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第二章解析函数1.用导数定义,求下列函数的导数:(1)()Re.fxzz解:因0()()limzfzzfzz0()Re()Relimzzzzzzzz0ReReRelimzzzzzzzz0Relim(ReRe)zzzzzz000Relim(Re)lim(Re),zxyzxzzzzzxiy当0z时,上述极限不存在,故导数不存在;当0z时,上述极限为0,故导数为0.2.下列函数在何处可导?何处不可导?何处解析?何处不解析?(1)2().fzzz解:22222222()||()()()(),fzzzzzzzzxyxiyxxyiyxy这里2222(,)(),(,)().uxyxxyvxyyxy2222222,2,2,2.xyyxuxyxvxyyuxyvxy要,xyyxuvuv,当且当0,xy而,,,xyxyuuvv均连续,故2().fzzz仅在0z处可导,处处不解析.(2)3223()3(3).fzxxyixyy解:这里322322(,)3,(,)3.33,xuxyxxyvxyxyyuxy226,6,33,yxyuxyvxyvxy四个偏导数均连续且,xyyxuvuv处处成立,故()fz在整个复平面上处处可导,也处处解析.3.确定下列函数的解析区域和奇点,并求出导数.(1)(,).azbcdczd至少有一不为零解:当0c时,()azbfzczd除dzc外在复平面上处处解析,dzc为奇点,222()()()()()()()()().()()azbfzczdazbczdczdazbczdaczdcazbadcbczdczd当0c时,显然有0d,故()azbfzd在复平面上处处解析,且()afzd.4.若函数()fz在区域D内解析,并满足下列条件之一,试证()fz必为常数.(1)()fz在区域D内解析;(2)2;vu(3)arg()fz在D内为常数;(4)(,,).aubvcabc为不全为零的实常数证(1)因为()fz在D中解析,所以满足CR条件,,uvuvxyyx又()fzuiv也在D中解析,也满足CR条件()(),.uvuvxyyx从而应有0uuvvxyxy恒成立,故在D中,uv为常数,()fz为常数.(2)因()fz在D中解析且有2()fzuiu,由CR条件,有2,2.uuuxyuuuyx则可推出0uuxy,即uC(常数).故()fz必为D中常数.(3)设()fzuiv,由条件知arctanvCu,从而22(/)(/)0,0,1(/)1(/)vuvuyxvuvu计算得2222()/0vuuuvuxxuv,2222()/0,vuuuvuyyuv化简,利用CR条件得0,0.uuuvyxuuuvxy所以0,uuxy同理0,vvxy即在D中,uv为常数,故()fz在D中为常数.(4)法一:设0,a则()/,ucbva求导得,,ubvubvxaxyay由CR条件,,ubuvbvxayxay故,uv必为常数,即()fz在D中为常数.设0,0,0abc则bvc,知v为常数,又由CR条件知u也必为常数,所以()fz在D中为常数.法二:等式两边对,xy求偏导得:00xxyyaubvaubv,由CR条件,我们有0,00xyxxyyaubuuabbuauuba即,而220ab,故0xyuu,从而u为常数,即有()fz在D中为常数.5.设()fz在区域D内解析,试证:222222()|()|4|()|.fzfzxy证:设222(),|()|,fzuivfzuv222(),|()|()().uuuufzifzxyxy而2222222222222222222222222()|()|()()2()()()(),fzuvuvxyxyuuvvuuvvuvuvxxxxyyyy又()fz解析,则实部u及虚部v均为调和函数.故222222220,0.uuvvuvxyxy则22222222()|()|4(()())4|()|.uufzfzxyxy6.由下列条件求解解析函数().fzuiv(1)22()(4);uxyxxyy解:因22363,uvxxyyxy所以22(363)vxxyydy22333(),xyxyyx又222263(),363,()3,vuxyyxxxyyxxxx而所以则3()xxC.故222233222222223()()(4)(33)(1)()(1)()2(1)2(1)(1)()2(1)(1)(2)(1)fzuivxyxxyyixyxyyxCixxiyyixiyxyixyiCizixyxyiiziCiizxyxyiCiizCi(2)23;vxyx解:因23,2,vvyxxy由()fz解析,有22,2().uvxuxdxxyxy又23,uvyyx而(),uyy所以()23,yy则2()3.yyyC故22()3(23).fzxyyCixyx(3)2(1),(2);uxyfi解:因2,2(1),uuyxxy由()fz的解析性,有2(1),vuxxy22(1)(1)(),vxdxxy又2,vuyyx而(),vyy所以2()2,(),yyyyC则22(1),vxyC故22()2(1)((1)),fzxyixyC由(2)fi得(2)(1),fiCi推出0.C即2222()2(1)(21)(21)(1).fzxyiyxxizziz7.设sin,pxvey求p的值使v为调和函数,并求出解析函数().fzuiv解:要使(,)vxy为调和函数,则有0.xxyyvvv即2sinsin0,pxpxpeyey所以1p时,v为调和函数,要使()fz解析,则有,.xyyxuvuv1(,)coscos(),1sin()sin.pxpxxpxpxyuxyudxeydxeyypueyypeyp所以11()()sin,()()cos.pxpxypeyypeyCpp即(,)cos,pxuxypeyC故(cossin),1,()(cossin),1.xzxzeyiyCeCpfzeyiyCeCp8.试解方程:(1)13;zei解:(2)3132(cossin)233ikzeiieln2(2)3,0,1,2.ikek故ln2(2),0,1,2.3zikk(2)ln;2iz解:2cossin.22izeii9.求下列各式的值。(1)cos;i解()()11cos.22iiiieeeei(2)(34);Lni解:(34)ln5(34)LniiArgi4ln5(2arctan).3ik(3)1(1);ii解:1(1)(1)(1)iiLniie(1)ln2(2)4ln22ln2244ln224cos(ln2)sin(ln2).44iikkikkeeei(4)33;i解:3(3)ln3(3)(ln32)3iiikiee(3)ln323ln32ln3227(cosln3sinln3).ikkiikeeeeei
本文标题:复变函数第二章答案
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