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当前位置:首页 > 建筑/环境 > 工程监理 > 工程数学复变函数习题课 (chapter1-3)
例1设复数满足,1||证明zz1.1||,11||,11||,1时当时;当时;当zzz证明))((||2zzz)(Re2||||22zz)(Re2||||1|1|222zzz222222||||1|||||1|||zzzz)||1)(1|(|22z.1||,01||,01||,0时当时;当时;当zzz一复数与复变函数的概念例2映射)1(2zzw将圆周4||z映射成什么?解设,iyxz则,ivuw)1(2zzw)(222yxiyxiyx即)(222yxxxu)(222yxyyv4||z的参数方程为,sin4,cos4yx因此)cos41cos4(2u,cos217sin215v所以4||z的像为.115417422vu例3zw1将下列曲线映射成什么曲线?1)1|1|z2)4argz解设.,ivuwiyxz则221yxiyxzw即2222,yxyvyxxu1)4argz的方程),0(xxy因此4argz的像为),0(vvu即.4argw2)1|1|z的方程为,222xyx因此1|1|z的像为,21u即.21Rew例4求下列复数的值1));21exp(i2));3cos(i3));43(Lniii)31(4)解1))2sin2(cos)21exp(1ieieiiyxzee)sin(cosyiyex2)xshyixchyiyxsincos)cos(13sin13cos)3cos(shichi123121ishch3)zizzArg||lnLn5|43|iki234arctan)43Arg(ikii)12(34arctan5ln)43(Ln4)abbeaLn)31(Ln)31(iiiei)232(2(lnkiie)2lnsin2ln(cos232iek例5复数acz与复数caz)(是否相等?说明理由。解不等由于zacacezLn)Lnexp()(acazcz)Lnexp(Lnzaec}Lnexp{)LnIm()LnRe(zaizaec]}2)LnIm([lnexp{)LnRe(kzaicecza}2)Ln(Im)Ln(Reexp{ckizaiczac)2Lnexp(ickzca所以仅当c为整数时acz与相等。caz)(二解析函数例6设考察下列函数的可导性、解析性.,iyxz并在可导的时,求其导数.1)2233)(yixyxzf解,33yxu22yxv为可微函数,由于,32xux,32yuy,22xyvx,22yxvy当,2322yxx2223xyy时,即当23,23yx或0,0yx时)(zf可导,因此)(zf在复平面上每一点处都不解析.所以由于,23)(22ixyxivuzfxx所以)2323(if,827427i.0)0(f2))sincos()sincos()(yxyyieyyyxezfxx解),sincos(yyyxeux为可微函数,)sincos(yxyyevx而)sinsin(cosyyyxyeuxxyv)cossin(sinyyyxyeuxyxv所以)(zf在复平面上每一点处都可导,从而解析。且xxivuzf)()sinsin(cosyyyxyex)cossin(sinyyyxyiexzez)1(例7已知),(),()(yxivyxuzf为iyxz的的解析函数,且,chcosyyxu求函数).(zf解xyuvyxchsin)(chsinxgdyyxv)(shsinxgyxyuyx1shcos)(shcosxgyxvx)(1)(xg即cxxg)(cxyxvshsin)sh(sinchcos)(cxyxiyyxzficizzcosc(为任意实常数)例8设),(),()(yxivyxuzf为iyxz的解析函数,且,22yxvu求函数).(zf解,2xvuxxyvuyy2,,xyyxvuvu,yxuyyxux)(22xgyxyuxuyx)(xgy,)(xxgcxxg2)(2,)(2122cyxxyu,)(2122cyxxyv))(21()(21)(2222cyxxyicyxxyzfcizi)1(212c(为任意实常数)例9计算积分,)(2Cdziyx其中C为1)从点0沿2xy到点;1i2)从点0沿实轴到点,1再沿1Rez到点.1i解1):C,2ixxz10xx原式10)(22ixxdxxi)21(i65612)21CCC,:1xzC10xx,1:2iyzC10yy原式21CC102xdx10)1(yiidyi61三复函数的积分例10计算积分Cxdzyxyyiyyyxe)]sincos(sincos[其中C为从原点沿上半圆周1|1|z到点.1i解法一)]sincos(sincos[yxyyiyyyxex)sin(cosyiyxex)sin(cosyiyiyex)sin(cos)(yiyeiyxxzze原式Czdzzeizdzze10izizdzeze1010izieei101)1(1cos1sin1iee解法二10dxxexCxdzyxyyiyyyxe)]sincos(sincos[10)sincos(sin[cosidyyyyiyyye10)(xxexe10cosyeyi10sinyey1cos1sin1iee例11计算围道积分.)1(22Czdzzze其中C为正向圆周;5.1|2|)1(z2||)2(z解(1)Czdzzze22)1(Czdzzze22)1(122zzzeiei2或22)1(1zz21zz212z2)1(1zCdzezCCCCdzezdzezdzezdzez0014zzei1)(2zzeiei2(2)作正向圆周;5.0|:|1zC5.0|1:|2zCCdzzzez22)1(dzzzez22)1(dzzzez22)1(1C2C02)1(2zzzei022zzzeii6ei2或22)1(1zz21zz212zCdzezCCCdzezdzezdzez2)1(1zCdzez04zzei0)(2zzei14zzei1)(2zzeii6ei2例12设)(zf在单连通域B解析,且1|)(1|zf证明:(1)在B内,恒有;0)(zf(2)对B内任意一条正向简单闭曲线,C有Cdzzfzfzf0)()(2)(证(1)由三角不等式得|)(1|1zf|)(|1zf所以,0|)(|zf即在B内,.0)(zf(2)恒有.0)(zf由于所以)()(2)(zfzfzf在B内解析,即Cdzzfzfzf0)()(2)(例13设)(zf解析,且,2)0(,1)0(ff计算1||22)()1(zdzzzfz并计算.)(2cos202defi解由高阶导数公式得1||22)()1(zdzzzfz02])()1[(2zzfzi02)]()1()()1(2[2zzfzzfzi)]0()0(2[2ffii81||22)()1(zdzzzfziez令20)()1(2iiefediei20222)()(defeeiiiii8202)(2cos4defii202)(2cosdefi2例14*)(zfRaz||),0(Rrr.)](Re[1)(20dereafrafii若在试证明对任都有内解析,一实数证明razdzazzfiaf||2)()(21)(ireaz令2021iiierreaf22)(direidereafrii20)(21razdzzf||)(0ireaz令20)(direreafii20)(210dereafrii20)(21dereafrii所以)(af20)(21dereafrii20)(21dereafrii.)](Re[120dereafrii例15*设)(zf在整个复平面上解析且有界,ba,为两个任意定复数,证明0))(()(lim||RzRdzbzazzf并由此证明)(zf为常数函数。证由于,|)(|Mzf取|},||,max{|baR则Rzdzbzazzf||))(()(Rzdsbzazzf|||||||)(|RzdsbRaRM|||)||)(|(|)||)(|(2bRaRRM0R0))(()(lim||RzRdzbzazzf又因当))(()(bzazzfba时[1abazzf)(bzzf)(Rz||dzRz||Rz||]dzdz)]()([2bfafabi所以令R得:),()(bfaf即)(zf为常数函数。
本文标题:工程数学复变函数习题课 (chapter1-3)
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