复变函数练习题习题3.3

习题3.3

1．计算下列积分，其中积分闭路取正向.

dz3（1）|z??z?1 1|?1解：

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

dz44（4）|z? z(z?2)|?1解：

2 1

dz1/(z?2)??dz444?z(z?2)|z|?1z|z|?1???2?i?1? ?4??3!?(z?2)?z?0?i?120 ??7 3(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：被积函数(z?1)(z?2)(z4?16)有6个奇点，

只有z?1在圆|z|?3/2的内部，于是函数

14|z|?3/2在闭圆域上解析，则由(z?2)(z?16)Cauchy积分公式得

|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?1 2?i ?51

3

4.用Cauchy积分公式计算函数f(z)?e/z沿正向圆周|z|?1的积分值，然后利用圆周|z|?1的参数方程z?e(??????)证明下面积分

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

z（1）解：函数f(z)?e/z的奇点z?0在积分

z|z|?1e路径的内部，而函数在闭区域|z|?1上解

析，于是由Cauchy积分公式得

ezdz?2?ie?2?i.?|z|?1z z?0（2）证明：圆周|z|?1的参数方程为

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?比较等式两边的虚部得

??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.由下面所给调和函数求解析函数f(z)?u?iv. （2）u?e(xcosy?ysiny),f(0)?0;

解：对u求偏导数有

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

x解法1：由Cauchy-Riemann条件得

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

对第一式两边对x积分得

v??e(siny?xsiny?ycosy)dxxx ?(x?1)esiny?e(siny?ycosy)?g(y)x ?e(xsiny?ycosy)?g(y)两边对y求导，并且与上面所得v?y比较有

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为任意实常数. 从而

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方程和解析函数的求导公式可得

z???f?(z)?u?x?ivx?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为任意实常数.

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：对v求偏导数有

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

x?y?2xy2xy?u?,u?2,x?y??22222222 (x?y)(x?y)(x?y)对第二式两边对y积分得

222xyu??2dy22(x?y)?x ?2?g(x) 2x?y两边对x求导，并且与上面所得u?x比较有

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为任意实常数. 从而

?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方程和解析函数的求导公式可得

???f?(z)?u?x?ivx?vy?ivx22x?y?2xy ?2+i22222(x?y)(x?y) 1 ?2z于是

9

11f(z)??2dz?c???1?c, 1zz其中c为任意实常数.

z由于f(2)?0,代入上式得c??1/2,所以

11f(z)??.2z

10.设f(z)和g(z)在简单闭路C上及其内部解析，试证：

（1）若f(z)在C上及其内部处处不为零，则有

?f(z)?g(z).

Cf?(z)dz?0; f(z)（2）若在C上有f(z)?g(z),则在C的内部有证明：（1）因为f(z)在简单闭路C上及其内

f?(z)部解析并且处处不为零，则f(z)在简单闭路C上

及其内部处处解析，于是由Cauchy积分定理得

10

?Cf?(z)dz?0; f(z)（2）若对于C上的任意一点?有f(?)?g(?),由于f(z)和g(z)在简单闭路C上及其内部解析，则对于C的内部的任意一点z，由Cauchy积分公

式得

1f(?)1g(?)f(z)?d??d??g(z),?? 2?iC??z2?iC??z所以在C的内部有f(z)?g(z).

11

本文来源：https://www.bwwdw.com/article/0bo8.html