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=>(<8)(5.3.1)n=0 n!The entire function 2e3= also has a Maclaurin series expansion. The simplest way toobtain it is to replace zby 3zon each side of equation (5.3.1) and then multiply through theresulting equation by z?:2e*-23m2(<8)h=o n!Finally,ifwe replace n by n-2 here, we have32F22e3 =(<8)=2 (n-2)Example 2.One can useexpansion (5.3.1)and the definition (Sec.3.6)e"-e-"sinz:P2ito find the Maclaurin series for the entire function f(-)=sin z.To give the details, we refer toexpansion(5.3.1)andwrite1[(iz)"(-iz)"=↓2[1-(-1)"4<8)sinz=>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+（-(<8),sinz =(5.3.2)(2n+1)!1=0Termby term differentiation will be justified in Sec.5.8 using that procedure here, we differentiateeach side of equation (5.3.2)and write(-1)"2n+14.2ml =2(-1)2斤COSz=(2n+1)!=o(2n+1)! dzn=0Thus,2hCOSZ=(<8)(5.3.3)(2n)!adExample3.Because sinhz=-isin(iz)(Sec.3.7),we need only replace =by iz oneachsideofequation(5.3.2)andmultiplythroughtheresultby-i to seethat, ≥2n+lsinh z =(=<8).(5.3.4)"=0(2n+1)!Likewise, since coshz = cos(iz), it follows from expansion (5.3.3) thatF22ncoshz=(=<8),(5.3.5)= (2n)!Observe that the Taylor series for coshz about the point zo=-2元i can be obtainedby replacing the variable z by z+2元i on each side of equation (5.3.5)and then recalling thatcosh(z+2元i)=coshzforallz:Seancoshz=(=<8).(2n)!Example 4.Another Maclaurin series representation is1=>"(=<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