復(fù)變函數(shù)第二章答案

1、第二章解析函數(shù)1用導(dǎo)數(shù)定義，求下列函數(shù)的導(dǎo)數(shù)：(1)()Re.fxzz?解:因0()()limzfzzfzz??????0()Re()Relimzzzzzzzz?????????0ReReRelimzzzzzzzz??????????0Relim(ReRe)zzzzzz????????000Relim(Re)lim(Re)zxyzxzzzzzxiy????????????????當(dāng)時(shí)上述極限不存在故導(dǎo)數(shù)不存在；當(dāng)時(shí)上述極限為0故導(dǎo)數(shù)為0

2、.0z?0z?2下列函數(shù)在何處可導(dǎo)何處不可導(dǎo)何處解析何處不解析(1)2().fzzz??解:22222222()||()()()()fzzzzzzzzxyxiyxxyiyxy??????????????這里2222()()()().uxyxxyvxyyxy????2222222222.xyyxuxyxvxyyuxyvxy????????要，當(dāng)且當(dāng)而均連續(xù)，故僅xyyxuvuv???0xy??xyxyuuvv2().fzzz??在處可導(dǎo)，

3、處處不解析.0z?(2)3223()3(3).fzxxyixyy????解:這里322322()3()3.33xuxyxxyvxyxyyuxy??????226633yxyuxyvxyvxy?????四個(gè)偏導(dǎo)數(shù)均連續(xù)且處處成立故在整個(gè)復(fù)平面上處處可導(dǎo)xyyxuvuv???()fz也處處解析.3確定下列函數(shù)的解析區(qū)域和奇點(diǎn)并求出導(dǎo)數(shù).(1)().azbcdczd??至少有一不為零(3)設(shè)由條件知從而()fzuiv??arctanvCu?2

4、2()()001()1()vuvuyxvuvu????????計(jì)算得，2222()0vuuuvuxxuv???????2222()0vuuuvuyyuv???????化簡(jiǎn)利用條件得CR?00.uuuvyxuuuvxy????????????????????所以同理即在中為常數(shù)故在中為常0uuxy??????0vvxy??????Duv()fzD數(shù).(4)法一：設(shè)則求導(dǎo)得0a?()ucbva??ubvubvxaxyay??????????

5、??由條件CR?ubuvbvxayxay??????????故必為常數(shù)即在中為常數(shù).uv()fzD設(shè)則知為常數(shù)又由條件知也必為常數(shù)所000abc???bvc?vCR?u以在中為常數(shù).()fzD法二：等式兩邊對(duì)求偏導(dǎo)得：，由條件，我們有xy00xxyyaubvaubv???????CR?，000xyxxyyaubuuabbuauuba?????????????????????即而，故，從而為常數(shù)，即有在中為常數(shù).220ab??0xyuu?