电大复变函数形成性考核册参考答案1

1、复变函数习题总汇与参考答案第1章 复数与复变函数一、单项选择题1、若z1=（a, b）,z2=(c, d),则z1z2=（c）a （ac+bd, a） b (ac-bd, b)c （ac-bd, ac+bd） d (ac+bd, bc-ad)2、若r0，则n（，r）= z：（d）a |z|r b 0|z|rc r|z|r3、若z=x+iy, 则y=(d)a b c d4、若a= ，则 |a|=（c）a 3 b 0 c 1 d 2二、填空题1、若z=x+iy, w=z2=u+iv, 则v=（ 2xy ）2、复平面上满足rez=4的点集为（ z=x+iy|x=4 ）3、（ 设e为点集，若它是开集，

2、且是连通的，则e ）称为区域。4、设z0=x0+iy0, zn=xn+iyn(n=1,2,)，则zn以zo为极限的充分必要条件是 xn=x0，且 yn=y0。三、计算题1、求复数-1-i的实部、虚部、模与主辐角。解：re(-1-i)=-1 im(-1-i)=-1|-1-i|=2、写出复数-i的三角式。解：3、写出复数 的代数式。解：4、求根式 的值。解：四、证明题1、证明若 ，则a2+b2=1。证明：而 3、证明：证明：第2章 解析函数一、单项选择题1若f(z)= x2-y2+2xyi,则2、若f(z)=u(x, y)+iv(x,y), 则柯西黎曼条件为（d）a bc d3、若f(z)=z+1

3、, 则f(z)在复平面上（c）a 仅在点z=0解析 b 无处解析c 处处解析 d 在z=0不解析且在z0解析4、若f（z）在复平面解析，g(z)在复平面上连续，则f(z)+g(z)在复平面上（c）a解析 b 可导c连续 d 不连续二、填空题1、若f(z)在点a不解析，则称a为f(z)的奇点。2、若f(z)在点z=1的邻域可导，则f(z)在点z=1解析。3、若f(z)=z2+2z+1，则 4、若 ，则 不存在。三、计算题：1、设f(z)=zre(z), 求解： =2、设f(z)=excosy+iexsiny,求解：f(z)=excosy+iexsiny=ez,z=x+iyu=excosy v=e

4、xsinyf(z)=u+ivf(z)在复平面解析，且 =excosy+iexsiny3、设f(z)=u+iv在区域g内为解析函数，且满足u=x3-3xy2，f(i)=0,试求f(z)。解：依c-r条件有vy=ux=3x2-3y2则v（x1y）=3x2y-y3+c(c为常数)故f(z)=x3-3xy2+i(3x2y-y3+c)=x3-3xy2+i(cx2y-y3)+ic =z3+ic，为使f(i)=0, 当x=0,y=1时，f(i)=0, 有f(0)=-i+ic=0c=1 f(z)=z3+i4、设f(z)=u+iv在区域g内为解析函数，且满足u=2(x-1)y,f(2)=-i,试求f(z)。解：

5、依c-r条件有vy=ux=2yv= =y2+(x) vx=(x)=v=y2-x2+2x+c(c为常数)f(z)=2(x-1)y+i(y2-x2+2x+c)为使f(z)=-i,当x=2 y=0时,f(2)=ci=-i c=-1f(z)=2(x-1)y+i(y2-x2+2x-1) =-(z-1)2i四、证明题1、试在复平面讨论f(z)=iz的解析性。解：令f(z)=u+iv z=x+iy则iz=i(x+iy)=-y+ixu=-y v=x于是ux=0 uy=-1vx=1 vy=0ux、uy、vx在复平面内处处连接又ux=vy uy=-vx。f(z)=iz在复平面解析。2、试证：若函数f(z)在区域g

6、内为解析函数，且满足条件（z）=0，zg，则f(z)在g内为常数。证：设f(z)=u+iv,z=x+iy,zgf(z)在g内解析，ux=vy, uy=-vx又（z）=0, （z）=ux+ivxux=0 vx=0uy=-vx=0 ux=vy=0u为实常数c1，v也为实常数c2，f(z)=c1+ic2=z0f(z)在g内为常数。复变函数课程作业参考解答2第3章 初等函数一、单项选择题1. z = ( a ) 是根式函数的支点. (a) 0 (b) 1 (c) (d) i2. z = ( d ) 是函数的支点. (a) i (b) 2i (c) -1 (d) 03. ei =( b ). (a) e

7、-1+e (b) cos1+isin1 (c) sin1 (d) cos14. sin1= ( a ) (a) (b) (c) (d) 二、填空题1. cosi = 2. = e(cos1+isin1)3. lni =4. ln(1+i) = k为整数.三、计算题1. 设z=x+iy，计算.解: = = 2. 设z = x+iy, 计算. 解: z = x+iy 3. 求方程的解.解: lnz = 由对数函数的定义有: z= 所给方程的解为z = i4. 求方程的解.解: =根据指数函数的定义有:z=n2+i 或z=n(1+)四、证明题1. 试证: . 证明：根据正弦函数及余弦正数定义有： s

8、in2z=2sinzcosz2. 证明: . 证明: 令a= b=sinx+sin2x+sinnx = 第4章 解析函数的积分理论一、单项选择题1. ( d ) , c为起点在0 , 终点在1+i的直线段. (a) 0 (b) 1 (c) 2i (d) 2(1+i)2. . (a) 0 (b) 10 (c) i (d) 3. (a) i (b) 10 (c) 10i (d) 04. =( a ). (a) (b) (c) (d) 二、填空题1. 若与沿曲线c可积，则.2. 设l为曲线c的长度, 若f(z)沿c可积, 且在c上满足,则.3. 4. 三、计算题1.计算积分,其中c为自0到2+i的直

9、线段. 解: c的方程为: 其次由得 = =2. 计算积分. 解: = 作区域d:积分途径在d内被积函数的奇点z=2与z=3均不在d内,所以被积函数在d内解析.由定理4.2得:=03. 计算积分. 解: 奇点z=1和z=-1不在区域d,内 的三个根也不在d内 由定理4.2 得 =04. 计算积分, . 解: 由定理4.6得 四、证明题1. 计算积分，并由此证明. 证明：在圆域 |z|1内解析 = 另一方面,在圆|z|= =(实部和虚部为0) = = = = =0 而为偶函数0= = 复变函数课程作业参考解答3第5章 解析函数的幂级数表示一、单项选择题1. 幂级数的收敛半径等于( b ) ( a

10、 ) 0 (b) 1 ( c ) 2 (d) 32. 点z=-1是f(z)=r ( b )级零点. ( a ) 1 (b)2 (c)3 (d)53. 级数的收敛圆为( d ). (a) | z-1| 3 (b) |z|1 (d) |z| 14. 设f(z)在点a解析, 点b是f(z)的奇点中离点a最近的奇点,于是,使f(z)=成立的收敛圆的半径等于( c ). (a) a+b+1 (b) b-a+1(c) |a-b| (d) |a+b|二、填空题1.级数1+z+的收敛圆r=+即整个复平面2.若f(z)= (k为常数),则z=m(m=0, )为f(z)的 1 级零点. 3.幂有数的收敛半径等于

11、0 . 4.z=0是f(z)=ez-1的 1 级零点. 三、计算题 1.将函数f(z)=在点z=0展开幂级数. 解: f(z)= =- 2.将函数f(z)=(1-z)-2在点z=0展开成幂级数. 解：而(1-z)-1= = 3将函数f(z)=(z+2)-1在点z=1展开成幂级数. 解：f(z)=(z+2)-1= = 4将函数f(z)=ez在点z=1展开成幂级数. 解： f(z)=ez f(n)=ez 四、证明题 1证明:1-ei2z=-2isinzeiz 证：eiz=cosz+isinze-iz=cos-isinz eiz-e-iz=2isinz -2isinz=-( eiz-e-iz) =

12、eiz-e-iz -2isinz eiz=( e-iz- eiz) eiz =e0- e2iz=1- e2iz2试用解析函数的唯一性定理证明等式: cos2z= cos2z-sin2z 证f1(z)=cos2z,则f1(z)复平面g解析设f2(z)coszsin2，则f2(z)也在整个复平面g解析取e=k为实数轴,则e在g内有聚点.当e为实数时,知cos2z=cos2z-sin2z,即f1(z)= f2(z)由解析函数唯一性定理,由以上三条知f1(z)= f2(z) 成立即cos2z= cos2z-sin2z 第6章 解析函数的罗朗级数表示 一、单项选择题 1函数f(z)=在点z=2的去心邻域

13、( d ) 内可展成罗朗级数. (a) 0 (b) 0 (c) 1 (d) 0 2设点为f(z)的孤立奇点，若=c，则点为f(z)的( c ). (a) 本性奇点 (b) 极点 (c) 可去奇点 (d) 解析点 3若点为函数f(z)的孤立奇点，则点为f(z)的极点的充分必要条件是( d ). (a) f(z)=c() (b) f(z)= (c) f(z)=c() (d) f(z)= 4若点为函数f(z)的孤立奇点，则点为f(z)的本性奇点的充要条件是（ b ）. (a) f(z)= c() (b) f(z)不存在 (c) f(z)=c() (d) f(z)= 二、填空题 1设为函数f(z)在点

14、的罗朗级数,称为该级数的主要部分. 2.设点为函数f(z)的奇点,若f(z)在点的某个 某个去心邻域内解析,则称点为f(z)的孤立奇点. 3.若f(z)=，则点z=0为f(z)的 0 级极点. 不是极点,若f(z)= 则z=0为f(z)的一个极点. 4.若f(z)=(sin)-1,则点z0为f(z)非孤立 奇点. 三、计算题1将函数f(z)=(z-2)-1在点z=0的去心邻域展成罗朗级数.解: f(z)= = - = - 2将函数f(z)在点的去心邻域展成罗朗级数. 解: f(z)= 3试求函数f(z)=z-3sinz3的有限奇点,并判定奇点的类别. 解: 解析,无奇点,f(z)的有限奇点为z

15、=0. 并且为3阶极点. 4试求函数f(z)=z-1的有限奇点，并判定奇点的类别. 解: f(z)的m阶奇点即的阶零点,而零点为z=0，z=1，z=-1，且均为1阶零点。的有限奇点为z=0，z=1,z=-1且均为1阶极点. 四、证明题 1设f(z)=,试证z=0为f(z)的6级极点. 证:要证z=0为f(z)的6级极点,只需证z=0为的6阶零点即可.而 =8z3 =8z6 令 则 为的6阶零点 z=0 为f(z)的6级极点.if we dont do that it will go on and go on. we have to stop it; we need the courage to

16、 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 med

17、ia.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 vis

18、iting 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

19、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

20、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 investigatio

21、n 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 pl

22、ays 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-t

23、o-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

24、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,

25、 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