復(fù)變函數(shù)練習題習題33

本文格式為Word版，下載可任意編輯——復(fù)變函數(shù)練習題習題33習題3.3

1．計算以下積分，其中積分閉路取正向.

dz3（1）|z??z?11|?1解：

dz1/(z?z?1)??dz3?z?1|z?1|?1z?1|z?1|?11?2?i2z?z?1z?12??i3

dz44（4）|z?z(z?2)|?1解：

21

dz1/(z?2)??dz444?z(z?2)|z|?1z|z|?1???2?i?1??4??3!?(z?2)?z?0?i?120??73(0?2)5?i?16

4sinzdz4n?1（6）|z?(z?i)|?2解：

sinzdz2?i(4n)?sinz??4n?1?z?i(z?i)(4n)!|z|?22?i?sinzz?i(4n)!2?i?sini(4n)!?2??sh1(4n)!

2

dz4?（8）|z|?3(z?1)(z?2)(z?16)

21解：被積函數(shù)(z?1)(z?2)(z4?16)有6個奇點，

只有z?1在圓|z|?3/2的內(nèi)部，于是函數(shù)

14|z|?3/2在閉圓域上解析，則由(z?2)(z?16)Cauchy積分公式得

|z|?dz?3(z?1)(z?2)(z4?16)?2|z|?1/(z?2)(z?16)dz?3z?1241?2?i4(z?2)(z?16)z?12?i?51

3

4.用Cauchy積分公式計算函數(shù)f(z)?e/z沿正向圓周|z|?1的積分值，然后利用圓周|z|?1的參數(shù)方程z?e(??????)證明下面積分

i?z??0ecos?cos(sin?)d???.

z（1）解：函數(shù)f(z)?e/z的奇點z?0在積分

z|z|?1e路徑的內(nèi)部，而函數(shù)在閉區(qū)域|z|?1上解

析，于是由Cauchy積分公式得

ezdz?2?ie?2?i.?|z|?1zz?0（2）證明：圓周|z|?1的參數(shù)方程為

zz?e(??????)，在它上有z?(?)?ie,于是

i?i?4

i??eiee2?i??dz??d?i?|z|?1z??e?cos??isin???eid?z??ei???e?????cos?[cos(sin?)?isin(sin?)]id?sin(sin?)?iecos???[?e????cos?cos(sin?)]d?cos?

???ecos?sin(sin?)d??i?e???cos(sin?)d?比較等式兩邊的虛部得

??e??cos?cos(sin?)d??2?

又

?e??????0?cos???e?0??0cos(sin?)d?cos(sin?)d???ecos?cos(sin?)d?0cos??cos(sin(??))d(??)??ecos(sin?)d???e??????ecos(sin?)d???ecos(sin?)d????????ecos(sin?)d???ecos(sin?)d????2?ecos(sin?)d?0coscos0coscos00cos0cos(??)?cos?

所以

5

?

?0ecos?cos(sin?)d???.

7.由下面所給調(diào)和函數(shù)求解析函數(shù)f(z)?u?iv.（2）u?e(xcosy?ysiny),f(0)?0;

解：對u求偏導(dǎo)數(shù)有

x?ux?e(xcosy?ysiny?cosy),x?uy??e(siny?xsiny?ycosy),

x解法1：由Cauchy-Riemann條件得

x??e(siny?xsiny?ycosy),x?vy?e(xcosy?ysiny?cosy),

對第一式兩邊對x積分得

v??e(siny?xsiny?ycosy)dxxx?(x?1)esiny?e(siny?ycosy)?g(y)x?e(xsiny?ycosy)?g(y)兩邊對y求導(dǎo)，并且與上面所得v?y比較有

xv?y?ex(xcosy?cosy?ysiny)?g?(y)x?e(xcosy?ysiny?cosy)

6

于是得g?(y)?0,即g(y)?c,其中c為任意實常數(shù).從而

v?e(xsiny?ycosy)?c，

即

xf(z)?e(xcosy?ysiny)?i[e(xsiny?ycosy)?c]?zez?ci由于f(0)?0,代入上式得c?0,所以

xxf(z)?ze.

解法2：由Cauchy-Riemann方程和解析函數(shù)的求導(dǎo)公式可得

z???f?(z)?u?x?i?ux?iuyx?e(xcosy?ysiny?cosy)x?i[?e(siny?xsiny?ycosy)]?ez(z?1)于是

zf(z)??e(z?1)dz?ic?ze?ic,

0zz其中c為任意實常數(shù).

由于f(0)?0,代入上式得c?0,所以

7

f(z)?ze.

（4）v?y/(x?y),f(2)?0.

解：對v求偏導(dǎo)數(shù)有

22z?2xyx2?y2?v?,v,x?y?222222(x?y)(x?y)解法1：由Cauchy-Riemann條件得

x?y?2xy2xy?u?,u?2,x?y??22222222(x?y)(x?y)(x?y)對其次式兩邊對y積分得

222xyu??2dy22(x?y)?x?2?g(x)2x?y兩邊對x求導(dǎo)，并且與上面所得u?x比較有

x?yu??g?(x)x?222(x?y)22x?y?222(x?y)

8

22于是得g?(x)?0,即g(x)?c,其中c為任意實常數(shù).從而

?xu?2?c2，x?y即

?xy1f(z)?2?c?i2???c22，x?yx?yz由于f(2)?0,代入上式得c?1/2,所以

11f(z)??.2z

解法2：由Cauchy-Riemann方程和解析函數(shù)的求導(dǎo)公式可得

???f?(z)?u?x?i?vy?i22x?y?2xy?2+i22222(x?y)(x?y)1?2z于是

9

11f(z)??2dz?c???1?c,1zz其中c為任意實常數(shù).

z由于f(2)?0,代入上式得c??1/2,所以

11f(z)??.2z

10.設(shè)f(z)和g(z)在簡單閉路C上及其內(nèi)部解析，試證：

（1）若f(z)在C上及其內(nèi)部四處不為零，則有

?f(z)?g(z).

Cf?(z)dz?0;f(z)（2）若在C上有f(z)?g(z),則在C的內(nèi)部有證明：（1）由于f(z)在簡單閉路C上及其內(nèi)

f?(z)部解析并且四處不為零，則f(z)在簡單閉路C上

及其內(nèi)部四處解析，于是由Cauchy積分定理得

10

?Cf?(z)dz?0;f(z)（2）若