20. 若z=x+iy，求证
(1) sinz=sinxchy+icosx?shy 证明：
eiz?e?izei?x?iy??e??x?yi??isinz??2i2i1?.?e?y?xi?ey?xi?2i?sinx?chy?icosx.shy (2)cosz=cosx?chy-isinx?shy
当y→+∞时，e-y→0，ey→+∞有|sinz|→∞． 当y→-∞时，e-y→+∞，ey→0有|sinz|→∞．
11cos?x?iy??e?y?xi?ey?xi≥?e?y?ey?22同理得 所以当y→∞时有|cosz|→∞． 习题三
1. 计算积分C的直线段.
解 设直线段的方程为y?x,则z?x?ix.
?(x?y?ix2)dz,其中C为从原点到点1+i
证明：
eiz?e?iz1?i?x?yi??i?x?yi??cosz???e?e221??e?y?xi?ey?xi?21??e?y??cosx?isinx??ey.?cosx?isinx??2?ey?e?y?e?y?ey??.cosx??isinx.?2?2??cosx.chy?isinx.shy (3)|sinz|2=sin2x+sh2y 证明： sinz?1??y?xie?ey?xi??sinx?chy?icosx?shy2i
0?x?1
故
22x?y?ixdz?x?y?ix????d(x?ix)??0101C11??ix2(1?i)dx?i(1?i)?x303ii?1?(1?i)?33
2. 计算积分C?(1?z)dz，其中积分路径C为
(1) 从点0到点1+i的直线段;
(2) 沿抛物线y=x2，从点0到点1+i的弧段. 解 (1)设z?x?ix. 0?x?1
sinz?sin2xch2y?cos2x.sh2y?sin2x?ch2y?sh2y???cos2x?sin2x?sh2y?sinx?shy(4)|cosz|2=cos2x+sh2y
证明：cosz?cosxchy?isinxshy
222
C??1?z?dz???1?x??ix(d01?x)ix?
i2z?x?ix(2)设. 0?x?1
cosz?cos2x.ch2y?sin2x.sh2y?cos2x?ch2y?sh2y???cos2x?sin2x?.sh2y?cos2x?sh2y
21. 证明当y→∞时，|sin(x+iy)|和|cos(x+iy)|都趋于无穷大． 证明： sinz?1?iz?iz?1??y?xie?e??e?ey?xi?2i2i
2C221?zdz?1?x?ixd(x?ix)???????012i3
3. 计算积分C?zdz，其中积分路径C为
(1) 从点-i到点i的直线段;
(2) 沿单位圆周|z|=1的左半圆周，从点-i到点i; (3) 沿单位圆周|z|=1的右半圆周，从点-i到点i. 解 (1)设z?iy. ?1?y?1
1sinz??e?y?xi?ey?xi2?y?xi?yy?xiye?ee?e∴ sinz≥C?zdz??ydiy?i?ydy?i?1?111
而
1?y?xi?e?ey?xi??1?e?y?ey?22
3??i?(2)设z?e. ?从2到2
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??i?C?i?2zdz??32?1de?i?3?de?2i22（4）在C4所围的区域内包含两个奇点z?0,z?i,
故
11111dz?(?????Cz(z2?1)??C4z2z?i2?z?i)dz?2?i??i??i
13??i?(3) 设z?e. ?从2到2
?10.利用牛顿-莱布尼兹公式计算下列积分. (1) ???2i0C?zdz??23?21de?2i
zCi?zcosdz2
e (2) ???i10?zdz(2?iz)dz (3) ?
1ii2z?e???6. 计算积分
?sinz?dzCln(z?1)dz?(4) 1z?1 (5)
i1?tanzdz?0z?sinzdz (6) ?1cos2z
,其中C为
Cz?a?0.
解 (1)
??z?e解 ?Czz?sinz?dz??zdz?e????sinzdz???2i0z1zcosdz?sin222??2i0?2ch1
∵e?sinz在
zz?a所围的区域内解析
2?(2)
?∴?CCez?sinzdz?0??e?i10?zdz??e?z20??ii??22
113i1从而
???z?e2?0zi??sinz?dz??zdz?adae??C0(2?iz)dz??(2?iz)d(2?iz)??(2?iz)?ii3(3)
1i1??11i?33
(4)
iln(z?1)121?2idz?ln(z?1)dln(z?1)?ln(z?1)??(?3ln22)1?1z?1?1284 i?a2i?ei?d??0z?e???故
Cz?sinz?dz?01z(z?1)2(5)
?为
1210z?sinzdz???zdcosz??zcosz10??coszdz?sin1?cos10011
7. 计算积分（1）（4）
C1:z???12Cdz,其中积分路径C32(6)
ii1?tanz122idz?seczdz?secztanzdz?tanz?tan2z1?1cos2z?1?12112?2????tan1?tan1?th1??ith1?22?ii1 （2）
32
C2:z? （3）
C3:z?i? C4:z?i?1解：（1）在奇点z?0.
z?122z(z?1)只有一个所围的区域内，
11. 计算积分(1)
z?i?1??Cezdzz2?1,其中C为
z?i?1 (2) (3)
z?2
解 (1)
??C11111dz??(????)dz?2?i?0?0?2?i?C1zz(z?1)2z?i2z?i21ezezez??Cz2?1dz???C(z?i)(z?i)dz?2?i?z?i(2)
ezezez??Cz2?1dz???C(z?i)(z?i)dz?2?i?z?iz?i??ei
（2）在所围的区域内包含三个奇点z?0,z??i.
C2故
??11111dz??(????)dz?2?i??i??i?0?Cz(z?1)C2z2z?i2z?i21z??i???e?i
(3)
ezezezi?i??Cz2?1dz???C1z2?1dz???C2z2?1dz??e??e?2?isin1
（3）在C2所围的区域内包含一个奇点z??i,故
11111dz?(?????Cz(z2?1)??C3z2z?i2?z?i)dz?0?0??i???i1 12 / 37
(1)??x3?6x2y?3xy2?2y3;(2)??excosy?1?i(exsiny?1).
解(1) 设∴
16. 求下列积分的值,其中积分路径C均为|z|=1.
ez??Cz5dz
w?u?i?,u?x?6xy?3xy?2y3223 ??0
?u?u22??6x2?6xy?6y2?3x?12xy?3y?x ?y
(1) (2)
??zcosz2dz,z?1dz02??CCz32 (3) (z?z0)tan?2u?2u??6x?12y?6x?12y22?x ?y
从而有
?2u?2u??0?x2?y2,w满足拉普拉斯方程,从而是调和函数.
解 (1)
ez2?iz(4)dz?(e)??Cz54!(2)
z?0??i12
cosz2?i(2)dz?(cosz)3??Cz2!(3)
(2)
z?0w?u?i?,
u?ex?cosy?1
???i
??ex?siny?1
?u?u??ex?siny?ex?cosy∴?x ?y
2?u?2uxx??e?cosy?e?cosy22?x ?y
z2dz?2?i(tanz)'??C(z?z0)2tanz?z0??isec2z02
1??C(z?1)3(z?1)3dz17. 计算积分,其中积分路径C为 (1)中心位于点z?1,半径为R?2的正向圆周 (2) 中心位于点z??1,半径为R?2的正向圆周
从而有
?2u?2u??0?x2?y2,u满足拉普拉斯方程,从而是调和函
数.
?????ex?cosy?ex?siny?y?x
解：(1)
C?2??2?xx??siny?e?e?siny2?y?x2
3?i8
?2??2??2?02?x?y,?满足拉普拉斯方程,从而是调和函
数.
22u?x?y20.证明:函数,
内包含了奇点z?1
z?1?12?i1dz?()(2)3??C(z?1)3(z?1)32!(z?1)∴
(2)
C内包含了奇点z??1,
12?i1(2)dz?()33??C2!(z?1)3∴(z?1)(z?1)19. 验证下列函数为调和函数.
3?iz??1??8
??xx2?y2都是调和函
数,但f(z)?u?i?不是解析函数 证明:
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2?2u?u?u?u??2y??2?2x?222?y?y?x ?x
????2u?2u?2?02?x?y∴,从而u是调和函数. ???2xy??y2?x2??2?x(x?y2)2 ?y(x2?y2)2
?2??6xy2?2x3?2?6xy2?2x3??2223223?x(x?y) ?y(x2?y)
2xx?u?u2yydx?dy)?C??4dx?x?02dy?C(1,0)1x?y?x(x?y2)21xxy??1?2?2?1?C20xx?yx?y2(x,y)(?f(z)?yx?i(?1?C)x2?y2x2?y2
由f(1)?0.，得C=0
1??f?z??i???1??z?
23.设
?2??2??2?02?y∴?x,从而?是调和函数.
?u???u??????x ?y ?y但∵?x∴不满足C-R方程,从而f(z)?u?i?不是解析函数.
22.由下列各已知调和函数,求解析函数
p(z)?(z?a1)(z?a2)?(z?an),其中
ai(?i1,?2,n各不相同，闭路C不通过a1,a2,?,an,证明积分
1p?(z)dz??2πiCp(z)
f(z)?u?i?
(1)u?x?y?xy (2)
22等于位于C内的p(z)的零点的个数.
证明: 不妨设闭路C内P(z)的零点的个数为k, 其零点分别为
u?y,f(1)?022x?y
a1,a2,...ak
?(z?a)?(z?a)?(z?a)?...(z?a)...(z?ak1k1k?2k?3nnn?1?u???u????2y?x???2x?y??x ?y ?y解 (1)因为 ?x所以
(x,y)???(0,0)?221P?(z)1dz????CC2πiP(z)2πi??)dz(z?a1)(z?a2)...(z?an)?u?u(x,y)x?xdx?y(2x?y)dy?Cdx?dy?C??(0,0)(2y?x)dx?(2x?y)dy?C??0?0?y?x111111dz?dz?...?dz??????CCC2πiz?a12πiz?a22πiz?ank个??xy??2xy?C22?1??1???...?1????k1111dz?...?dz????CC2πiz?ak?12πiz?anx2y222f(z)?x?y?xy?i(???2xy?C)22
令y=0,上式变为
24.试证明下述定理(无界区域的柯西积分公式): 设
f(z)在闭路C及其外部区域D内解析，且
limf(z)?A??z??，则
x2f(x)?x?i(?C)2
2??f(z)?A,z?D,1f(?)???zd???A,z?G.2πiC?
其中G为C所围内部区域.
证明：在D内任取一点Z，并取充分大的R，作圆CR:
从而
z2f(z)?z?i??iC2
2?u2xy?ux2?y2??2?22222?x(x?y)?y(x?y) (2)
用线积分法，取（x0,y0）为(1,0)，有
z?R，将C与Z包含在内
CR为边界的区域内解析，依柯西积
则f(z)在以C及分公式，有
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1f(?)f(?)f(z)?[?d?-d?]???CCR2πi??z??z f(??z)??R因为??z 在上解析，且
???1?i2n?1?1?(?1)n?i?1(?1)n??????i ?nnnnn?1n?1n?1??11?i2n?1因为?发散,所以?发散
nn?1nn?1?lim??f(?)1?limf(?)??limf(?)?1z?????z???1??
?1?5i26n?()发散 (2)??22n?1n?1?n所以，当Z在C外部时，有
f(z)?A?1f(?)?C??zd?2πi?
1?5in15nlim()?lim(?i)?0 又因为n??n??2221f(?)??C??zd???f(z)?A2πi即
设Z在C内，则f(z)=0，即
1?5i()发散 所以?2n?1(3)
?n?n?1??e1??nn?1nπin发散,又因为
0?1f(?)f(?)[?d????C??zd?]2πi?CR??z
1f(?)??C??zd??A2πi故有：
习题四
1. 复级数?an与?bn都发散,则级数?(an?bn)n?1n?1n?1?e???nn?1n?1?iπncosππ?isin?1ππnn?(cos?isin)收?nnnnn?1敛,所以不绝对收敛. (4)
??n?1????in1?? lnnn?1lnn和
?ab发散.这个命题是否成立?为什么?
?nnn?1答.不一定．反例: ???11?11an???i2,?bn????i2发散 ?nn?1nnn?1n?1nn?1但?(an?bn)??i?n?1n?1????11?因为
lnnn?1
所以级数不绝对收敛. 又因为当n=2k时, 级数化为
(?1)kln2k2收敛 2n?k?1?收敛
2(an?bn)??发散 ?n?1n?1n??11anbn??[?(2?4)]收敛. ?nnn?1n?12.下列复数项级数是否收敛,是绝对收敛还是条件
收敛?
??n1?i2n?11?5in) (3) ?e (1)? (2)?(n2nn?1n?1n?1k?当n=2k+1时, 级数化为?(?1)也收敛
k?1ln(2k?1)所以原级数条件收敛 (5)
?iπcosin?1en?e?n1?en1?1n??n???()??() ?n2222n?022n?02en?0n?0??incosin?(4) (5) ?n
2n?1lnnn?0?1ne其中?()n 发散,?()收敛
n?02en?02??所以原级数发散.
(1)
?3.证明:若Re(an)?0,且?an和?an2收敛,则级数
?解
n?1n?1?an?1?2n绝对收敛.
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