成果綜合綜合

第四復(fù)變第四復(fù)變函數(shù)的本章介紹復(fù)變函數(shù)的級(jí)數(shù)概念重點(diǎn)是Taylor級(jí)數(shù)、Laurent級(jí)數(shù)及其展1復(fù)數(shù)列的1復(fù)數(shù)列的2級(jí)數(shù)的3典型1.復(fù)數(shù)列的極n1.復(fù)數(shù)列的極nanibn(n1,2,L為一串復(fù)數(shù)，稱{n為一復(fù)數(shù)列簡稱數(shù)設(shè){n}為一數(shù)列，aib為一確定的復(fù)數(shù)若對(duì)任意給定0,相應(yīng)地都能找到一個(gè)正數(shù)N(nNnn時(shí)，n以為極limn此時(shí)也稱復(fù)數(shù)列{n}收斂于復(fù)數(shù)列{nn12,L收斂于liman復(fù)數(shù)列{nn12,L收斂于limanlimbnb那末對(duì)于任意給定的如果limn證能找到一個(gè)正整數(shù)N,當(dāng)nN時(shí)有(anibn)(aib)a)i(bnb)analimanlimbn即limbn反之limaabblimbn反之limaabb那末當(dāng)nN時(shí)nn22n(anibn)(a(ana)i(bnanabnblimn所下列數(shù)列是否收斂如果收斂求出其極限1ni(1)n1下列數(shù)列是否收斂如果收斂求出其極限1ni(1)n1i(2) (1)n;nn1n(3)zne.22.復(fù)數(shù)項(xiàng)級(jí)設(shè){2.復(fù)數(shù)項(xiàng)級(jí)設(shè){n{anbnn1,2,L)為一復(fù)數(shù)列n12LnLnnSnk12Lk則稱級(jí)n收斂如果部分則稱級(jí)n收斂如果部分和數(shù){Sn收斂limSnS則稱級(jí)n發(fā)散若部分和數(shù){Sn}不收斂說明：與實(shí)數(shù)項(xiàng)級(jí)數(shù)相同,limSn例如,級(jí)數(shù)zn1 1z例如,級(jí)數(shù)zn1 1zz2Lzn-(zn1lim lim11z1時(shí),由于n11z1時(shí)級(jí)數(shù)收斂級(jí)數(shù)n(an級(jí)數(shù)n(anibn收斂的充要條件是an和bn都收斂Sn12L證(a1a2Lan)i(b1b2Lbnnin根據(jù)Sn極限存在的充根據(jù)Sn極限存在的充要條件{n}和{n}的極限存在收斂的充要條件是an即和都收斂cn1n1發(fā)散n1n1發(fā)散因?yàn)閍n解n1收斂2n因?yàn)閷?shí)數(shù)項(xiàng)級(jí)數(shù)an和bn收斂因?yàn)閷?shí)數(shù)項(xiàng)級(jí)數(shù)an和bn收斂的必要條limanlimbn0和n收斂的必要條件所以復(fù)數(shù)項(xiàng)limnlimn0級(jí)數(shù)n發(fā)散limn判別級(jí)數(shù)的斂散性時(shí)nlimlimelimn判別級(jí)數(shù)的斂散性時(shí)nlimlimeein因例如:收斂n為絕對(duì)收斂如果如果收斂,那末n也收斂n成立22bannn2222b如果收斂,那末n也收斂n成立22bannn2222bbaaba,而nnnnnn及都收斂根據(jù)實(shí)數(shù)項(xiàng)級(jí)數(shù)的絕對(duì)收斂性ann及是收斂的也都收斂nnkk,根據(jù)實(shí)數(shù)項(xiàng)級(jí)數(shù)的絕對(duì)收斂性ann及是收斂的也都收斂nnkk,knnlimknkkkkk即.b由,nnnnnnnkb2知,kkkkkkb由,nnnnnnnkb2知,kkkkkkan與bn絕對(duì)收斂時(shí)n絕對(duì)收斂n絕對(duì)收an與bn絕對(duì)收斂1nin(1例n11nin(1)(cosisin因?yàn)?1解nnnn所以 (11nin(1例n11nin(1)(cosisin因?yàn)?1解nnnn所以 (11)cosπ(11)sinbnnnnnnliman1lim而i所以數(shù)列 (11)en收斂1nnn1i級(jí)是否收斂n1i解n1i1(1)n但n13n1i級(jí)是否收斂n1i解n1i1(1)n但n13n12121311n(1L)i(1n111雖(1)n因?yàn)榧?jí)數(shù)收斂發(fā)散n1n原級(jí)數(shù)仍發(fā).(8i)n級(jí)數(shù)是否絕對(duì)收例n8(8i)n(8i)n級(jí)數(shù)是否絕對(duì)收例n8(8i)nn!解收斂故原級(jí)數(shù)收斂且為絕對(duì)收斂例4級(jí)數(shù)[1in1因?yàn)槔?級(jí)數(shù)[1in1因?yàn)橐彩諗拷馐諗縩2n(1)n但為條件收斂n冪1冪級(jí)數(shù)的冪1冪級(jí)數(shù)的2冪級(jí)數(shù)的斂3冪級(jí)數(shù)的1.冪級(jí)數(shù)的概設(shè)fn(z1.冪級(jí)數(shù)的概設(shè)fn(z)}D上的復(fù)變函數(shù)列,fn(z)f1(z)f2(z)Lfn(z)Sn(z)f1(z)f2(z)Lfn(z)Dz0limSn(zDz0limSn(z0S(z0存在，稱fnz)z0收斂且Sz0)為它的和z的一個(gè)函數(shù)SS(z)f1(z)f2(z)Lfn(z)當(dāng)fn(z)cn1(z當(dāng)fn(z)cn1(zfn(z) 時(shí)或c(za)c(za)c(za)2Lc(za)nnn012ncczcz2L或n012n2.冪級(jí)數(shù)的斂散znz2.冪級(jí)數(shù)的斂散znz(0收斂n0z如果在zz,級(jí)數(shù)必絕對(duì)收斂z因?yàn)榧?jí)數(shù)czn收斂lim有 n有c zzznqcz,c則0n因?yàn)榧?jí)數(shù)czn收斂lim有 n有c zzznqcz,c則0n nzz00czcLcznncn012n故級(jí)cnn絕對(duì)收斂對(duì)于一個(gè)冪級(jí)數(shù)其收斂半徑的情況有三種級(jí)數(shù)1zz2LL對(duì)于一個(gè)冪級(jí)數(shù)其收斂半徑的情況有三種級(jí)數(shù)1zz2LLx1則從某個(gè)n開始n212,x均收斂 z=0外都發(fā)散z=0外都發(fā)散1z22z2Lnnznz0時(shí)(3)既存在使級(jí)數(shù)發(fā)散的正實(shí)數(shù)z時(shí),級(jí)數(shù)收斂;z時(shí),級(jí)數(shù)發(fā)散.如圖ycnn的收斂范圍ycnn的收斂范圍是以原點(diǎn)為中心的圓域R .. xc(zac(za)nnza為中心的圓域一般的結(jié)論要對(duì)具體級(jí)數(shù)進(jìn)行具體分析n,, 2nn收斂半徑R均為1,zn,, 2nn收斂半徑R均為1,z2n在點(diǎn)z1發(fā)散在其它點(diǎn)都收斂nlim方法1（比值法如cnlim方法1（比值法如cnn在復(fù)平面內(nèi)處處收斂RcnnRz0均發(fā)散R1(3)0方法2（根值法cnnR方法2（根值法cnnR在復(fù)平面內(nèi)處處收斂cnnRz0均發(fā)散R1(3)0的收斂半徑pn1因?yàn)?nnplim的收斂半徑pn1因?yàn)?nnplimn1)pn(11)nn1R3.冪級(jí)數(shù)的性f(z)azn,Rrg(z)bz,Rrnn1n23.冪級(jí)數(shù)的性f(z)azn,Rrg(z)bz,Rrnn1n2b)znnbf(z)g(z)annnnf(z)g(z)(a)nnzbznn(anb0La0bn)znR(1,r2z其f(z)anznz時(shí)f(z)anznz時(shí)zzRg(z)rf[g(z)]an[g(z)]n時(shí)c(z則n0f(z)cn(zc(z則n0f(z)cn(zzaR在zaRf(z)ncn(za)n1.設(shè)C為zaR內(nèi)的一條(可求長)f(z)dzcn(za)nzccf((z.a1zz2LznL例1求111 1zz2LznL例1求111 1zzL,(z解n1zlimsnlimznzz絕對(duì)收斂收斂半徑為111zz2LL例(z3nn(1i)n(cosin)zn例(z3nn(1i)n(cosin)znlimn1,R解nn11Rlimcn3n3nn cosin1enn2c1elim故R.ec cosin1enn2c1elim故R.ecnn4cn(1limn2)n;limc(n12R221zcn(zn.1z解1zcn(zn.1z解1z1(za)(b111zbb11g(z)(zzb當(dāng)1(za)(za)2L(za)nzb當(dāng)1(za)(za)2L(za)n11zbbbb1111(za)(z故b(b(bz1L(za)n(bbzR時(shí)設(shè)1z.求級(jí)數(shù)(nlimlimn2解nn求級(jí)數(shù)(nlimlimn2解nnzzz(n1)z.n(n1)zn100z1(n.nz1(1(2n1)zn1的收斂半徑與和函數(shù)52c12R.解cn112,z2z時(shí)1(2n1)zn1的收斂半徑與和函數(shù)52c12R.解cn112,z2z時(shí)122nzz1121故(2n.(12z)(111問題的Taylor級(jí)數(shù)函數(shù)的1.問題的引f1.問題的引fxx0Ux0,)內(nèi)具有直到n1階的導(dǎo)數(shù)，則xUx0有f(x0)f(x0)(xx0)Lf(x)(n)(f) (xx0)Rn(nRnx)是余項(xiàng)，且Rnxoxx0)n(xx0ffxx0Ux0,)內(nèi)有各階導(dǎo)數(shù)fx)在Ux0)內(nèi)能展開成Taylor級(jí)數(shù)Ux0)內(nèi)，fx的Taylor公式Rnx0(n如果函數(shù)fz在區(qū)D內(nèi)解析f(z)在內(nèi)有任一階導(dǎo)數(shù)f0z),f1(zLfnf0z),f1(zLfn(z),L在(可求長)滑曲線C上連續(xù)，fn(z)在C上收斂于f(z)fn(z)Mn且Mn存在Mnf(z)dzfn(z)dzfn(z)dzn1CCn f(z)dz(z)dz]f0kk1Cnf(z)n f(z)dz(z)dz]f0kk1Cnf(z)dzfk(z)dzfk(z)dzk1kn1Cfk(z)fk(z)kn1kn1MkdsLMk(nkn1k2.Taylor級(jí)數(shù)D內(nèi)解析z0為Df(z)設(shè)d為z02.Taylor級(jí)數(shù)D內(nèi)解析z0為Df(z)設(shè)d為z0到D的邊界上各點(diǎn)的最短距離zd時(shí)當(dāng)f(z)cn(zz0nd.1(n)cf)其中n0Dn0,1,fzD內(nèi)解析記0fzD內(nèi)解析記0Dz0為中心的任一圓周它與它的內(nèi)部全包含D,K.dr.z0K0Df()d1f(z)2πf()d1f(z)2πiKK上zK的內(nèi)部z所111則z01z(zz0)(zz1L()n00001(zz0)n(zz0)(zz1L()n00001(zz0)nn0(z0f(1Kf(z))n1(zz0n[(0f(1Kf(z))n1(zz0n(0f(1)n1d](zz0n[ 0K由高階導(dǎo)數(shù)公式(n)f)f由高階導(dǎo)數(shù)公式(n)f)f(z)(zz0n00rdf(zDKD)內(nèi)解析f(K上也連續(xù)f(K上有界ff(zDKD)內(nèi)解析f(K上也連續(xù)f(K上有界f()MKnf(f(z1(z(2π0000Mqn1nrzzzzrq是與積分變量無關(guān)的量且0qfzz0已被展開成冪f(z)fzz0已被展開成冪f(z)a(z)a(zL01n0200nf(z0)a0 f(z0)a11 f(n))即n0泰勒級(jí)數(shù)因而解析函數(shù)的泰勒級(jí)數(shù)唯一3.將函數(shù)展開成Taylor由Taylor3.將函數(shù)展開成Taylor由Taylor展開定理計(jì)算1 (n)),n0,1,fn0fzz0展開成冪級(jí)數(shù)z0的泰勒展開式例如，因?yàn)?ez)(n)ezz0的泰勒展開式例如，因?yàn)?ez)(n)ezz01,(n0,1,(ez)(L 1zzL因?yàn)閑z在復(fù)平面內(nèi)處處解析所以級(jí)數(shù)的收斂半徑Rsinz與coszz0的泰勒展開式z2n1(2nsinzsinz與coszz0的泰勒展開式z2n1(2nsinzzLLn,(Rz2nLcosz1Ln,(R(代換等求函數(shù)的.sinzz0的Taylor展開式1eizsinz1sinzz0的Taylor展開式1eizsinz1(iz)n(iz)nz2n1(2n(1)n4.典型例 把函例展開成z的冪級(jí)4.典型例 把函例展開成z的冪級(jí)1z1上有一奇點(diǎn)z由解(1z1內(nèi)處處解析z的冪級(jí)數(shù)可展開11zz2L(1)nznz1111z111z12z3z2L(1)n1zln(1zz0例ln(1zz0例泰勒展開式ln(1z在從1向左沿負(fù)實(shí)軸剪開平面內(nèi)是解析的,1是它的一個(gè)奇點(diǎn)yxz1z的冪級(jí)數(shù)所以它1[ln(1z)]解11zz2L(1)nznL(1)n1[ln(1z)]解11zz2L(1)nznL(1)n(zCz10z的曲線1zzdz(1)n1z00nz2ln(1z)LnLz即231把函數(shù)f(z)例展開成z的冪級(jí)3z111解13z221[13z)L1把函數(shù)f(z)例展開成z的冪級(jí)3z111解13z221[13z)L L]2n22222n2nLz13z323n21,z,n122求arctanz在z0.zarctanz解,101(1)且n(z2)nz求arctanz在z0.zarctanz解,101(1)且n(z2)nz1zz(zarctanzn2n所)100z2n1(1),zn2n求cos2z的冪級(jí)因?yàn)閏os2z1(1cos2z),例解2cos2z1(2z)2(2z)4(2z)6L求cos2z的冪級(jí)因?yàn)閏os2z1(1cos2z),例解2cos2z1(2z)2(2z)4(2z)6L22446621 2Lzcos2z1(1cos2z)2234561 2 2Lze1在z0Taylor級(jí)數(shù)將例e1解e1在z0Taylor級(jí)數(shù)將例e1解z因1內(nèi)進(jìn)行展開所以收斂半徑為z可eze令f(z) 1zf(z) 1z對(duì)fz(1z)f(z)zf(z)(1z)f(z)(1z(1z)f(z)(1z)f(z)f(z)(1z)f(z)(2z)f(z)LLf(01,f(00,f(01,f(0)所以fz)的Taylor級(jí)數(shù)111z21zz23附1)ez1zLznL(z11附1)ez1zLznL(z111zz2LznLzn(z111zz2L(1)nznL(1)nzn,(zz2n1z34)sinzLL,(zn((2n5)cosz1z2nL(1)(zn6)ln(15)cosz1z2nL(1)(zn6)ln(1z)zLn23(1)n(zn7)(1z)1z(1)(1)(2)L(1)L(n1)(zL,5.函數(shù)的零定f(5.函數(shù)的零定f(z)在區(qū)域D內(nèi)的一點(diǎn)z0處的設(shè)解析函值為零，則稱z0為解析函數(shù)f(z)的零點(diǎn).若函數(shù)f(z)在點(diǎn)z0的某個(gè)鄰域O(z0內(nèi)解定f(z0；且除了點(diǎn)z0外，在O(z0)f(z)處處不為零，則稱z0為f(z)孤立零點(diǎn)z0,z1f(z)z(z1)3的零點(diǎn)例定如果f(z)定如果f(z)在點(diǎn)z0的鄰域內(nèi)解析，且有f(z)(zz0)(z)m其中(z)在點(diǎn)z0解析，且(z00m則稱z0f(z)m級(jí)零點(diǎn)，m1不恒為零的解析函數(shù)的零點(diǎn)必是孤事實(shí)上，設(shè)z0為f(z)m級(jí)零z0的一個(gè)鄰域Oz0,1f(z)(zz0)(z)m其中z)在其中z)在點(diǎn)z0解析，且z0從而(z)在點(diǎn)z0必為連續(xù).由例2.1可知存在z0的一個(gè)鄰域O(z0 )，(z)恒不為零2f(z)在鄰域O(z0)（min(1,2)）內(nèi)z0f(z)的孤立零點(diǎn).除z0外，再無零點(diǎn)，即:這是解析函數(shù)區(qū)別于實(shí)可微函數(shù)的又一特性.例如1xx2例如1xx2f(x)sinx,1x0fx)的零fx)可微，但nfx)的零點(diǎn)，且limxn所以x0是零點(diǎn)xn的極限點(diǎn)，不是孤立的.推論：f(z)在區(qū)推論：f(z)在區(qū)域D內(nèi)解析znf(z)在D內(nèi)的一列零點(diǎn)，且znz0z0Df(z)在D中必恒為零.解析函數(shù)的唯一性定理：設(shè)f(z)與g(z)在區(qū)域D內(nèi)解析，{zn (n=1,2,…)是Dm≠n時(shí)，zmzn z0Dnfzngzn)，則zDf(z)g(z).解析函數(shù)的惟一解析函數(shù)的惟一性定理說明了解析函數(shù)一個(gè)非常重要的特性：在區(qū)域DDDmf(z在z0解析z0f(z)mmf(z在z0解析z0f(z)m)0,(n0,1,2,Lm)(n)(zff(m)(z00z0f(z)m證f(z)(zz0(z)設(shè)(z)在z0的Taylor級(jí)數(shù)展開為(z)c(z)c(zL01020c0(z0)0,從而f(z)在z0的Taylor級(jí)數(shù)展c0(z0)0,從而f(z)在z0的Taylor級(jí)數(shù)展開式)mf(z)c(zc(z(zL001020)0,(n0,1,2,Lm(n)(zf0(m)f)c0 例f(z)z3f(z例f(z)z3f(z)sinzf(1)3z2z13解z1知.f(0)cos1z0f(z的一級(jí)零點(diǎn)f(zz5(z21)2.z0是五級(jí)零點(diǎn)zi是二級(jí)零點(diǎn)§4.4Laurent級(jí)1§4.4Laurent級(jí)1234問題的級(jí)數(shù)函數(shù)的Laurent級(jí)數(shù)展典型例1問題的引c(za)c1問題的引c(za)cc(za)c(za)2nn012nfn(z)f1(z)f2(z)Lfn(z)fn(z)的：cn(zz0X(z)x(n)zny(n)x(n)*Y(z)X(z)H(z)cn(zz0n考慮雙邊冪級(jí)cn(zz0n考慮雙邊冪級(jí)cn(zz0ncn(zz0 c(zz 令(zz0nRR令(zz0nRR時(shí),若(1)R1R2(2)R1R2兩收斂域有公共部分zR2雙邊冪級(jí)數(shù)cn(zz0)n的收斂區(qū)域Rz雙邊冪級(jí)數(shù)cn(zz0)n的收斂區(qū)域RzR10200zR1z0z2Laurent級(jí)數(shù)展開定f(z在圓環(huán)域R1z2Laurent級(jí)數(shù)展開定f(z在圓環(huán)域R1zR2內(nèi)處處解級(jí)f(z在D內(nèi)可展開f(z)cn(zz0)nf(12πi( (n0,n0CC為圓環(huán)域內(nèi)繞z0ff11df(z)證K111(z0)(z因ff11df(z)證K111(z0)(z因z1zz1000(zz1 0n0(z0n0z0f(112(z0(zf(112(z0(zz0nfn(z)Mn且Mn存在Mnf(z)dzfn(z)dzfn(z)dzn1CCf(1K1z0z11(z0)(z1f(1K1z0z11(z0)(z11zz(z01(zz 0(zzn1nz00f(11 1 Mnf(1nf(11 1 Mnf(1n1d(zz0？1(z0n1)n(z0f(f(11df(z)則12π2cc(f(f(11df(z)則12π2cc(z(zn0n0cn(zz0)ncn與c可用一個(gè)式子表示為f(1c(n0,1,znn1C0注f(z)f(z)的Laurent注f(z)f(z)的Laurentbf(z)(zn0bn(n2,3函數(shù)的Laurent展開f(1d(n3函數(shù)的Laurent展開f(1d(n0,1,cznn12πi0Ccn(zz0)nf(z)1f(z)在z0及z1都不解析z(1而在圓環(huán)域0z1及01f(z)在z0及z1都不解析z(1而在圓環(huán)域0z1及0z1內(nèi)都解析0z1內(nèi)1z11f(z)z(11111zz2LznL,z1f(z)1zz2LznLz(1z11內(nèi)，在圓01f(z)z11內(nèi)，在圓01f(z)z(1111(1z)(1z)2L(1z)n11(1z)11(1z)(1z)2(1z)n1f(z)f(z)4典型例fz)ez展成Laurent級(jí)數(shù)0z內(nèi)例解1zLze14典型例fz)ez展成Laurent級(jí)數(shù)0z內(nèi)例解1zLze121zL2zz2z1z11z1f(z)函在圓環(huán)域(z1)(z2)1z1f(z)函在圓環(huán)域(z1)(z2)1z3)2z1)0zf(z在這些區(qū)域內(nèi)展成Laurent級(jí)數(shù)11f(z),解(1(21)0z1內(nèi)z2z從y111zz2Lzn則ox121111221z 1Ly111zz2Lzn則ox121111221z 1L L2222121zfzL(1z 222413z7z2248y2在1z2內(nèi)11z由2ox12z1z111111Lz1z1z12y2在1z2內(nèi)11z由2ox12z1z111111L