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1、复变函数考试试题与答案各种总结复变函数考试试题(一)一、判断题(20 分):1、若 f(z)在 z0的某个邻域内可导,则函数 f(z)在 z0解析、()2、有界整函数必在整个复平面为常数、()3、若nz收敛,则Renz与Imnz都收敛、()4、若 f(z)在区域 D内解析,且0)(zf,则Czf)(常数)、()5、若函数f(z)在 z0处解析,则它在该点的某个邻域内可以展开为幂级数、()6、若 z0就是)(zf的 m阶零点,则 z0就是 1/)(zf的 m阶极点、()7、若)(lim0zfzz存在且有限,则 z0就是函数f(z)的可去奇点、()8、若函数f(z)在就是区域D内的单叶函数,则)(
2、0)(Dzzf、()9、若 f(z)在区域 D 内解析,则对 D 内任一简单闭曲线C0)(Cdzzf、()10、若函数f(z)在区域 D内的某个圆内恒等于常数,则 f(z)在区域 D内恒等于常数、()二、填空题(20 分)1、1|00)(zznzzdz_、(n为自然数)2、zz22cossin _、3、函数zsin的周期为 _、4、设11)(2zzf,则)(zf的孤立奇点有_、5、幂级数0nnnz的收敛半径为_、6、若函数f(z)在整个平面上处处解析,则称它就是 _、7、若nnzlim,则nzzznn.lim21_、8、)0,(Renzzes_,其中 n 为自然数、9、zzsin的孤立奇点为_
3、、复变函数考试试题与答案各种总结10、若0z就是)(zf的极点,则_)(lim0zfzz、三、计算题(40 分):1、设)2)(1(1)(zzzf,求)(zf在 1|0:zzD内的罗朗展式、2、.cos11|zdzz3、设Cdzzf173)(2,其中3|:|zzC,试求).1(if4、求复数11zzw的实部与虚部、四、证明题、(20 分)1、函数)(zf在区域D内解析、证明:如果|)(|zf在D内为常数,那么它在D内为常数、2、试证:()(1)f zzz在割去线段0Re1z的z平面内能分出两个单值解析分支,并求出支割线0Re1z上岸取正值的那支在1z的值、复变函数考试试题(一)参考答案一判断题
4、1.2.6.10.二.填空题1、2101inn;2、1;3、2k,()kz;4、zi;5、1 6、整函数;7、;8、1(1)!n;9、0;10、三.计算题、1、解因为01,z所以01z111()(1)(2)12(1)2f zzzzz001()22nnnnzz、2、解因为文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1
5、B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编
6、码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J
7、1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A
8、6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B
9、10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3
10、C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6
11、S6M4J1B10文档编码:CC5J1B6G5A6 HO4B10K5I3C5 ZX6S6M4J1B10复变函数考试试题与答案各种总结22212Re()limlim1cossinzzzzs f zzz,22212Re()limlim1cossinzzzzs f zzz、所以22212(Re()Re()0coszzzdzis f zs f zz、3、解 令2()371,则它在z平面解析,由柯西公式有在3z内,()()2()cf zdzizz、所以1(1)2()2(13 6)2(613)zifiiziii、4、解 令zabi,则222222122(1)2(1)211111(1)(1)(1)zabia
12、bwzzababab、故2212(1)Re()11(1)zazab,2212Im()1(1)zbzab、四、证明题、1、证明 设在D内()f zC、令2222(),()f zuivf zuvc则、两边分别对,x y求偏导数,得0(1)0(2)xxyyuuvvuuvv因为函数在D内解析,所以,xyyxuv uv、代入(2)则上述方程组变为00 xxxxuuvvvuuv、消去xu得,22()0 xuvv、1)若220uv,则()0f z为常数、2)若0 xv,由方程(1)(2)及.CR方程有0,xu0yu,0yv、所以12,uc vc、(12,c c为常数)、所以12()fzcic为常数、文档编码
13、:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编
14、码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档
15、编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文
16、档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10
17、文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P1
18、0文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P
19、10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10复变函数考试试题与答案各种总结2、证明()(1)f zzz的支点为0,1z、于就
20、是割去线段0Re1z的z平面内变点就不可能单绕0 或 1 转一周,故能分出两个单值解析分支、由于当z从支割线上岸一点出发,连续变动到0,1z时,只有z的幅角增加、所以()(1)f zzz的幅角共增加2、由已知所取分支在支割线上岸取正值,于就是可认为该分支在上岸之幅角为0,因而此分支在1z的幅角为2,故2(1)22ifei、复变函数考试试题(二)一.判断题、(20 分)1、若函数),(),()(yxivyxuzf在 D 内连续,则 u(x,y)与 v(x,y)都在 D 内连续、()2、cos z与 sin z在复平面内有界、()3、若函数 f(z)在 z0解析,则 f(z)在 z0连续、()4、
21、有界整函数必为常数、()5、如 z0就是函数 f(z)的本性奇点,则)(lim0zfzz一定不存在、()6、若函数 f(z)在 z0可导,则 f(z)在 z0解析、()7、若 f(z)在区域 D 内解析,则对 D 内任一简单闭曲线C0)(Cdzzf、()8、若数列nz收敛,则Renz与Imnz都收敛、()9、若 f(z)在区域 D 内解析,则|f(z)|也在 D 内解析、()10、存在一个在零点解析的函数f(z)使0)11(nf且,.2,1,21)21(nnnf、()二、填空题、(20 分)1、设iz,则_,arg_,|zzz2、设Ciyxzyxixyxzf),sin(1()2()(222,则
22、)(lim1zfiz_、3、1|00)(zznzzdz_、(n为自然数)4、幂级数0nnnz的收敛半径为 _、5、若 z0就是 f(z)的 m阶零点且 m0,则 z0就是)(zf的_零点、文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5
23、K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN
24、5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 Z
25、N5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9
26、ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9
27、 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D
28、9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1
29、D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10复变函数考试试题与答案各种总结6、函数 ez的周期为 _、7、方程083235zzz在单位圆内的零点个数为_、8、设211)(zzf,则)(zf的孤立奇点有 _、9、函数|)(zzf的不解析点之集为 _、10、_)1,1(Res4zz、三、计算题、(40 分)1、求函数)2sin(3z的幂级数展开式、2、在复平面上取上半虚轴作割线、试在所得的区域内取定函数z在正实轴取正实值的一个解析分支,并求它在上半虚轴左沿的点及右沿的点iz处的值、3、计算积分:iizzId|,积分路径为(1)
30、单位圆(1|z)的右半圆、4、求dzzzz22)2(sin、四、证明题、(20 分)1、设函数 f(z)在区域 D 内解析,试证:f(z)在 D 内为常数的充要条件就是)(zf在D 内解析、2、试用儒歇定理证明代数基本定理、复变函数考试试题(二)参考答案一.判断题、1.6.10.、二、填空题1、1,2,i;2、3(1sin 2)i;3、2101inn;4、1;5、1m、6、2k i,()kz、7、0;8、i;9、R;10、0、三、计算题文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G
31、6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7
32、G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K
33、7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5
34、K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN
35、5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 Z
36、N5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9
37、ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10复变函数考试试题与答案各种总结1、解3212163300(1)(2)(1)2sin(2)(21)!(21)!nnnnnnnzzznn、2、解 令izre、则22(),(0,1)kif zzrek、又因为在正实轴去正实值,所以0k、所
38、以4()if ie、3、单位圆的右半圆周为ize,22、所以22222iiiizdzdeei、4、解dzzzz22)2(sin2)(sin2zzi2cos2zzi=0、四、证明题、1、证明 (必要性)令12()f zcic,则12()f zcic、(12,c c为实常数)、令12(,),(,)u x yc v x yc、则0 xyyxuvuv、即,u v满足.CR,且,xyyxuvuv连续,故()f z在D内解析、(充分性)令()f zuiv,则()f zuiv,因为()f z与()f z在D内解析,所以,xyyxuvuv,且(),()xyyyxxuvvuvv、比较等式两边得0 xyyxuvu
39、v、从而在D内,u v均为常数,故()f z在D内为常数、2、即要证“任一n次方程101100(0)nnnna za zazaa有且只有n个根”、证明令1011()0nnnnf za za zaza,取10max,1naaRa,当z在:CzR上时,有111110()()nnnnnnza RaRaaaRa R、()f z、由儒歇定理知在圆zR内,方程10110nnnna za zaza与00na z有相同个数的根、而00na z在zR内有一个n重根0z、因此n次方程在zR内有n个根、复变函数考试试题(三)一、判断题、(20 分)、文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN
40、5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 Z
41、N5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9
42、ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9
43、 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D
44、9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1
45、D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L
46、1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10复变函数考试试题与答案各种总结1、cos z 与 sin z 的周期均为k2、()2、若 f(z)在 z0处满足柯西-黎曼条件,则 f(z
47、)在 z0解析、()3、若函数 f(z)在 z0处解析,则 f(z)在 z0连续、()4、若数列nz收敛,则Renz与Imnz都收敛、()5、若函数 f(z)就是区域 D内解析且在 D内的某个圆内恒为常数,则数 f(z)在区域D内为常数、()6、若函数 f(z)在 z0解析,则 f(z)在 z0的某个邻域内可导、()7、如果函数 f(z)在 1|:|zzD上解析,且)1|(|1|)(|zzf,则)1|(|1|)(|zzf、()8、若函数 f(z)在 z0处解析,则它在该点的某个邻域内可以展开为幂级数、()9、若z0就 是)(zf的m阶 零 点,则z0就 是 1/)(zf的m阶 极 点、()10
48、、若0z就 是)(zf的 可 去 奇 点,则0),(Res0zzf、()二、填空题、(20 分)1、设11)(2zzf,则 f(z)的定义域为 _、2、函数 ez的周期为 _、3、若nnninnz)11(12,则nznlim_、4、zz22cossin_、5、1|00)(zznzzdz_、(n为自然数)6、幂级数0nnnx的收敛半径为 _、7、设11)(2zzf,则 f(z)的孤立奇点有 _、8、设1ze,则_z、9、若0z就是)(zf的极点,则_)(lim0zfzz、文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI
49、10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 H
50、I10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3 HI10D8F9L1D9 ZN5K7G6F6P10文档编码:CF10K6R2O9B3