电大《复变函数》作业集答案

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专业好文档第一章练习题参考答案一、1 ; 2,;3 ;4; 5 ;6 ; 7. ;二、1 ; 2 ; 3; 4 56 三、1 ; 2 ; 3 ; 4四、1 ;2 (为实数);3 (为任意实数);4; 五、1直线;2 以 (-3,0), (-1,0)为焦点,长半轴为2,短半轴为的椭圆:;3 直线; 4 以为起点的射线;六、1上半平面,无界单通区域;2由直线及所构成的带形区域(不含两直线),无界单连通区域;3以为圆心,以4为半径的圆的内部(不含圆心),有界多连通区域;4由射线逆时针旋转到射线构成的半平面,无界单连通区域七、证明: 八、由 即可证明。几何意义:平行四边形两对角线的平方和等于两邻边平方和的两倍。 (33)九、多项式的系数是实数, 故十、当沿实轴趋于0时,极限值为1;当沿虚轴趋于0时,极限值为-1故 当时,的极限不存在十一、证明:令则又因 是实系数方程的根,那么于是 所以 于是方程的根十二、 .第二章练习题参考答案一、1充分条件 2充分必要条件 3)在处可微; ) 在处成立45 (2,-3,2) 二、1C 2C 3D 4. D 5.A 6.D 7.D 8.A (34)三、1解: 故在上可导,没有解析点2解:故 在全平面内可导,在全平面内解析 3解: 仅当时,C-R条件成立,故此函数在直线上处处可导,而在复平面上处处不解析.4解: 因此仅在两相交直线上处处可导,在平面处处不解析5解: C-R条件处处成立,且偏导数处处连续,因而处处可微,即处处解析6解:令,则在z平面上处处可微且 从而要使 , 只需:,从而在直线上,可导,在z 平面上处处不解析7解:设,则(35)=,由于在z平面上处处可微,且 若,则必须要,解得 ,函数在z=0点可导,平面上处处不可微四、证明: (1) 又 得 常数同理可得常数 常数(2) 在区域内解析 (36)又 时得 即结论成立当时 得 即常数同理可得常数 常数(3 ) 在区域内解析 得 得 常数同理可得 常数 常数五、解:1 2 3 (37)六、解 : 主值 主值 七、解:(1) (2) (3) (4) 八、解: ,由C-R 条件我们可以得到: 九、解:因为且 在平面上处处连续,所以在平面上处处可微;又因为处处成立,从而在平面上处处解析,且=(38)第三章练习题参考答案一、1 ; 20; 3;4 -1; 5; 6 0; 7; 8; 9 ; 10 .二、单项选择题 1D 2B 3.B 4.B 5.A 6.C 7.B 8.A 9.B 10.C三、1证明: 2解:原式=- + 3解:显然被积函数的奇点为0和1(1)在内,被积函数有唯一奇点0,故(2)在内,被积函数有唯一奇点1,故(3)对,由复合闭路定理 其中4解:当在的内部时, (39)当在的内部时,原式=当不在的内部时,原式=5 解:当0,1均不在C 的内部时,被积函数在C 上及其内部解析,由Cauchy-Gourssat定理,当点0 在C 的内部而点1在C 的外部时,由柯西积分公式得:当1在C 的内部,而点0在C 的外部时,由高阶导数公式得:当0,1 均在C 的内部时,在C 的内部作分别以0,1为圆心半径充分小的圆周,使得他们互不包含也互不相交,由复合闭路定理,有3证明:当 则 ,是解析函数且当,则 也是解析函数且6证明:令,则(40) , = =故 7解: 因为 ,由C-R方程可得 用偏积分法 因此 由 所以 8解: 用偏积分法(41) 故 (C为实常数)或 其中 (C为实常数)9解: 由 + 得 - 得 从而 C为复常数10解:,, 一般情况下不相等,可能相等的情况:是简单闭曲线,是的连续函数,且与无关;是平行于实轴的线段;第四章练习题参考答案一、1复数列收敛的充分必要条件是实数列与均收敛。2复数项级数收敛的充分必要条件是实级数与均收敛。3复数项级数绝对收敛的充分必要条件是实级数与均绝对收敛。4幂级数收敛域为圆域:,而洛朗级数的(42)收敛域为圆环域。二、填空题1,0; 23,+,0; 4 7i 8三、判断题1 ; 2. ; 3. ; 4. ; 5. 四、证明:级数收敛,相当于幂级数在处收敛。因此该幂级数的收敛半径。但若,则幂级数在收敛圆内绝对收敛,特别在处绝对收敛,即级数收敛。这与题设矛盾。从而幂级数的收敛半径。五、1=2因为 ,故 3(43)4因为,而所以 5因为而故 6因为且,从而可得:=六、(1)(2)因为,而 所以 (3)因为,而(44)所以 (4)用公式求展开式的系数;故 函数距最近的奇点为,所以级数的收敛半径 (5)因为,而所以 七、(1)在内,故(45)(2)在内,从而有(3)(4)因为所以 (5)(6) (46) 第五章 参考答案一、1 是一级极点,是二级极点;2 是二级极点;3 是本性奇点; 4 是三级极点,是一级极点;5 是一级极点;6 一级极点二、1 解 ,可去奇点,是一级极点 2 解 因为 所以 3 解 是二级极点,是一级极点 4 解 (47)5 解 6 解 为二级极点,为的一级极点, 7 解 是的二级零点,故为的二级极点,由于其为偶函数,洛朗展开式的奇次项的系数为零,所以 8 解 0是二级极点,-2是一级极点, 用极点处留数计算公式得 三、1 解 在 内,是函数一级极点 2 解 为一级极点,为二级极点 3 解 为一级极点, (48)4 解 在内,是被积函数一级极点5 解 在内,被积函数有三个一级极点 6 解 在内,被积函数有一个一级极点 7 解 因为 所以 , 8 解 因为 所以 , 四 证明: 因为 是的级极点,故有解析函数,使得 (49)所以 为的级极点五、解 因为是的级零点,故有在某邻域内解析的函数,使 所以 六、解 因为是的级极点,故有在某邻域内解析的函数,使模拟试题(一)参考答案一、1 D; 2 D ; 3 A ; 4 B; 5 A. 二、1 ;2; 3 1 ; 4 三、 证: 因为 在区域内解析,且从而 (3分)(50)所以 (5分)系数行列式 所以 ,同理 (7分) 即 在内为常数 (8分)四、解 (2分) (4分) (6分)由 得 (8分)五、解 : (2分)而 (4分) (6分)所以 (8分)(51)级数的收敛半径为 (10分)六、 解: 因为 (4分)所以 (8分) (10分)七、 (1) 解 : 在内,是二级极点,是一级极点 (1分) (3分) (5分) (6分)(2)解: (6 分)(3) 解 : 在内,均为函数的一级极点 (2分) (4分) (6分)(4) 解 : (4分) (6分)(52)模拟试题(二)参考答案一、1B; 2C; 3A; 4D; 5B二、1; 2;3; 40; 5三、证明:在区域D内解析,且 对求偏导 对求偏导 (2分)由 得 (6分)得 (7分)若 得 常数同理可得 常数故 (8分)四、解: (2分) (4分) (6分) (8分)由 得 (53) (10分) (10分)五、解: (2分) () (8分)根据1到最近奇点0的距离得收敛半径为1 (10分)六、解:: 利用 (2分) (6分) (8分)(36)七、1解:原式 (2分) (4分)2解:原式= (2分) (4分)3解:原式= (2分)(54)= (4分)4解:原式= (4分)5解:原式= (2分) (4分)6解:1) 当时 , 原式=0 (2分)2) 当时, 原式= (4分)(55)If we dont do that it will go on and go on. We have to stop it; we need the courage to do it.His comments came hours after Fifa vice-president Jeffrey Webb - also in London for the FAs celebrations - said he wanted to meet Ivory Coast international Toure to discuss his complaint.CSKA general director Roman Babaev says the matter has been exaggerated by the Ivorian and the British media.Blatter, 77, said: It has been decided by the Fifa congress that it is a nonsense for racism to be dealt with with fines. You can always find money from somebody to pay them.It is a nonsense to have matches played without spectators because it is against the spirit of football and against the visiting team. It is all nonsense.We can do something better to fight racism and discrimination.This is one of the villains we have today in our game. But it is only with harsh sanctions that racism and discrimination can be washed out of football.The (lack of) air up there Watch mCayman Islands-based Webb, the head of Fifas anti-racism taskforce, is in London for the Football Associations 150th anniversary celebrations and will attend Citys Premier League match at Chelsea on Sunday.I am going to be at the match tomorrow and I have asked to meet Yaya Toure, he told BBC Sport.For me its about how he felt and I would like to speak to him first to find out what his experience was.Uefa hasopened disciplinary proceedings against CSKAfor the racist behaviour of their fans duringCitys 2-1 win.Michel Platini, president of European footballs governing body, has also ordered an immediate investigation into the referees actions.CSKA said they were surprised and disappointed by Toures complaint. In a statement the Russian side added: We found no racist insults from fans of CSKA.Baumgartner the disappointing news: Mission aborted.The supersonic descent could happen as early as Sunda.The weather plays an important role in this mission. Starting at the ground, conditions have to be very calm - winds less than 2 mph, with no precipitation or humidity and limited cloud cover. The balloon, with capsule attached, will move through the lower level of the atmosphere (the troposphere) where our day-to-day weather lives. It will climb higher than the tip of Mount Everest (5.5 miles/8.85 kilometers), drifting even higher than the cruising altitude of commercial airliners (5.6 miles/9.17 kilometers) and into the stratosphere. As he crosses the boundary layer (called the tropopause),e can expect a lot of turbulence.The balloon will slowly drift to the edge of space at 120,000 feet ( Then, I would assume, he will slowly step out onto something resembling an Olympic diving platform.Below, the Earth becomes the concrete bottom of a swimming pool that he wants to land on, but not too hard. Still, hell be traveling fast, so despite the distance, it will not be like diving into the deep end of a pool. It will be like he is diving into the shallow end.Skydiver preps for the big jumpWhen he jumps, he is expected to reach the speed of sound - 690 mph (1,110 kph) - in less than 40 seconds. Like hitting the top of the water, he will begin to slow as he approaches the more dense air closer to Earth. But this will not be enough to stop him completely.If he goes too fast or spins out of control, he has a stabilization parachute that can be deployed to slow him down. His team hopes its not needed. Instead, he plans to deploy his 270-square-foot (25-square-meter) main chute at an altitude of around 5,000 feet (1,524 meters).In order to deploy this chute successfully, he will have to slow to 172 mph (277 kph). He will have a reserve parachute that will open automatically if he loses consciousness at mach speeds.Even if everything goes as planned, it wont. Baumgartner still will free fall at a speed that would cause you and me to pass out, and no parachute is guaranteed to work higher than 25,000 feet (7,620 meters).cause there
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