《复变函数论 Theory of Complex Variable Functions》课程授课教案（教材讲义，复分析 Complex Analysis）Chapter Ⅴ Series

Chapter VSeriesThis chapter is devoted mainly to series representations of analytic functions. Wepresent theorems that guarantee the existence of such representations, and wedevelop somefacilityinmanipulatingseries.$5.1.ConvergenceofSeriesDefinition 5.1.1. Let (=,} cC be a sequence, then the following formalnotation(5.1.1)is called a series of complex numbers, its nth partial sum is defined asna(5.1.2)Definition 5.1.2. The series (5.1.1) is said to be convergent to the sum S ifthe sequence (S,) of partial sums converges to S; we then writeNote that, since a sequence has at most one limit, a series has at most one sum.When a series does not converge, we say that it divergesTheorem5.1.1.Supposethat2,=x,+iy,(n=1,2,..)andS=X+iywhere xn,yn,X,Y are all real mumbers. Then2.-s(5.1.3)n=lif and only if

Chapter Ⅴ Series This chapter is devoted mainly to series representations of analytic functions. We present theorems that guarantee the existence of such representations, and we develop some facility in manipulating series. §5.1. Convergence of Series Definition 5.1.1. Let be a sequence, then the following formal notation zn }{ ⊂ C ∑ ++++= LL (5.1.1) ∞ = n n n zzzz 21 1 is called a series of complex numbers, its nth partial sum is defined as n n k n k ∑ +++== zzzzS = 21 L 1 n = K)2,1( . (5.1.2) Definition 5.1.2. The series (5.1.1) is said to be convergent to the sum if the sequence of partial sums converges to ; we then write S }{Sn S Sz n ∑ n = ∞ =1 . Note that, since a sequence has at most one limit, a series has at most one sum. When a series does not converge, we say that it diverges. Theorem 5.1.1. Suppose that nnn += iyxz n = K),2,1( and = + iYXS where YXyx are all real numbers. Then nn , Sz n ∑ n = ∞ =1 (5.1.3) if and only if

=Yand(5.1.4)X12n=nProof.To prove the theorem,wefirst writethepartial sums(5.1.2)asS,=X,+iy,(5.1.5)whereX. :andY,-ZykXkk=lL.Then from Proposition 4.2.1,we see thatlimS, =S- limX,=X and limY,=Yn→>αSSince X, and Y, are the nth partial sums of the seriesy,, theandZxn=lnsltheorem is provedThis theorem tells us that one can writeZ(x, +iy,)=>Ex,+iEy.=he/-whenever it is known that the two series on the right converge or that the one on theleftdoes.By recalling from calculus that the nth term of a convergent series of realnumbers approacheszero asntendsto infinity,wecan see immediately fromTheorem 5.1.1that the same is true for a convergent series of complex numbers.That is, a necessary conditionfor the convergence of series (5.1.1)is thatlimz, =0(5.1.6)Definition 5.1.3. Series (5.1.1) is called absolutely comvergent if the series2-2元+n=ln=1of realnumbers convergesSincex≤x,+y,=zIandy≤/x,+y,==/forallneNweknowfromthecomparisontestincalculusthat iftheseriesisabsolutely2n=l/x,| andconvergent, then two seriesy,mustconverge.Moreover,sincen=ln=the absolute convergence of a series of real numbers implies the convergence of theseries itself, it follows that there are real numbers X and Y

Xx and . (5.1.4) n ∑ n = ∞ =1 Yy n ∑ n = ∞ =1 Proof. To prove the theorem, we first write the partial sums (5.1.2) as nnn = + iYXS , (5.1.5) where ∑= = n k n k xX 1 and ∑ . = = n k n k yY 1 Then from Proposition 4.2.1, we see that = ⇔ ∞→ n SS n lim n XX n = ∞→lim and n YY n = ∞→lim . Since and are the nth partial sums of the series and , the theorem is proved. Xn Yn ∑ ∞ n=1 n x ∑ ∞ n=1 n y This theorem tells us that one can write ∑∑∑ ∞ = ∞ = ∞ = +=+ 1 1 1 )( n n n n n n n yixyix whenever it is known that the two series on the right converge or that the one on the left does. By recalling from calculus that the nth term of a convergent series of real numbers approaches zero as tends to infinity, we can see immediately from Theorem 5.1.1 that the same is true for a convergent series of complex numbers. That is, a necessary condition for the convergence of series (5.1.1) is that n = 0lim∞→ n n z . (5.1.6) Definition 5.1.3. Series (5.1.1) is called absolutely convergent if the series ∑∑ ∞ = ∞ = += 1 22 1 n nn n n yxz of real numbers converges. Since || 22 n nnn =+≤ zyxx and || 22 n nnn =+≤ zyxy for all n∈N , we know from the comparison test in calculus that if the series is absolutely convergent, then two series ∑ ∞ n=1 n z ∑ ∞ n=1 n x and ∑ ∞ n=1 n y must converge. Moreover, since the absolute convergence of a series of real numbers implies the convergence of the series itself, it follows that there are real numbers X and Y

to which series (5.1.4)converge. According to Theorem 5.1.1,then, series (5.1.1)converges.Consequently,wegetthefollowingresult.Theorem5.1.2.Everyabsolutelyconvergentseriesofcomplexnumbers isconvergent.In establishing the fact that the sum of a series is a given number S , it is often convenient todefine the remainder p, after n terms:P,=S-S.(5.1.7)SinceS= S, + p, and S, -S=pn-0,we see that the following conclusion holds.Theorem 5.1.3.A series converges to a number Sif and only if the sequence of remainderstendstozeroWe shall make considerableuse of this theorem in our treatment of power series.They areseriesoftheformNZa,(z-zo)" =ao +ai(z-zo)+a(z-zo)"+..+a,(z-zo)"+.",n=0where zo and the coefficients a are complex constants and z may be any point in a statedregion containing zo.In such series, involving a variable z, we shall denote sums, partial sumsand remainders by S(z), S,(=) and P,(=), respectivelyExample. With the aid of remainders, it is easy to verify that1N"=wheneverz<1(5.1.9)1-z1=0We need only recall the identity1+2+2*+.+*"_1-*+(z±1)1- ztowritethepartial sums=1+2+2++*-+_1-2"1AkS,(z) =(z±1)1- 2k=0112"If S(z), then Ip,(=)/HS(=)-S,(2)/=(z ±1)1-2[1- z]It is clear from this that the remainder P(2) tend to zero when <1 but not when ≥1Summationformula(5.1.9)isestablished

to which series (5.1.4) converge. According to Theorem 5.1.1, then, series (5.1.1) converges. Consequently, we get the following result. Theorem 5.1.2. Every absolutely convergent series of complex numbers is convergent. In establishing the fact that the sum of a series is a given number , it is often convenient to define the remainder S ρn after n terms: ρn = − SS n . (5.1.7) Since = SS + ρnn and −=− 0 n n SS ρ , we see that the following conclusion holds. Theorem 5.1.3. A series converges to a number if and only if the sequence of remainders tends to zero. S We shall make considerable use of this theorem in our treatment of power series. They are series of the form ∑ L +−++−+−+=− L ∞ = n n n n n zzazzazzaazza )()()()( 0 2 020 0 0 10 , where and the coefficients are complex constants and may be any point in a stated region containing . In such series, involving a variable , we shall denote sums, partial sums and remainders by , and 0 z an z 0 z z zS )( zS )( n z)( ρn , respectively. Example. With the aid of remainders, it is easy to verify that z z n n − ∑ = ∞ = 1 1 0 whenever z <1. (5.1.9) We need only recall the identity z z zzz n n − − =++++ + 1 1 1 1 2 L z ≠ )1( to write the partial sums 2 1 1 0 1)( − − = ∑ ++++== k n k k n L zzzzzS z z n − − = 1 1 z ≠ )1( . If z zS − = 1 1 )( , then |1| || |)()(||)(| z z zSzSz n n n − ρ =−= z ≠ )1( . It is clear from this that the remainder z)( ρ N tend to zero when z <1 but not when z ≥1. Summation formula (5.1.9) is established

$5.2.TaylorSeriesWeturn nowtoTaylor'stheorem,which is oneofthemost importantresults ofthechapterTheorem 5.2.1(Taylor). Suppose that f is analytic in a disk-zo<Ro (Fig. 5-1).Thenf(=)hasthepowerseriesrepresentationf(2)=a,(z-20)" (z-20l<Ro),(5.2.1)ne0wheref(n(o)(n=0,1,2,...)(5.2.2)a,=n!OxFig.5-1Proof.We first prove the theorem when z=0, in which case series (5.2.1) becomes()=≥(0)(=<Ro)(5.2.3)n!Aaand is called a Maclaurin seriesTo prove (5.2.3), we let IzkR and write=r.Let Codenote any positively orientedcircle =ro, where r<ro < Ro (See Fig. 5-2). Since f is analytic inside and on the circleC and the point z is interior to Co, the Cauchy integral formula yields thatf(s)ds10f(=)=(5.2.4)2元iJS-2y0HR.XFig. 5-2Nowthefactor 1/(s-z)intheintegrand herecanbeput in theform

§5.2. Taylor Series We turn now to Taylor’s theorem, which is one of the most important results of the chapter. Theorem 5.2.1(Taylor). Suppose that f is analytic in a disk <− Rzz 00 (Fig. 5-1). Then zf )( has the power series representation n n n zzazf )()( 0 0 ∑ −= ∞ = ( ) <− Rzz 00 , (5.2.1) where ! )( 0 )( n zf a n n = n = K),2,1,0( . (5.2.2) Fig. 5-1 Proof. We first prove the theorem when z0 = 0 , in which case series (5.2.1) becomes ( ) ∑ ∞ = = 0 ! (0) ( ) n n n z n f f z ( ) R0 z < (5.2.3) and is called a Maclaurin series. To prove (5.2.3), we let and write 0 | z |< R z = r . Let denote any positively oriented circle C0 0 z = r , where (See Fig. 5-2). Since is analytic inside and on the circle and the point 0 R0 r < r < f C0 z is interior to , the Cauchy integral formula yields that C0 ∫ π − = 0 ( ) 2 1 ( ) C s z f s ds i f z . (5.2.4) Now the factor 1/(s − z) in the integrand here can be put in the form Fig. 5-2

111(5.2.5)1-(=/s)S-zandweknowfromtheexampleinSec.5.2that17(5.2.6)1-z1-zwhen=isanycomplexnumberotherthanunity.Replacing zbyz/s inexpression (5.2.7)then,wecanrewriteequation(5.2.5)as111P-k +="(5.2.8)ka05f+l(s - 2)s"S-zMultiplying through this equation by f(s) and then integrating each side with respect to saround Co, we find thatf(s)dsf(s)ds_f(s)dsk+ 5k+0(s-z)s"-In view of expression (5.2.4) and the fact (Sec. 4.13) that二[ ()d _ () (k = ,1..),54+2元k!this reduces,after we multiply throughby 1 /(2元 i),to(+ P(),f(2)=(5.2.9)kok!whereP,(=)=二[ (s)ds(5.2.10)2元iJco(s-z)s"Representation (5.2.3)nowfollows once it is shown thatlim p.(=) = 0(5.2.11)To accomplish this, we recall that == r and that Co has radius ro, where ro >r. Then, ifs is a point on Co, we can see that s-z≥ s-z= r-r. Consequently, if M denotesthemaximumvalue off(s)on Co,thenMle,(2)≤.Mro72元=>0(n→0),2元（n-)rro-rrbecause of (r /r)<1. Limit (5.2.11) clearly holds.To verify the theorem when the disk of radius Rois centered at an arbitrarypoint =o,wesuppose that f is analytic when =-zo<R。 and note that the composite functionf(z+zo) must be analytic when [(z+zo)-zo<R. This last inequality is, of course, just[=|< R.; and, if we writ g(2)= f(z+ zo), the analyticity of g in the disk |=|<R。 ensuresthe existence ofa Maclaurin series representation:g(a)=g((=<R)n!n=0That is,(+20)=2(2 (<R),n!2=0

)/(1 111 − szszs ⋅= − ; (5.2.5) and we know from the example in Sec. 5.2 that z z z z n n k k − += − ∑ − 1 = 1 1 1 0 (5.2.6) when is any complex number other than unity. Replacing by in expression (5.2.7), then, we can rewrite equation (5.2.5) as z z /sz n n n k k k szs zz szs )( 11 1 1 0 1 − = + − ∑ − = + . (5.2.8) Multiplying through this equation by and then integrating each side with respect to around , we find that sf )( s C0 ∫ ∑∫ ∫ − = + − = + − 1 0 1 0 0 0 )( )( )( )( n k C C n nk k C szs dssf zz s dssf zs dssf . In view of expression (5.2.4) and the fact (Sec. 4.13) that ! )0()( 2 1 )( 1 0 k f s dssf i k C k = ∫ + π k = K),2,1,0( , this reduces, after we multiply through by π i)2/(1 , to ∑ − = = + 1 0 )( )( ! )0( )( n k n k k zz k f zf ρ , (5.2.9) where ∫ − = 0 )( )( 2 )( C n n n szs dssf i z z π ρ . (5.2.10) Representation (5.2.3) now follows once it is shown that = 0)(lim∞→ z n n ρ . (5.2.11) To accomplish this, we recall that = rz and that has radius , where . Then, if is a point on , we can see that C0 0r > rr0 s C0 −=−≥− rrzszs 0 |||| . Consequently, if M denotes the maximum value of sf )( on , then C0 2 )(0 )(2 )( 0 0 0 0 00 ⎟ ⎟ ∞→→ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = − ⋅≤ n r r rr Mr r rrr Mr z n n n n π π ρ , because of 1)/( . Limit (5.2.11) clearly holds. rr 0 < To verify the theorem when the disk of radius is centered at an arbitrary point , we suppose that is analytic when R0 0 z f <− Rzz 00 and note that the composite function )( must be analytic when 0 + zzf 000 )( <−+ Rzzz . This last inequality is, of course, just < Rz 0 ; and, if we writ )()( 0 = + zzfzg , the analyticity of g in the disk < Rz 0 ensures the existence of a Maclaurin series representation: ∑ ∞ = = 0 )( ! )0( )( n n n z n g zg )( < Rz 0 . That is, ∑ ∞ = =+ 0 0 )( 0 ! )( )( n n n z n zf zzf )( < Rz 0

After replacing z by z-zo in this equation and its condition of validity, we have the desiredthe expansion (5.2.1). The proof is completed.Note that the series (5.2.1) can be written(a)= (c0)+ ()(2(-)()R),-1!2!(5.2.12)which is called the Taylor expansion of f(-) about the point zo:And the series on the righthand side of (5.2.12) is called the Taylor series about zo-It is the familiar Taylor series fromcalculus,adaptedtofunctionsofacomplexvariableFrom the Taylor's theorem above, we know that any function which is analytic at a point zocan be expanded as a Taylor series about zoFor, if f is analytic at zo, then it is analyticthroughout some neighborhood -zo< R。 of that point (Sec. 2.13) and then has the Taylorexpansion. Also, if f is entire, R can be chosen arbitrarily large; and the condition ofvaliditybecomes-z<oo.The series then converges to f() at each point z in the finite plane

After replacing by in this equation and its condition of validity, we have the desired the expansion (5.2.1). The proof is completed. z 0 − zz Note that the series (5.2.1) can be written ()( ) !2 )( )( !1 )( )()( 00 2 0 0 0 0 0 Rzzzz zf zz zf zfzf <−+− ′′ +− ′ += L , (5.2.12) which is called the Taylor expansion of about the point . And the series on the right hand side of (5.2.12) is called the Taylor series about . It is the familiar Taylor series from calculus, adapted to functions of a complex variable. zf )( 0 z 0 z From the Taylor’s theorem above, we know that any function which is analytic at a point can be expanded as a Taylor series about . For, if is analytic at , then it is analytic throughout some neighborhood 0 z 0 z f 0 z <− Rzz 00 of that point (Sec. 2.13) and then has the Taylor expansion. Also, if is entire, can be chosen arbitrarily large; and the condition of validity becomes f R0 zz 0 ∞<− . The series then converges to zf )( at each point z in the finite plane

$5.3.ExamplesWhen it is known that f is analytic everywhere inside a circle centered at zo,the convergenceofits Taylor series about zo to f(z) for each point z within that circle is ensured, no test forthe convergence of the series is required. In fact, according to Taylor's theorem, the seriesconverges to f(z) within the circle about zo whose radius is the distance from zo to thenearest point z, where f fails to be analytic. In Sec. 5.8, we shall find that this is actually thelargest circle centeredat zo such that the series converges tof(z)for all z interior to itAlso,inSec.5.9,weshall seethatif thereareconstants a.(n=0,1,2...)such thatf(=)=Za,(z-zo)"n=(for all points z interior to some circle centered at =o,then the power series here must be theTaylor series for f about zo, regardless of how those constants arise. This observation oftenallows us to find the coefficients a, in Taylor series in more efficient ways that by appealingdirectly to theformula a, =f(n)(zo)/n! in Taylor's theorem.In the following expansions, we usetheformula in Taylor's theorem to find the Maclaurinseries expansions of somefairly simplefunctions,and weemphasizetheuseofthose expansionsin finding other representations.In our example, we shall freely use expected properties ofconvergent series, such as those verified in Exercises 3 and 4, Sec. 5.1.Example1. For f(z)=e", we have f(n)(=)=e= and f(n)(O)=1. Thus,52"ei=>(2in!急n!2i0n!But 1-(-1)" =0 when n is even, and so we can replace n by 2n+1 in this last series:;2n+1_2n+11P1Z[1-(-1 2n+(<8)sinz=2i0(2n+1)!Inasmuchas1-(-1)2m+1 = 2, 2n+1 =(i2)"i=(-1)"i,this reduces to

§5.3. Examples When it is known that is analytic everywhere inside a circle centered at , the convergence of its Taylor series about to for each point f 0 z 0 z zf )( z within that circle is ensured; no test for the convergence of the series is required. In fact, according to Taylor’s theorem, the series converges to within the circle about whose radius is the distance from to the nearest point where fails to be analytic. In Sec. 5.8, we shall find that this is actually the largest circle centered at such that the series converges to for all zf )( 0 z 0 z 1 z f 0 z zf )( z interior to it. Also, in Sec. 5.9, we shall see that if there are constants na = K)2,1,0( n such that ∑ ∞ = −= 0 0 )()( n n n zzazf for all points interior to some circle centered at , then the power series here must be the Taylor series for about , regardless of how those constants arise. This observation often allows us to find the coefficients in Taylor series in more efficient ways that by appealing directly to the formula in Taylor’s theorem. z 0 z f 0 z an !/)( 0 )( nzfa n n = In the following expansions, we use the formula in Taylor’s theorem to find the Maclaurin series expansions of some fairly simple functions, and we emphasize the use of those expansions in finding other representations. In our example, we shall freely use expected properties of convergent series, such as those verified in Exercises 3 and 4, Sec. 5.1. Example 1. For , we have and . Thus, z )( = ezf n z )( = ezf )( 1)0()( = n f )( ! 0 = ∑ ∞< ∞ = z n z e n n z . (5.3.1) The entire function also has a Maclaurin series expansion. The simplest way to obtain it is to replace by on each side of equation (5.3.1) and then multiply through the resulting equation by : z ez 32 z 3z 2 z 2 0 32 ! 3 + ∞ = = ∑ n n n z z n ez z ∞< )( . Finally, if we replace n by n − 2 here, we have n n n z z n ez ∑ ∞ = − − = 2 2 32 !)2( 3 z ∞< )( . Example 2. One can use expansion (5.3.1) and the definition (Sec. 3.6) i ee z iziz 2 sin − − = to find the Maclaurin series for the entire function = sin)( zzf . To give the details, we refer to expansion (5.3.1) and write ! ])1(1[ 2 1 ! )( ! )( 2 1 sin 0 0 0 n zi in iz n iz i z nn n n n n n n ∑∑∑ ∞ = ∞ = ∞ = −−=⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − = − z ∞< )( . But =−− 0)1(1 when is even, and so we can replace by n n n n +12 in this last series: !)12( ])1(1[ 2 1 sin 1212 0 12 + −−= ∞ ++ = + ∑ n zi i z nn n n z ∞< )( . Inasmuch as iiii n n n n )1()(,2)1(1 12 212 −===−− + + , this reduces to

2n+（-("(=<1),(5.3.6)1-2oThe derivatives of the function f(=)=1/(1-z), which fails to be analytic at z=1, aren!f(n)(z)=(n=0,1,2...);(1-2)and, in particular, f(")(0)= n!. Note that expansion (5.3.6) gives us the sum of an infinitegeometric series,where =is the common ratioofadjacentterms:11+z+22+23+..(<1)二This is, of course, the summation formula that was found in another way in the example is Sec.5.1.If we substitute -z for z in equation (5.3.6) and its condition of validity, and note that[=<1 when |- z|<1, we see that 1(-1)"2"(=<1).1+z1=If, on the other hand, we replace the variable z in equation (5.3.6) by 1-z,we have theTaylorseriesrepresentation1-2(-1)(=-1)(z-1/<1)2=0This condition of validity follows from the one associated with expansion (5.3.6)since 1-z<1is the same as [=-1<1

∑ ∞ = + + −= 0 12 !)12( )1(sin n n n n z z z ∞< )( . (5.3.2) Term by term differentiation will be justified in Sec. 5.8 using that procedure here, we differentiate each side of equation (5.3.2) and write ∑ ∑ ∞ = + ∞ = + + −= + − = 0 12 2 0 !)12( 12 )1( !)12( )1( cos n n n n n n z n n z dz d n z . Thus, ∑ ∞ = −= 0 2 !)2( )1(cos n n n n z z z ∞< )( . (5.3.3) Example 3. Because sinh = − iziz )sin( (Sec. 3.7), we need only replace by iz on each side of equation (5.3.2) and multiply through the result by z − i to see that ∑ ∞ = + + = 0 12 !)12( sinh n n n z z z ∞< )( . (5.3.4) Likewise, since cosh = izz )cos( , it follows from expansion (5.3.3) that ∑ ∞ = = 0 2 !)2( cosh n n n z z z ∞< )( . (5.3.5) Observe that the Taylor series for cosh z about the point 2π iz0 = − can be obtained by replacing the variable z by + 2π iz on each side of equation (5.3.5) and then recalling that =π+ cosh)2cosh( ziz for all z : ∑ ∞ = + = 0 2 !)2( )2( cosh n n n iz z π z ∞< )( . Example 4. Another Maclaurin series representation is ∑ ∞ = = 1− 0 1 n n z z z < )1( . (5.3.6) The derivatives of the function = − zzf )1/(1)( , which fails to be analytic at z =1, are 1 )( )1( ! )( + − = n n z n zf n = K)2,1,0( ; and, in particular, . Note that expansion (5.3.6) gives us the sum of an infinite geometric series, where is the common ratio of adjacent terms: !)0()( nf n = z z zzz − =++++ 1 1 1 32 L z < )1( . This is, of course, the summation formula that was found in another way in the example is Sec. 5.1. If we substitute − z for z in equation (5.3.6) and its condition of validity, and note that z < 1 when z <− 1, we see that ∑ ∞ = −= + 0 )1( 1 1 n nn z z z < )1( . If, on the other hand, we replace the variable z in equation (5.3.6) by , we have the Taylor series representation 1− z ∑ ∞ = −−= 0 )1()1( 1 n n n z z z <− )11( . This condition of validity follows from the one associated with expansion (5.3.6) since z <− 11 is the same as z <− 11

Example5.Forourfinal example,letusexpandthefunction1+2-2_12(1+=2)-1_1 (f(2) =2-+"W231+ =21+ 2into a series involving powers of z. We cannot find a Maclaurin series for f(=) since it is notanalytic at z = 0. But we do know from expansion (5.3.6) that=-2*+24-2+2-.()1+22Hence, when 0<<1,(2-1+22-24+-6--*+..)f(2) = --11-2+23-2+.LWe call such terms as 1/23 and 1/z negative powers of z since they can be written -3and z',respectively.The theory of expansions involving negative powers of z-zowill bediscussedinthenext section

Example 5. For our final example, let us expand the function ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + −= + −+ ⋅= + + = 2 3 2 2 353 2 1 1 2 1 1 1)1(2121 )( zzz z zzz z zf into a series involving powers of . We cannot find a Maclaurin series for since it is not analytic at . But we do know from expansion (5.3.6) that z zf )( z = 0 −+−+−= L + 8642 2 1 1 1 zzzz z z < )1( . Hence, when z << 10 , . 11 12( ) 1 )( 53 3 8642 3 L L +−+−+= +−+−+−= zzz z z zzzz z zf We call such terms as and negative powers of since they can be written and , respectively. The theory of expansions involving negative powers of will be discussed in the next section. 3 /1 z /1 z z −3 z −1 z 0 − zz

$5.4.LaurentSeriesIf afunction f fails to be analytic at point zo,we cannot apply Taylor's theorem at that pointIt is often possible,however,to finda seriesrepresentationfor f()involvingboth positive andnegative powers of z -Zo, see Example 5, Exercises 10,11, Sec. 5.3. We now present the theoryofsuch representations,and webegin withLaurent's theorem.Theorem 5.4.1(Laurent). Suppose that a fiunction f is analytic in an annular domainR,<-zo<R, where O≤R,<R, ≤oo and let C denote any positively oriented simpleclosed path around =o and lying in that domain (Fig.5-3) Then f(=) has the seriesrepresentationb2a.(=-=0)+2-f(2)=(R, <z-zo<R,),(5.4.1)(z-z0)"n=0wheref(s)ds1(n=0,1,2...)(5.4.2)anC($-z0)n+2元iandf(s)ds1b, =(n=0,1,2...)(5.4.3)2元/Jc(s-2)-VRx0Fig. 5-3Proof. We shall prove Laurent's theorem first when zo =O, in which case the annulus iscentered at the origin.The verification ofthe theorem when zo is arbitrary will follow readilyyRCFig. 5-4

§5.4. Laurent Series If a function fails to be analytic at point , we cannot apply Taylor’s theorem at that point. It is often possible, however, to find a series representation for involving both positive and negative powers of , see Example 5, Exercises 10,11, Sec. 5.3. We now present the theory of such representations, and we begin with Laurent’s theorem. f 0 z zf )( 0 − zz Theorem 5.4.1(Laurent). Suppose that a function f is analytic in an annular domain 1 <−< RzzR 20 where 0 <≤ RR 21 ≤ ∞ and let C denote any positively oriented simple closed path around and lying in that domain (Fig. 5-3). Then has the series representation 0 z zf )( ∑ ∑ ∞ = ∞ = − +−= 0 1 0 0 )( )()( n n n n n n zz b zzazf ( ) 1 <−< RzzR 20 , (5.4.1) where ∫ + − = C n n zs dssf i a 1 0 )( )( 2 1 π n = K)2,1,0( (5.4.2) and ∫ +− − = C n n zs dssf i b 1 0 )( )( 2 1 π n = K)2,1,0( . (5.4.3) Proof. We shall prove Laurent’s theorem first when z0 = 0 , in which case the annulus is centered at the origin. The verification of the theorem when is arbitrary will follow readily. 0 z Fig. 5-3 Fig. 5-4