電大復變函數(shù)形成性考核冊參考答案1

1、專業(yè)好文檔復變函數(shù)習題總匯與參考答案第1章 復數(shù)與復變函數(shù)一、單項選擇題1、若z1=（a, b）,z2=(c, d),則z1z2=（c）a （ac+bd, a） b (ac-bd, b)c （ac-bd, ac+bd） d (ac+bd, bc-ad)2、若r0，則n（，r）= z：（d）a |z|r b 0|z|rc r|z|r3、若z=x+iy, 則y=(d)a b c d4、若a= ，則 |a|=（c）a 3 b 0 c 1 d 2二、填空題1、若z=x+iy, w=z2=u+iv, 則v=（ 2xy ）2、復平面上滿足rez=4的點集為（ z=x+iy|x=4 ）3、（ 設e為點集，若

2、它是開集，且是連通的，則e ）稱為區(qū)域。4、設z0=x0+iy0, zn=xn+iyn(n=1,2,)，則zn以zo為極限的充分必要條件是 xn=x0，且 yn=y0。三、計算題1、求復數(shù)-1-i的實部、虛部、模與主輻角。解：re(-1-i)=-1 im(-1-i)=-1|-1-i|=2、寫出復數(shù)-i的三角式。解：3、寫出復數(shù) 的代數(shù)式。解：4、求根式 的值。解：四、證明題1、證明若 ，則a2+b2=1。證明：而 3、證明：證明：第2章 解析函數(shù)一、單項選擇題1若f(z)= x2-y2+2xyi,則2、若f(z)=u(x, y)+iv(x,y), 則柯西黎曼條件為（d）a bc d3、若f(z

3、)=z+1, 則f(z)在復平面上（c）a 僅在點z=0解析 b 無處解析c 處處解析 d 在z=0不解析且在z0解析4、若f（z）在復平面解析，g(z)在復平面上連續(xù)，則f(z)+g(z)在復平面上（c）a解析 b 可導c連續(xù) d 不連續(xù)二、填空題1、若f(z)在點a不解析，則稱a為f(z)的奇點。2、若f(z)在點z=1的鄰域可導，則f(z)在點z=1解析。3、若f(z)=z2+2z+1，則 4、若 ，則 不存在。三、計算題：1、設f(z)=zre(z), 求解： =2、設f(z)=excosy+iexsiny,求解：f(z)=excosy+iexsiny=ez,z=x+iyu=excos

4、y v=exsinyf(z)=u+ivf(z)在復平面解析，且 =excosy+iexsiny3、設f(z)=u+iv在區(qū)域g內(nèi)為解析函數(shù)，且滿足u=x3-3xy2，f(i)=0,試求f(z)。解：依c-r條件有vy=ux=3x2-3y2則v（x1y）=3x2y-y3+c(c為常數(shù))故f(z)=x3-3xy2+i(3x2y-y3+c)=x3-3xy2+i(cx2y-y3)+ic =z3+ic，為使f(i)=0, 當x=0,y=1時，f(i)=0, 有f(0)=-i+ic=0c=1 f(z)=z3+i4、設f(z)=u+iv在區(qū)域g內(nèi)為解析函數(shù)，且滿足u=2(x-1)y,f(2)=-i,試求f(

5、z)。解：依c-r條件有vy=ux=2yv= =y2+(x) vx=(x)=v=y2-x2+2x+c(c為常數(shù))f(z)=2(x-1)y+i(y2-x2+2x+c)為使f(z)=-i,當x=2 y=0時,f(2)=ci=-i c=-1f(z)=2(x-1)y+i(y2-x2+2x-1) =-(z-1)2i四、證明題1、試在復平面討論f(z)=iz的解析性。解：令f(z)=u+iv z=x+iy則iz=i(x+iy)=-y+ixu=-y v=x于是ux=0 uy=-1vx=1 vy=0ux、uy、vx在復平面內(nèi)處處連接又ux=vy uy=-vx。f(z)=iz在復平面解析。2、試證：若函數(shù)f(z

6、)在區(qū)域g內(nèi)為解析函數(shù)，且滿足條件（z）=0，zg，則f(z)在g內(nèi)為常數(shù)。證：設f(z)=u+iv,z=x+iy,zgf(z)在g內(nèi)解析，ux=vy, uy=-vx又（z）=0, （z）=ux+ivxux=0 vx=0uy=-vx=0 ux=vy=0u為實常數(shù)c1，v也為實常數(shù)c2，f(z)=c1+ic2=z0f(z)在g內(nèi)為常數(shù)。復變函數(shù)課程作業(yè)參考解答2第3章 初等函數(shù)一、單項選擇題1. z = ( a ) 是根式函數(shù)的支點. (a) 0 (b) 1 (c) (d) i2. z = ( d ) 是函數(shù)的支點. (a) i (b) 2i (c) -1 (d) 03. ei =( b ).

7、(a) e-1+e (b) cos1+isin1 (c) sin1 (d) cos14. sin1= ( a ) (a) (b) (c) (d) 二、填空題1. cosi = 2. = e(cos1+isin1)3. lni =4. ln(1+i) = k為整數(shù).三、計算題1. 設z=x+iy，計算.解: = = 2. 設z = x+iy, 計算. 解: z = x+iy 3. 求方程的解.解: lnz = 由對數(shù)函數(shù)的定義有: z= 所給方程的解為z = i4. 求方程的解.解: =根據(jù)指數(shù)函數(shù)的定義有:z=n2+i 或z=n(1+)四、證明題1. 試證: . 證明：根據(jù)正弦函數(shù)及余弦正數(shù)定

8、義有： sin2z=2sinzcosz2. 證明: . 證明: 令a= b=sinx+sin2x+sinnx = 第4章 解析函數(shù)的積分理論一、單項選擇題1. ( d ) , c為起點在0 , 終點在1+i的直線段. (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、填空題1. 若與沿曲線c可積，則.2. 設l為曲線c的長度, 若f(z)沿c可積, 且在c上滿足,則.3. 4. 三、計算題1.計算積分,其中c為自0到

9、2+i的直線段. 解: c的方程為: 其次由得 = =2. 計算積分. 解: = 作區(qū)域d:積分途徑在d內(nèi)被積函數(shù)的奇點z=2與z=3均不在d內(nèi),所以被積函數(shù)在d內(nèi)解析.由定理4.2得:=03. 計算積分. 解: 奇點z=1和z=-1不在區(qū)域d,內(nèi) 的三個根也不在d內(nèi) 由定理4.2 得 =04. 計算積分, . 解: 由定理4.6得 四、證明題1. 計算積分，并由此證明. 證明：在圓域 |z|1內(nèi)解析 = 另一方面,在圓|z|= =(實部和虛部為0) = = = = =0 而為偶函數(shù)0= = 復變函數(shù)課程作業(yè)參考解答3第5章 解析函數(shù)的冪級數(shù)表示一、單項選擇題1. 冪級數(shù)的收斂半徑等于( b

10、) ( a ) 0 (b) 1 ( c ) 2 (d) 32. 點z=-1是f(z)=r ( b )級零點. ( a ) 1 (b)2 (c)3 (d)53. 級數(shù)的收斂圓為( d ). (a) | z-1| 3 (b) |z|1 (d) |z| 14. 設f(z)在點a解析, 點b是f(z)的奇點中離點a最近的奇點,于是,使f(z)=成立的收斂圓的半徑等于( c ). (a) a+b+1 (b) b-a+1(c) |a-b| (d) |a+b|二、填空題1.級數(shù)1+z+的收斂圓r=+即整個復平面2.若f(z)= (k為常數(shù)),則z=m(m=0, )為f(z)的 1 級零點. 3.冪有數(shù)的收斂

11、半徑等于 0 . 4.z=0是f(z)=ez-1的 1 級零點. 三、計算題 1.將函數(shù)f(z)=在點z=0展開冪級數(shù). 解: f(z)= =- 2.將函數(shù)f(z)=(1-z)-2在點z=0展開成冪級數(shù). 解：而(1-z)-1= = 3將函數(shù)f(z)=(z+2)-1在點z=1展開成冪級數(shù). 解：f(z)=(z+2)-1= = 4將函數(shù)f(z)=ez在點z=1展開成冪級數(shù). 解： f(z)=ez f(n)=ez 四、證明題 1證明:1-ei2z=-2isinzeiz 證：eiz=cosz+isinze-iz=cos-isinz eiz-e-iz=2isinz -2isinz=-( eiz-e-i

12、z) = eiz-e-iz -2isinz eiz=( e-iz- eiz) eiz =e0- e2iz=1- e2iz2試用解析函數(shù)的唯一性定理證明等式: cos2z= cos2z-sin2z 證f1(z)=cos2z,則f1(z)復平面g解析設f2(z)coszsin2，則f2(z)也在整個復平面g解析取e=k為實數(shù)軸,則e在g內(nèi)有聚點.當e為實數(shù)時,知cos2z=cos2z-sin2z,即f1(z)= f2(z)由解析函數(shù)唯一性定理,由以上三條知f1(z)= f2(z) 成立即cos2z= cos2z-sin2z 第6章 解析函數(shù)的羅朗級數(shù)表示 一、單項選擇題 1函數(shù)f(z)=在點z=2

13、的去心鄰域( d ) 內(nèi)可展成羅朗級數(shù). (a) 0 (b) 0 (c) 1 (d) 0 2設點為f(z)的孤立奇點，若=c，則點為f(z)的( c ). (a) 本性奇點 (b) 極點 (c) 可去奇點 (d) 解析點 3若點為函數(shù)f(z)的孤立奇點，則點為f(z)的極點的充分必要條件是( d ). (a) f(z)=c() (b) f(z)= (c) f(z)=c() (d) f(z)= 4若點為函數(shù)f(z)的孤立奇點，則點為f(z)的本性奇點的充要條件是（ b ）. (a) f(z)= c() (b) f(z)不存在 (c) f(z)=c() (d) f(z)= 二、填空題 1設為函數(shù)f

14、(z)在點的羅朗級數(shù),稱為該級數(shù)的主要部分. 2.設點為函數(shù)f(z)的奇點,若f(z)在點的某個 某個去心鄰域內(nèi)解析,則稱點為f(z)的孤立奇點. 3.若f(z)=，則點z=0為f(z)的 0 級極點. 不是極點,若f(z)= 則z=0為f(z)的一個極點. 4.若f(z)=(sin)-1,則點z0為f(z)非孤立 奇點. 三、計算題1將函數(shù)f(z)=(z-2)-1在點z=0的去心鄰域展成羅朗級數(shù).解: f(z)= = - = - 2將函數(shù)f(z)在點的去心鄰域展成羅朗級數(shù). 解: f(z)= 3試求函數(shù)f(z)=z-3sinz3的有限奇點,并判定奇點的類別. 解: 解析,無奇點,f(z)的有

15、限奇點為z=0. 并且為3階極點. 4試求函數(shù)f(z)=z-1的有限奇點，并判定奇點的類別. 解: f(z)的m階奇點即的階零點,而零點為z=0，z=1，z=-1，且均為1階零點。的有限奇點為z=0，z=1,z=-1且均為1階極點. 四、證明題 1設f(z)=,試證z=0為f(z)的6級極點. 證:要證z=0為f(z)的6級極點,只需證z=0為的6階零點即可.而 =8z3 =8z6 令 則 為的6階零點 z=0 為f(z)的6級極點.if we dont do that it will go on and go on. we have to stop it; we need the coura

16、ge 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 britis

17、h 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 th

18、e 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-b

19、ased 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

20、 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 investi

21、gation 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 weath

22、er 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

23、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 expe

24、ct 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. s

25、till, 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