天津理工大学复变复习题解（留）

课后习题参考解答

第一部分 复变涵数 课后习题一 1、解（1）z?虚部Im?z???213?2i3?2i32实部Re?z??3,????i,133?2i?3?2i??3?2i?9?413132

13321322 ?3??2?z??i,z????????,tan???,argz??arctan13131333?13??13? ?2?z?1?3i??i?3i?1?i???i?3i?3??i?3?3i?3?5i.

i1?i?1?i??1?i?2222235Re?z??,Im?z???,22359251?3??5?z??i,z??????????34. 22442?2??2?2255tan???,argz??arctan

33 2、解 ?1??1?cos??isin??e

i?i????? z?1?i3,r?z?12??3?2?2,tan??3,argz?,z?2?cos?isin??2e3

?3?33? 3、解 ?1??3?i?5,记z?3?i,z?????????z?2?cos????isin????,?6????6??3?2???1??2,tan??2?13??53?,argz?? 36??????????3?i??2?cos????isin??????6??? ???6????5??31??5?????25?cos?????isin?????32??2?2???163?16i66??????????5 ?2?6?1?6cos??isin??cosk?0,w0?cos??2k?6?isin??2k?6,k?0,1,?,5

31;3?3??ik?1w1?cos?isin?i;

6622665?5?31;7?7?31

k?2,w2?cos?isin???ik?3,w0?cos?isin???i66226622?isin?k?4,w0?cos9?9??isin??i;k?5,66??w5?cos11?11?31

?isin??i66224、解?1?x3?8?0,??2k???2k??，k=0,1, 2 ?x3??8,x?3?8?38?cos??isin???2?cos?isin?33??????3?3??k?0时,w0?2?cos?isin??1?3i;k?1时,w1?2??isin?cos???2

33?33???5?5??k?2时,w1?2?cos?isin33????1?3i ?3 ?2?微分方程y????8y?0的特征方程为??8?0，由上题的结果知

此特征方程的根为?1??2,?2?1?3i,?3?1?3i。故微分方程y????8y?0的通解为

y?c1e?2??exc2cos3x?c3sin3x.

5、不等式?1?argz??1??确定的区域为由射线

?? 39

不???1及???1??构成的角区域，内。即为一半平面。是无界的单连通区域。 （2）设z?x?iy，不等式

包括两射线在

如图所示。

z?1?4z?1即

?x?1?2?y2?4?x?1?2?y，两边平方得，

2?x?1?2?y2?16?x?1??y2

x2?2x?1?y2?16x2?32x?16y2?16,2215x2?34x?15y2??15

17??8?。此不等式所确定的区 经配方得?2?x???y????15??15?域为中心在z??17半泾为8的圆的外部区域（不包括圆周在内）是无界的多连通的。如图所示。

1515 （3）设z?x?iy,代入不等式z?2?z?2?6得

?x?2?2?y2??x?2?2?y2?6,移项得?x?2?2?y2??6??x?2?2?y2

两边平方得?x?2?2?y2?36??x?2?2?y2?12?x?2?2?y2

2x2?4x?4?y2?36?x2?5x?4?y2?12?x?2??y2

12?x?2??y2?36?8x,3?x?2??y2?9?2x,两边平方得

229?x?2??y2?81?4x2?36x,2??

9x2?36x?36?9y2?81?4x2?36x

5x?9y?45,22x2y2??1 95x2y2??1及其围成95所以此不等式所确定的区域为椭圆。的区域，有界单连通区域。如图所示。

（4）设z=x+iy Rez2?Rex2?y2?2xyi?x2?y2, x2?y2<1所确定的区域为等轴双曲线x2?y2?1两支之

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????间的部分。

6、解（1）z?acost?ibsint,x?acost,y?bsint,则

xy。即椭圆2?x??y?cost?,sint?,cos2t?sin2t???????1aab?a??b?22x2y2?2?1 b ?2?z?t?,it令x?t,y?,xy?1，即映射成等轴双曲线xy?1。

x?iyx?iyx?y?2?2?i222?x?iy??x?iy?x?yx?yx?y21t 7、解 记z=x+iyw?1?1?zx?iy

?1?y?x,w?x2?iy2?1?i1,u?x,y???v?x,y??1

2x2x2x2x2y即直线y=x映射成直线u= —v。 ?2??x?1??y2?1,2x2?2x?1?y2?1,x2?y2?2x

w?xyxy即映射成直线u=1..

?i??i,2x2x2?y2x2?y22xz?z0 8、证明定理三 证 由定理一可知limf?z??f?z0??u?x0,y.0??iv?x.,y0? 的充要条件为limu?x,y??u?x0,y0?， limv?x,y??v?x0,y0?

x?x0x?x0y?y0y?y0再根据复变函数与二元实变量连续的定义可知，定理三成立。

9、证 因为f?z?在z0连续，则limf?z??f?z0??0。故f?z??0取

z?z0??f?z2?，由极限定义可知，必存在一正数?????0使得当0?z?zf?z??f?z0??f?z0?20?????

f?z0?2时有

?，由不等式

f?z0? 从而 2f?z??f?z0??f?z??f?z0??可得

f?z0?2?f?z??f?z0??0?f?z0?2?f?z??3f?z0? 2于是，在邻域z?z0????? 内f?z??0。 10、证 f?z??z?x?iy,u?x,y??x,v?x,y???y。因为

u?x,y??x,v?x,y?在全平面处处连续，故由定理三可知，f?z??z在复平面处处连续。

课后习题二 1、 解 f?z??1，由导数定义知

z??z11?2f?z??z??f?z??11f??z??lim?limz??zz?limz?z?z?lim2??2?z?0?z?0?z?0?z?0z?z?z?z?z?zz

2、解 ?1?f?z??x2?iy,u?x,y??x2,v?x,y???y

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?u?2x,?x?u?0,?y?v?0,?x?u?v?u?v?v??成立，要使?成立只要2x?1，??1，

?y?x?x?y?y11x??,故在直线z?上函数可导。而在复平面上处处不解析。

22

?2?f?z??2x3?3y2i,u?x,y??2x3,v?x,y??3y3

?u?u?v?v?u?v?6x2,???0,?9y2要使?成立，只要 ?x?y?x?y?x?y6x2?9y2,即2x??3y，故此函数只在直线2x??3y上可导，但在复平面上处处

不解析。

3、解 ?1?f?z???z?1?5,f??z??5?z?1?4,故f?z?在复平面上处处解析。 ?2??1,2z?1f??z???2z?z2?1?2,z2?1?0,z??1,故除了点

z??1外，此函数在复平面内处处解析。

4、解 f?z??z?1,2zz?1??由zz?1=0得z?0,z??1,故

?2?f?z?的奇点为0,?1。

5、解 因为f?z??my3?nx2y?ix3?lxy2,为解析函数

??u?x,y??my3?nx2y,v?x,y??x3?lxy2?u?u?v?v?2nxy,?3my2?nx2,?3x2?ly2,?2lxy. ?x?y?x?y，

由于柯西－黎曼方程成立，故

?2nxy?2lxy?2222?3my?nx??3x?ly?1?，由于对任薏的x,y值两等式成立，故 ?2?由（1）知n=l,由（2）知3m=-l,n=--3, 故n=l=-3。

6、解 ?1?sinz?sin?x?iy??sinxchy?icosxshy?0，于是

?sinxchy?0??cosxshy?0?1? ?2?因为chy?0，由（1）式得sinx?0?z?k??k?0,?1,?2,?? 代入（2）得cosk?shy?0

cosk???1,?k?0,?1,?2,??,shy?0?y?0。

?2?1?ez

?0?ez??1,zk?Ln??1??ln1?i?arg??1??2k????2k?1??i

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（k?0,?1,?2,?）

??7、?1?Ln??i??ln?i?i?arg??i??2k???0?i???2k??

?2?1?=??2k???i?k?0,?1,2,?? ?2??4? ?2?Ln??3?4i??ln?3?4i?i?arg??3?4i???ln5?i?arctan?????2k?? ???3??4???ln5?i???arctan?2k??3?????k?0,?1,?2,??

?????? ?3?e1?i2?e?e?i2?e?cos???isin???????e?0?i???ie ?22???????4?3i

?eiLn3?ei?ln3?i?arg3?2k????eiln3?6k??e?2k??cosln3?isinln3?

?5??1?i?i?ei?1?i??ei?ln1?i?i?ar?g1?i??2k???

?e???iln2???2k???4??e??????2k???4??cosln2?isinln2 ?e???????2k???4?ln2ln2? ??isin?cos?22??z?z8、?1?因为shz?0,故e?e?0,ez?e?z,e2z?1,

22z=Ln??1??ln?1?i?arg??1??2k????2k?1??i

1??z??k???i2???k?0,?1,?2,??

ew?e?w9、设z?shw?,2z?ew?e?w2zew?e2w?1,

2e2w?2zew?1?0,记y?ew,得二元一次方程y2?2zy?1?0

2z?4z2?4?z?z2?1,即ew?z?z2?1,取对数得 其根为y=

2w?Lnz?z2?1

课后习题三

1、 解 （1）先写出直线段的参数方程，显然其直角坐标方程为y?1x,令y?t,则x?3t,直

3线段的参数方程为 ?x?3t0?t?1,z?t??3t?it,z??t??3?i ??y?t?? 由积分算法2得

43

3?i3131?3?i?26

zdz???3t?it??3?i?dt??3?i??tdt??3?i?t??6?i?30330001222122（2）设曲线c1为从原点到点（3，0）的直线段c2

?x?3其参数方程为?0?t?1z?3?ti

y?t?由积分算发法2得

3?i?3,0?2?3,1?2?z0dz?z???0,0dz?z???3,02dz??z2dz??z2dz

c1c222?zdz???3t??3dt??27tdt?9t22c1001110?9

126 231zdz=????????3?tiidt?3?tid3?ti?3?ti??3?i?c??222110030326?26 故z2dz?z2dz?z2dz?9???3?i?6?i?????c1c233??03?i沿两条路线的积分值相等。 2、抛物线y?x的参数方程为?2?x?t0?t?1,复参数方程为 2?y?tz?t?it2，dz=(2ti+1)dt t. 由积分算法2得

??x012?iy?dz???t?it20112??2ti?1?dt??1?i??t?2ti?1?dt

201?1111?15??1?i??2t3i?t2dt??1?i??t4i?t3????i

?2030?660????3、解 （1）

?zc2dzdz??,被积函数的奇点为 2c?2z?4?z?1??3z1??1?3i,z2??1?3i由于z1?z2?2,所以两奇点均在圆z?1外，由柯西－古萨基

本定理可知，

dz?z2?2z?4?0，

（2）z?1在圆内，f?z??1.由柯西积分公式得

2?dz1z?2C?2?i?1?2?i。

z（3）由于f?z??ze,在复平面上处处解析，故由柯西－古萨基本定理可知zedz?0

z?c（4）设

i??A?z?2??B?z???A?B?z?2A?iB1AB2?44 ?2f?z??????iiiiz?2???????z???z?2?z??z???z?2??z???z?2?22?2?2?????0??A?B， 解得A?1?2,B??2, i?2A?B?1i4?i4?i?2?z?2??dz2?dzdz?，

?????c?i?i?cz?2?4?i??cz??z???z?2???2??2??由于i2在圆,z?1内，所以?1z?i2cdz?2?i，由于-2不在圆,z?1内

，从而

dzdz24?i

??2?i?.?cz?2?0，所以?c?i?4?i4?i?z???z?2?2??z4、解（1）由于f?z??e在复平面上处处解析，故有柯西积分公式可得

ezz2dz?2?ie?2?ei ?c:z?2?1z?2z?2 （2）因为被积函数f?z??z 的奇点z?3在圆 z?2;之外，故由柯西－古z?3萨基本定理可知

z?cz?3dz?0

（3）被积函数f?z???1的奇点为z??i,z??2i，由于曲线.c为圆 z?32z2?1z2?4???由于z??i在c内，z??2i 在c外，，根据复合闭路定理及柯西积分公式得

??zc2dz??1z2?4???z?i?r1??dzz?4?z?i??z?i2?z?i?r2??dzz?4?z?i? z?i2???11?2?i?2???0 2??i?4??i?i???i??4??i?i????注：适当取r1,r2的大小，使两圆互不包含又互不相交。

5、解 （1）因为e在复平面上处处解析，且其原函数之一为它自己。故由定理三知，

z???23?i?ie2zdz?12z3?i16?ie?e?e?2?i??i22??

1?cos6??isin6??cos2??isin2???02（2）因为sinz在复平面上处处解析，故由定理三得

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?????i?isin2zdz??1?cos2z1?1??idz??z?sin2z???i22?2???i?i

1?11??????i?sin2?i???i?sin?2?i?2?22??11??2?i?sin2?i???i?1ish2?i?????sh2??i222??

（3）被积函数f?z??1ABA?z?1??Bz?A?B?z?A

????z?z?1?zz?1z?z?1?z?z?1??A?B?0???A?1解得A??1,B?1，故

111

???zz?1z2?z被积函数的奇点在，z=0,z=1均在包含单位圆的闭曲线内，由柯西积分公式可得

?Cdzdzdz?????2?i?2?i?0 2??Czz?1z?zC(4) 因为被积函数f?z??sinz的奇点z??i在曲C：z?i?3内，故由柯西积分公式得 z?isinzei??i??e?i??i???e?e?1 （5）被积函?Cz?idz?2?i?sinzz??i?2?i?sin??i??2?i2i??数f?z??1?z2?9?22的奇点z= -3i在曲线：z?2?2之外，z=3i在C之内，由高阶导数公式得

dz??zCdz2?9?2?z?3i????C?z?3i?2?z?3i?2?C?z?3i?2dz??1????2?i 2???2?3i??z?3i=?2?24??

?2?i??2?i??.21654?z?3i?3z?3i?6i?3 6、解 ?1?v?y,f?2??0 2x?y2方法一：?v???x2xy?x2?y2?2x2?y2?y??2yx2?y2 ?v??22?yx2?y2x2?y2??????由于函数f?z?解析，所以满足柯西－黎曼方程。从而

?u?v2xy????y?xx2?y2??2

对y 进行偏积分，u?x,y????x2xy2?y2?2dy?x?d?x2?y2??x2?y2?2??x?g?x? 22x?y2222由此式得?u???x?y??x?2x?g??x??x?y?g??x?，又因为

?x?x2?y2?2?x2?y2?2?u?vx2?y2x2?y2????g??x??x?y?x2?y2?2?x2?y2?2

g??x??0,?g?x?=C

46

所以f?z???xyx?iy11

?C?i?C??C??C?x?itzx2?y2x2?y2x2?y2因为f?2??0,所以C?方法二：

1，故所求函数为f?z??1?1 22z?v?u?v?vx2?y22xyf??z???i??i??i2?y?y?y?xx2?y2x2?y2????2??x?x?iy?22?y22??1?1???x?iy???z2??2

因为f?2??0,所以C?1，故所求函数为f?z??1?1。 22z（3）u?2(x?1)y f(2)??i

方法一：

?u?u=2y, =2(x-1)由柯西-黎曼方程知

?y?x?v?u?v ?u?? ??y?x?x?yv=2ydy?y2?g(x)则?v?g'(x) （1）

?x?v?u????2(x?1) （2） ?x?y?由（1）（2）得g'(x)?2?2x

g?x????2?2x?dx?2x?x?C v(x'y)?y2?2x?x2?c

?f(2)?u?iv?2(x?1)y?i(y2?2x?x2?c)

2??i[i2xy?i2y?x2?2z?y2?c] ??i[(x?iy)2?2(x?iy)?c]??i[z2?2z?c]??i[(z?1)2]?(1?c)i

?f(2)??i??i??i[(2?1)2?(1?c)i]??i[1?1?c] ?2?c?1?c??1

从而f(z)??i(z?1) 方法二：

2?u?u?2(x?1) ?2y ?y?xf'(z)?

?u?u?i?2y?i?2(x?1)???i(2x?2yi)?2i ?x?y47

??2i(x?iy)?i??2iz?2i 所以f(z)??iz?2iz?c 因为f(z)??1

2所以c??i从而f(z)??iz?2iz?i??i(z?2z?1)??i(z?1) 注：方法二过程简单，但要将导函数凑成z的表达式，这一般有困难。

课后习题四

222(6?5i)n51．（1）令zn? 因为 6?5i?61(cosn??isin?),??arctgn86??所以z??61?(cosn??isin?)

?8???nn61?且61又因为z??zn收敛 ???1所以n?8?8n?0??n??(6?5i)n所以?绝对收敛 n8n?0?cosin1eiin?e?iin1e?n?enen?e?n（2）令zn? ?n?n?nn?12222z21? ?1(e)n?1???

222?2e??1?收敛，故原级数发散 因为?1(e)n发散,这由于e?1 ；?1????n?0nn222n?02?2e?p2．（1）cn?1

pn cn?11?n???limn?lim??lim?1?pn??cn??n?1n?????1??1???n?故R?1?1

?2(2).cn??n!?

nncn?1[(n?1)!]2nn

??lim?limn??cn??(n?1)n?1?n!?2n22n ?lim(n?1)(n!)n?limn?1?limn?1

nnn??(n?1)n(n?1)?n!?2n??n???n?1??1????1???n??n? ??

故R?0

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(3)cn?en,cn?1故lim3.(1)因为

所以

i?nn??cn?1 从而R?1

1?1?z?z2?z3????z?1 1?z1369n3n3?1?z?z?z?????(?1)z????z?1 1?z3所以z?1 从而R?1

1?1?z?z2_z3?????(?1)nzn????z?1 1?z1所以?1?z2?z4?z6?????(?1)nz2n????z2?1 21?z(2)因为

1??2z 即z?1 由于????2?1?z??1?z2?2/所以

1?1?z?22??11/

()2z1?z2 ??1[1?z2?z4?z6?????(?1)nz2n????]/ 2z1?2z?4z3?6z5?????(?1)2nz2n?1???? 2z?1?2z2?3z4?????(?1)n?1nz2(n?1)????????z2z4z6因为cosz?1???????z???

2!4!6!z4z8z12??????z2??? 所以cosz?1?2!4!6!2故z???从而R???

4.(1)①0?z?1由于圆环的圆心为z=0故展开的各项为z的幂

111 ?z(1?z)2z(1?z)2?因为1??zn?1?z?z2?????zn????z?1

1?zn?0所以(1)/??1?z2?11

?2(1?z)?1?z?2n?1?1?2z?3z?????nz

??????nzn?1

n?1?49

???11n?1n?2所以??nz??nz??(n?2)zn 0?z?1 2z(1?z)n?1zn?1n??1②0?z?1?1,由于圆环的圆心为z?1,故展开的级数的各项为(z?1)的幂

?111，因为1??(?1)nzn ?22z(1?z)(1?z)z1?zn?0z?1

?11故???(?1)n(z?1)n z?1?1 z1?z?1n?0从而

??1n?2n??(?1)(z?1)(z?1)??(?1)n(z?1)n?2 2z(1?z)n?0n?0 ?n??1?(?1)?n(z?1)n 0?z?1?1

⑶ 因为1?z???所以

1?1 zzz(1?) 1?zzzz2zz2z31???(1??????)??(??????)1?1111111e11?z11111111111?1?(?2?3????)?(?2?3????)2?(?2?3????)3????zz2!zz3!zzzzz

?1?1111??????? z2!z23!z34!z4ww2w3wn注：因为e?1?w???????????

2!3!n!作代换w?1111??(?2?3????)即得 1?zzzz课后习题五

1．⑴

1 22z(z?1)方法一： 12?z(z?i)2?z?i?因z?0是1的一级零点z??i是1的二级零点。 f(z)f(z)f(z)所以z?0是f(z)的一级极点z??i是f(z)的二级极点。 方法二：

f(z)?111??(z)且?(0)?0 故z?0是f(z)的一级极点

z(z2?1)2z 50

f(z)?1?z?i?211??(z)且?(i)?0，故z?i是f(z)的二级极点。同理22z(z?i)?z?i?z??i是f(z)的二级极点。

⑵sinzz?0是奇点

z3sinz1z3z511z2?3(z?????)?2?????

3!5!3!5!z3zz故z?0是f(z)的二级极点。但若不仔细一分析可能误认为是f(z)的三级极点。 ⑶f(z)?ln(z?1)因为 zln(z?1)1z2z3zz2?(z??????)?1?????? zz2323所以不包含负幂项z?0为f(z)的可去奇点 2.⑴f(z)?z?12 z?0、z?2是f(z)的一级极点，所以由规则Ⅲ可得

z?2zz?1z?11Res[f(z),0]?2|z?0?|z?0?? 由规则Ⅰ可得

2z?22(z?2z)'Res[f(z),2]?lim(2?z)z?2z?1z?13?lim? 2z?2z2z?2z1?e2z⑵f(z)?

z4?1?e2z12z?(2z)3122z223z3方法一：4?4{1?[1?2z??????]}?4[?2z??????]

2!3!2!3!zzz2 ??22412??????? 323z3zz1?e2z4故z=0是f(z)的三级极点，从而Res[4,0]??

3z方法二：

1?e2z4f(z)?，z?0是z的四级零点 4z2z2z2z又(1?e)|z?0?0 , (1?e)'|z?0??2e|z?0?0故z?0是f(z)的三级零点，根据规则

32z1d34 2z132z1Ⅱ可得Res[f(z),0]?limd3[z41?e????]?lim1?e?lim?2e??z?0dz3!z?03!3z43!z?0dz31⑶f(z)?cos

1?zw2w4因为cosw?1??????作代换运算w?1得

2!4!1?z

51

cos111?1?????? 241?z2!(1?z)4!(1?z)1?z因为级数中含无穷多个负幂项，所以z?1为cos1的本性奇点。 ⑷f(z)?1 zsinzz3z5z2z42?????)?z(1?????)故z?0是f(z)的二级极点 因为zsinz?z(z?3!5!3!5!sinz?0?z?k?,k??1,?2,???? 因为(zsinz)|z?k??0

(zsinz)'|z?k??(sinz?zcosz)|z?k??sink??k?cosk??k?(?1)k?0

所以z?k?,k??1,?2,???是f(z)的一级极点

Res[f(z),k?]?lim(z?k?)z?k?1z?k?罗必达(z?k?)'1 ?lim?lim?limz?k?z?k?z?k?zsinzzsinz(zsinz)'sinz?zcoszk??1? 11???sink??k?cosk?k?(?1)kk?Res[f(z),0]?1d1dzsinz?zcosz

lim[z2]?lim()?limz?0dzsinzz?0(2?1)!z?0dzzsinzsin2zsinz?zcos罗必达cosz?cosz?zsinzsinz ?lim?lim?lim?0 2z?0z?0z?02z2z3、.函数f(z)?ezz?12在C内有两个一级极点1和?1,所以由留数定理得

ez?c:z?2z2?1dz?2?i{Res[f(z),1]?Res[f(z),?1]}

ezeze而Res[f(z),1]?lim(z?1)2?lim?

z?1z?1z?1z?12ezez?e?1 Res[f(z),?1]?lim(z?1)2 ?lim?z??1z??1z?12z?1ezee?12?i(e2?1)故? dz?2?i[?]?c:z?2z2?1zz2e4. 函数f(z)?5z?2在C内有两个极点,它们是z?0为函数的一级极点,z?1为函数的二

2(z?1)2级极点。因为Res[f(z),0]?limzz?05z?2??2 2Z(1?z) 52

Res[f(z),1]?limd5z?2d?5z?2?5z?5z?22[(z?1)2?]?lim??lim?lim2?2所以由留数定理得?22z?1dzz?1dzz?1zZ(z?1)z?z?z?1?3c:z?25z?2dz?2?i(?2?2)?0

Z(z?1)2sin?z , cos?z?0 , ?z???k? z?1?k,(k?0,?1,?2,???)为函数仅cos?z22有的一级极点，由规则Ⅲ得

5. 因为tg?z?1sin?z1Res[tg?z1k?]?|1??,(k?0,?1,???) 于是由留数定理得

2(cos?z)'z?k?2??c:z?ntg?zdz?2?i1k??n2?12nRes[tg?z,k?]?2?i(?)??4ni

2?满足k?1?n的正整数k有n?1个，而负整数k有n个，另外k=0也满足k?21?n，故满2足k?1?n的k有n?1?n?1?2n个。 2 53

第二部分?tt?11.f(t)?{1 t2?1022积分变换课后习题一11F(?)?10??1(1?t2)e?i?tdt?10??1(1?t2)(cos?t?isin?t)dt10?2?(1?t)cos?tdt?2?cos?tdt?2?t2cos?tdt???22?2sin?t?2[sin??2[sin??210t2??sin?t10?2??211t0?sin?tdt]sin??2??tsin?tdt]0??0224cos?sin??sin??sin??[??]2???????{sin??2[?t?cos?t10?11cos?tdt]}?4?3(sin???cos?)f(t)的积分表达式为1??4??sin???cos?i?ti?tf(t)?F(?)ed??ed?3??????2?2??2??sin???cos???(cos?t?isin?t)d?3??????4???sin???cos?0?3cos?td?A2.(1)f(t)?{0其它0?t??F(?)??A(??????f(t)e?i?tdt???0Ae?i?tdt1A?i??)e?i?t???(e?1)0?i?i?A(1?e?iw?)i?t????t(2).f(t)?{sin0t??F(?)?????f(t)e?j?tdt?????sinte?j?tdt????sint(cos?t?jsin?t)dt??0??(?jsintsin?t)dt????2j?sintsin?tdt1[cos(1??)t?cos(1??)t]dt02sin(1??)tsin(1??)t???j[?]01??1??(1??)sin(1??)??(1??)sin(1??)t??j1??2(1??)(sin?cos???cos?sin??)?(1??)(sin?cos???cos?sin??)??j 54 1??2??2j??(1??)(sin??)?(1??)(sin??)1??22jsin????1??2sin?3.F(?)???j?????sin?11i?ti?tF(?)ed??ed???????2?2????sin?1?(cos?t?jsin?t)d?2???????sin?11??sin?s?t?cos?td??d?????02??????sin(1?t)??sin(1?t2)?1?d??02????sin(??sin(1?t)?1?t)?11?d??d???002??2???1??t?1???2?t??1?t?1?2?2?1?1??1?0t??1?0t?1?t?1???2???2??? ?4t??1t?1t?1?2?2?0????f(t)?

55

4.证：F(???0)??????f(t)e?i(???0)tdt???_?f(t)e?i?te??0tdt??????f(t)e??0te?i?tdt?F[e??0tf(t)]5.证：因F(?)?F[f(t)],要证dF(?)]??jtf(t)d?由于???时，F（?）?0即要证F?1[并用分布积分法可得F?1dF(?)?F[?jtf(t)]d??F（??1’?）2???i?tF'（?）edt?????1?[F(?)ei?t2????????jt??i?tF(?)ed?]?1?(?jt?F(?)ei?td?)2?????jtf(t)6.f?t?1?{0,其他f1(t)?f2?t??故f1?f2?0,(2)当0?t?sint,0?t??2,t?0，f2(t)?{0?te,t?0?????f1(?)f2(t??)d?(0????2t???0)

(1)当t?0,??0时t???0,f2(t??)?0?2时,?(t??)?t?sin?ed??esin?ed???00ttf1(t)*f2(t)?t由于?sin?ed??sin?e0??t0??cos?e?d?0t?sinte?[cos?e故,2t?t0??sin?ed?]?sinte?coste?1??sin?e?d?tt00t?t?sin?e0t?d??sintet?costet?1 1(sintet?costet?1)，从而 2?t0sin?e?d??f1?t??f2?t??（3）当t??11sintet?costet?1e?t?sint?cost?e?t 22?????2时，f1?t??f2?t???2sin?e0???t????d??e?t?20sin?e?d?，由于

?

20sin?e?d?56

=sin?e?2-

??0

?20cos?e?d?

?=e2?{cos?e?2???0?20sin?e?d?}

??=e?{?1?2?20sin?ed?}=e+1??2sin?e?d?

0??2??由此结果可知

?20?sin?ed?=1?1?e22??????1??,从而f1?t??f2?t???1?e2?2?????t?e ??综合以上结果可得

??0t?0?? ?1f1?t??f2?t????sint?cost?e?t?0?t?2?2??1????1?e2?e?tt???2?2???课后习题二

1、?1?f?t??sin解F?s????t 2?0??1??tt?1?t1?stsine?stdt?sin???e?st??cosedt

0022?s?22s =

1??t?st1t?1??st??1??t?st] [cosedt?cos?e?sinedt????2s022s02s02?s?212st?st?=1?1?11???sinedt?2?s2s?04s22??2s =

???0tsine?stdt2?Re?s??0?

1???t?st1 故 ?1?sinedt???2?22s2?4s?0 F?s?????0t?st14s22 sinedt?2??22s1?4s21?4s2?st 注：设s=x+iy,当Re(s)=x>0 时，e ?2?f?t??e解

?2t?e?xteiyt?0?t????

，

57

F?s?????e?2te?stdt????0e??s?2?tdt??1s?2e??s?2?t??10?s?2.,?Re?s?2??0? 0?3,0?t?2?3?f?t?????1,2?t?4，

??0,t?4解

F?s?????240f?t?e?stdt??03e?stdt??2e?stdt

??3se?st20?1se?st42?1s?3?4e?2s?e?4s?

?4?f?t??e2t?5??t?

解 由于L?e2t??1s?2,L???t????1,由拉氏变换的线性性质可得 L?f?t???L?e2t??5L???t???1s?2?5

2、?1?f?t??t2?3t?3，解 由拉氏变换的线性性质及幂级数的拉氏变换可得

L?f?t???L?t2??3L?t??3L?1??2s3?3s2?3s?1s3?3s2?3s?2? ?2?f?t??5sin2t?3cos2t 解 根琚L?sinkt??ks2?k2,L?coskt??ss2?k2 及拉氏变换的线性性质可得

L?f?t???5L?sin2t??3L?cos2t??103s10?3ss2?4?s2?4?s2?4 3、?1?F?s??1s2?a2,解由于L?1??a??s2?a2???sinat,故L?1??1?1?s2?a2???asinat?2?F?s??s2?2s?1s?s?1?2 解法一：用赫维赛德展开式法

s?s?1?2?0,s?0为s?s?1?2的一级零点，s?1为s?s?1?2的二级零点 s2?2s?1s?0??1?0,s2?2s?1s?0??1?0,

故s?0,s?1,分别为F?s?一级和二级极点。根据情况二下的赫维赛德展开式可得

f?t??s2?2s?1est?s?1?2?2s2s?0?1?2?1?!limd?s?1ds???s?1?2s2?2s?1s?s?1?2est?? ? 58

而

222=?1?limd?s?2s?1est???1?lim??2s?2?s??s?2s?1?est?s?2s?1test?

????2s?1ds?s?s?1?ss???1?2et?2tet

解法二：用化分式为部分式法：设

s2?2s?1s?s?1?2ABCA?s?1??Bs?s?1??Cs?A?B?s2???2A?B?C?s?A ?????22ss?1?s?1?2s?s?1?s?s?1?2比较两端分子中s的同次幂的系数可得方程组

A?B?1?? 解得A= -1, B=2 C=2,故 ??2A?2B?C?2?A??1?2 s?2s?1??1?2?2,由于L?1??1,Let?1

2ss?1?s?1?2ss?1s?s?1???由赫维赛德展开式可得

?1??d?1dst2st??L?1??lims?1e?lime?limtest?tet ??2?2s?1dss?1?s?1??s?1ds??s?1????s2?2s?1?tt故 L? ??1?2e?2te?2?s?s?1???1 注：在化分式为部分式法中，如果遇到非单极点，也要用到赫维赛德展开式法 ?3?F?s??s?1?2

s?2s?2?1??2tL?1?F?s???L?1?1??2L?1?????t??2e

?s?2??4?F?s???1111 ??s2s2?1s2s2?1??解 由于L?1??1??1??t,L????sint 22?s??s?1?根据卷积定理可知 L?t?sint??11?，由拉氏积分定义知 s2s2?1tttt?sint??sin??t???d???t?????cos????cos?d??t?sint

000??1故 L?1???t?sint?t?sint 22ss?1????4.⑴t?t

tt11111t?t???(t??)d???(t???2)d??t?2??3|t0?t3?t3?t3

0023236 59

⑵．sint?cost

解：由拉氏卷积定义可得

sint?cost??sintcos(t??)d??0t1t1tsintcos(t??)d??[sint?sin(2??t)]d? ??0022 ?t1111tsint?cos2(??t)|t0?tsint?0?tsint 2422⑶．t?e

解法一：由卷积定义知

ttt?et???et??d???(?et??)|t0??et??d???t?et??|t0??t?(1?et)?et?t?1

00解法二：由于L(t)?111ttL(t?e)? 由卷积定理知 L(e)?22s(s?1)s?1s

由情况二下的赫维赛法展开式得

t?et?lim[(s?1)s?111d?2est?estdestst??e]?lims?lim?lim2?s?1s2s?0ds(s?1)(2?1)!s?0ds?s2(s?1)s(s?1)??test(s?1)?est?t?1tt?e??e?t?1 =e?lim22s?0(s?1)(?1)t5.⑴y\?4y'?3y?e?t y(0)?y'(0)?1

解：记Y(s)?L[y(t)]根据拉氏变换的线性性质，微分性质及初始条件在方程两边取拉氏变换得：

s2Y(s)?s?1?4sY(s)?4?3Y(s)?(s2?4s?3)Y(s)?Y(s)?1 s?11?s?5 s?11s?5 ?2(s?1)(s?3)(s?1)(s?3)根据赫维赛法展开式得

1estdest2L[]?lim[(s?3)]?lim[(s?1)]s??1ds(s?1)2(s?3)s??3(s?1)2(s?3)(s?1)2(s?3)?1test(s?3)?est1?3t2te?t?e?t1?3tdest1?3t?e?lim()?e?lim?e?s??1dss?3s??14444(s?3)21?3t2te?t1?t?e??e 444A(s?3)?B(s?1)As?3A?Bs?B(A?B)s?3A?B s?5AB?????(s?1)(s?3)s?1s?3(s?1)(s?3)(s?1)(s?3)(s?1)(s?3) 60

{A?B?1(1)由(1)A?1?B代入(2)3A?B?5 (2) 3(1?B)?B?5

3?3B?5 , 3?2B?5,2B??2,B??1 A?1?B?1?(?1)?2

s?521s?5 L?1[??]?2e?t?e?3t (s?1)(s?3)s?1s?3(s?1)(s?3)故微分方程的解： y(t)?L?1[1s?5?1]?L[] 2(s?1)(s?3)(s?1)(s?3)1?3t2te?t1?t1?e??e?2e?t?e?3t?[(7?2t)e?t?3e?3t]4444?2?y????3y???3y??3?1,?解：记y(s)?L[y(t)],y?0??y??0??y???0??0根据拉氏变换的线性性质，微分性质及初始条件，在方程两边取拉氏变换得111s3y(s)?3s2y(s)?3sy(s)?y(s)?,(s3?3s2?3s?1)y(s)?,(s?1)3y(s)?,sss1y(s)?s?0,s??1分别是y(s)的一级和3级极点，所以微分方程的解3s(s?1)est1d2est3y(l)?L[y(s)]?lim?lim2[(s?1)]3s?0(s?1)3s??1(3?1)!dss(s?1)?11d2est1dtests?est?1?lim2()?1?lim()2s??1s??122dsdsss?1??1??1??1??1?1d(ts?1)estlim[]2s??12dss1[test?t(ts?1)est]s2?(ts?1)est2slim2s??1s41 [te?t?t(?t?1)e?t?2(?t?1)e?t]21[te?t?(t2?t)e?t?(2t?2)e?t]21[?t2e?t?(2t?2)e?t]2t2?t?1?[?e?te?t?e?t]

2t2?1?[?t?1]e?t2

61

?x??x?y?et(3)?,,x(0)?y(0)?1t?y?3x?2y?2e?解记x(s)?L[x(t)],y(s)?L[y(t)]两个方程两边取拉氏变换得1?sx(s)?1?x(s)?y(s)??s?1,?2?sy(s)?1?3x(s)?2y(s)?s?1?1?(s?1)x(s)?y(s)?1?,(1)?s?1?2?(s?2)y(s)?3x(s)?1?,(2)s?1?1由（1）得y(s)?(s?1)x(s)?1?(3)s?1(3)代入（2）得12(s?2)[(s?1)x(s)?1?]?3x(s)?1?s?1s?1s?22[(s?2)(s?1)?3]x(s)?(s?2)??1?s?1s?12s?2s(s2?s?1)x(s)??1??s?2??s?1s?1s?1s?1s?(s?1)22(s?s?1)x(s)?s?1s?s2?2s?12(s?s?1)x(s)?s?1s2?s?12(s?s?1)x(s)?s?11x(s)?s?1x(t)?:L?1[x(s)]?et代入（3）得s?1sy(s)??s?1s?11y(s)?s?1y(t)?L?1[y(s)]?et故微分方程组的解为：x(t)?et,y(t)?et6.解：设电阻R和电容C的电流分别为i1(t)和i2(t) 62

由电学原理知u(t)i1(t)?Rdui2(t)?cdt由电学上的基尔霍夫定律（在一节点A处的电流的代数和为0）d(t)U(t)有c???(t),U(0)?0dtR这就是该电路的电压U(t)应满足的微分方程记U(s)?L[u(t)].并对微分方程两边取拉氏变换U(s)得，CSU(s)??1R111所以U(s)??1Cs?1?CSRRC取拉氏逆变换有1?RCtu(t)??[U(s)]?eC其物理意义是：由于在一瞬间电路受单位脉冲电流的作用，1把电容的电压由零跃变到。然后由电容C向电阻R按指数C衰减规律放电。?117.解：设电路中电流强度为i?t?，由电学原理知电阻两端电压di(t)uR(t)?i(t)R，电感两端电压uc(t)?L。由电学中基尔霍dt夫定律（一个回路中的电压的代数和为零）得：?dt?t??L?Ri?t??E?dt?i?0??0?在微分方程两边取拉氏变换得ELSI(s)?RI(s)?SE(LS?R)I(s)?SEELI(s)??RS(LS?R)S(S?)LELi(t)?L?1[I(s)]?L?1[]RS(S?)LERA(S?)?BSABLL令???RRRSS(S?)S?S(S?)LLL

63

AS?A?RR?BS(A?B)S?ALL?RR

S(S?)S(S?)LL??A?B?0A??B?AR?EEELLA?,B???RR?EE?R?RRRSS(S?)S?LL故微分方程的解i(t)?

EL

EE?eRRR?tL?E(1?eRR?tL

) 64

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