎠ ⎞ ⎜ ⎝ ⎛ −= − − ⋅ − = − − zz zzz z zz z ; and, by identifying the coefficient of in the product on the right here, we find that . Also, since /1 z 2 B1 = ,])1()1(1[ 1 3 5 )1(1 1 1 3)1(5 )1( 25 ⎟ 2 −−+−− L ⎠ ⎞ ⎜ ⎝ ⎛ − += −+ ⋅ − +− = − − zz z z z z zz z when z <−< 1|1|0 , it is clear that 3 B2 = . Thus iBBidz zz z C π 10)(2 π )1( 25 21 =+= − − ∫ . Fig. 6-6
s6.3.UsingaSingleResidueIf the function f in Cauchy's residue theorem (Sec. 6.2) is, in addition, analytic at each point ofthe finite plane exterior to C,it is sometimes more efficient to evaluatethe integral offaround C by finding a single residue of a certain related function. We present the method as atheorem.Theorem6.3.1.Ifafunctionfisanalyticeverywhereinthecomplexplaneexceptforafinite number of singular points interiorto apositively oriented simple closedpath C,thenJ t= 2mrg[(](6.3.1)Proof.Webegin the derivation of expression (6.3.1)by constructing a circlez=R,as inFig.6-7.Thenif Cdenotesapositivelyoriented circle==R,whereR>R,.weknowfromLaurent'stheorem (Sec.5.4)thatEc,2"f(2) =(R <zk0)(6.3.2)1=-00wheref(=)dz(n=0,±1,±2,...)(6.3.3)2+2元iyCoC-1x0/R.!R。Fig. 6-7By writing n = -1 in expression (6.3.3), we find thatf(=)dz=2元ic-1(6.3.4)Observe that, since the condition of validity with representation (6.3.2) is not of the typeO=kR2, the coefficient C_ is not necessarily the residue of f at the point z=O,whichmaynot even bea singular point of f.But, if we replace=by 1/zin representation (6.3.2)and its condition of validity, we see that1-2%-2%k-RZand hencec.-Rs[ (9](6.3.5)
§6.3. Using a Single Residue If the function in Cauchy’s residue theorem (Sec. 6.2) is, in addition, analytic at each point of the finite plane exterior to C , it is sometimes more efficient to evaluate the integral of around by finding a single residue of a certain related function. We present the method as a theorem. f f C Theorem 6.3.1. If a function is analytic everywhere in the complex plane except for a finite number of singular points interior to a positively oriented simple closed path , then f C ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ∫ = z f z idzzf C z 11 Res2)( 2 0 π . (6.3.1) Proof. We begin the derivation of expression (6.3.1) by constructing a circle as in Fig. 6-7. Then if denotes a positively oriented circle 1 || = Rz C0 0 || = Rz , where , we know from Laurent’s theorem (Sec. 5.4) that > RR 10 )( )||( , (6.3.2) = ∑ 1 ∞<< ∞ −∞= zRzczf n n n where ∫ = ±±= + 0 ),2,1,0( )( 2 1 1 C n n n z dzzf i c K π . (6.3.3) Fig. 6-7 By writing n −= 1 in expression (6.3.3), we find that ∫ = − 0 2)( 1 C π icdzzf . (6.3.4) Observe that, since the condition of validity with representation (6.3.2) is not of the type , the coefficient is not necessarily the residue of at the point , which may not even be a singular point of . But, if we replace 2 ||0 << Rz −1 c f z = 0 f z by in representation (6.3.2) and its condition of validity, we see that /1 z ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ = = << ⎠ ⎞ ⎜ ⎝ ⎛ ∑∑ ∞ −∞= − ∞ −∞= + 1 2 2 2 1 ||0 11 R z z c z c z f z n n n n n n and hence ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = = − z f z c z 11 Res 2 0 1 . (6.3.5)
Then, in view of equations (6.3.4) and (6.3.5),1, ()d = 2 / e[1 (]SFinally, since f is analytic throughout the closed region bounded by C and Co,the principleof deformation of paths (Corollary 4.11.2, Sec. 4.11) yields the desired result (6.3.1).Example1.Inthe example in Sec.6.2, we evaluated the integral of5z-2f(=)=-(z - 1)around the circle =2, described counterclockwise, by finding the residues of f() atz=0 and z=1.Sincewhen 0=k1,1(1)5-2z5-2z1Z1- 2z(1- 2)(Z)5(1+z+22+..1(252+3+3z+··,Zwe see from Theorem 6.3.1 that, where the desired residue is 5,5z-2dz=2元i×5=10元Jc z(z -1)where C is the circle in question. This is just the result obtained in Example 1 in Sec. 6.2
Then, in view of equations (6.3.4) and (6.3.5), ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ∫ = z f z idzzf C z 11 Res2)( 2 0 0 π . Finally, since is analytic throughout the closed region bounded by and , the principle of deformation of paths (Corollary 4.11.2, Sec. 4.11) yields the desired result (6.3.1). f C C0 Example 1. In the example in Sec. 6.2, we evaluated the integral of )1( 25 )( − − = zz z zf around the circle , described counterclockwise, by finding the residues of at and . Since when z = 2|| zf )( z = 0 z = 1 < z < 1||0 , ,33 5 1(2 ) 5 1 125 )1( 2511 2 2 L L +++= ⎟ +++ ⎠ ⎞ ⎜ ⎝ ⎛ −= − ⋅ − = − − ⎟ = ⎠ ⎞ ⎜ ⎝ ⎛ z z zz z zz z zz z z f z we see from Theorem 6.3.1 that, where the desired residue is 5, iidz zz z C 1052 ππ )1( 25 =×= − − ∫ , where C is the circle in question. This is just the result obtained in Example 1 in Sec. 6.2
s6.4.TheThreeTypesof IsolatedSingularPointsWe saw in Sec. 6.1 that the theory of residues is based on the fact that if f has an isolatedsingularpoint zo,thenf()canberepresentedbya Laurentseriesb,bb,f(=)=a,(z-z0)" +(6.4.1)(z - 20)3(2-=0)"=-20n=0in a punctured disk O < z-zk R,. The portionb,b,b.(z -z0)2(z - z0)"2-20of the series, involving negative powers of z-zo, is called the principal part of f at o.Wenow use the principal part to identify the isolated singular point zo as one of three special types.This classification will aid us in the development of residue theory that appears in followingsections.Definition 6.4.1. If the principal part of f at zo contains at least one nonzero term butthenumberof suchterms isfinite,thenthereexistsapositiveinteger m suchthatbm￥0andbm+l=bm+2=...=0.In this case, the isolated singular point zo is called a pole of order m of f. A pole of orderm =1 of f is usually referred to as a simple pole of fThus, the isolated singular point zo is a pole of order m of f if and only if the Laurentexpansion takestheformbb,b,(6.4.2)f(2)=a,(-zo(z - 20)(-z0))n=0wherebm±0and0z-zkRExample1.Observe that thefunction3322 2z +32(z - 2)+3(04z-2k8)2+(z-2)--2Z-22-2has a simple pole at zo = 2. Its residue b, there is 3.Example2.ThefunctionN10sinhz1112+...(0<=k)hasapole++24357oforder m=3at z=0, with residue b, =1/6Thereremaintwoextremes,thecaseinwhichall of thecoefficients intheprincipalpart arezeroandtheoneinwhichaninfinitenumberofthemarenonzeroDefinition 6.4.2.When all of the b,'s are zero, the point =o is called a removablesingular point offThus, the isolated singular point zois a removable singular point of f if and only if theLaurentexpansiontakestheform(2)=a,(z-z0)" =a +a(z-z0)+a2(z-z0)2 +-(0/z-z k R). (6.4.3)=0Note that the residue at a removable singular point is always zero.If we define,or possiblyredefine,f at z。so thatf(=)=ao,expansion (6.4.3)becomes validthroughouttheentiredisk Iz-zok R,.Since a power series always represents an analytic function interior to itscircle of convergence (Sec.5.8), it follows thatf is analytic at z when it is assigned the
§6.4. The Three Types of Isolated Singular Points We saw in Sec. 6.1 that the theory of residues is based on the fact that if has an isolated singular point , then can be represented by a Laurent series f 0 z zf )( ∑ ∞ = + − ++ − + − +−= 0 0 2 0 2 0 1 0 )()( )()( n n n n n zz b zz b zz b zzazf L L (6.4.1) in a punctured disk . The portion 20 ||0 <−< Rzz L +L − ++ − + − n n zz b zz b zz b )()( 0 2 0 2 0 1 of the series, involving negative powers of 0 − zz , is called the principal part of at . We now use the principal part to identify the isolated singular point as one of three special types. This classification will aid us in the development of residue theory that appears in following sections. f 0 z 0 z Definition 6.4.1. If the principal part of at contains at least one nonzero term but the number of such terms is finite, then there exists a positive integer such that f 0 z m bm ≠ 0 and m+1 = bb m+2 = L = 0 . In this case, the isolated singular point is called a pole of order of . A pole of order of is usually referred to as a simple pole of . 0 z m f m = 1 f f Thus, the isolated singular point is a pole of order of if and only if the Laurent expansion takes the form 0 z m f ∑ ∞ = − ++ − + − +−= 0 0 2 0 2 0 1 0 )()( )()( n m n m n zz b zz b zz b zzazf L , (6.4.2) where ≠ 0 and . bm 20 ||0 <−< Rzz Example 1. Observe that the function )|2|0( 2 3 )2(2 2 3 2 3)2( 2 32 2 ∞<−< − +−+= − += − +− = − +− z z z z z z zz z zz has a simple pole at 2 . Its residue there is 3. z0 = 1 b Example 2. The function )||0( !7!5 1 !3 11 !7!5!3 1sinh 3 3 753 4 4 ∞<<+++⋅+=⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ++++= z zz z z zzz z zz z L L has a pole of order m = 3 at 0 , with residue z0 = 6/1 b1 = . There remain two extremes, the case in which all of the coefficients in the principal part are zero and the one in which an infinite number of them are nonzero. Definition 6.4.2. When all of the are zero, the point is called a removable singular point of . n 'sb 0 z f Thus, the isolated singular point is a removable singular point of if and only if the Laurent expansion takes the form 0 z f )()( .)||0()()( (6.4.3) 0 20 2 ∑ 0 10 20 0 ∞ = <−<+−+−+=−= n n n zzazzaazzazf L Rzz Note that the residue at a removable singular point is always zero. If we define, or possibly redefine, f at so that 0 z 00 )( = azf , expansion (6.4.3) becomes valid throughout the entire disk . Since a power series always represents an analytic function interior to its circle of convergence (Sec. 5.8), it follows that is analytic at when it is assigned the 20 || <− Rzz f 0 z