复变函数习题答案第2章习题详解 下载本文

10. 证明:如果函数f?z??u?iv在区域D内解析,并满足下列条件之一,那末f?z?是常数。 1) f?z?恒取实值;

证明:f?z?恒取实值,即v?x,y??0。

?f?z??u?iv是解析函数,所以

?u?v?u?v??0,???0 ?x?y?y?x??u?u??0 即u?x,y?为常数,故f?z?是常数。 ?x?y2) f?z?在D内解析;

证明:因为f?z??u?iv在区域D内解析,所以

?u?v?u?v???, ?x?y?y?x?u???v??u???v????, ?x?y?y?x 又为f?z??u?iv在区域D内解析,所以

??u?u?v?v????0,故f?z?是常数。 ?x?y?x?y3) f?z?在D内是一个常数;

?v?v??u??u2u?2v?0u?v?0???x?x?x?x??22证明:设u?v?c ? ? ? ?

?u?v?u?v?2u?u?2v?0?v?0???y?y?y?y???u?x ??u?y ??v22?u?v?u?v?u?v?????x?0 同时

???, 成立。所以??????0 ?v?x?y?y?x??x???x??y?u?u?v?v?u?v????0 即u,v均为常数,故f?z?是常数。 ??0 ?

?x?y?x?y?x?x4) argf?z?在D内是一个常数; 证明:设??argf?z?,则??????。 ○1如果????2,则u?0,从而

?u?u?u?u?v?v??0,????0,又f?z?在D内解析, ?x?y?x?y?x?y所以v为常数,故f?z?是常数。

?v??u?v?u?0???v?x??x ○2如果????,则??arctg,于是有?

?u?v22u??v?u?0??y?y??u?x ??u?y ??v22?u?v?u?v?u?v?????x?0 同时

???, 成立。所以??????0 ?v?x?y?y?x?x????x??y?u?u?v?v?u?v????0 即u,v均为常数,故f?z?是常数。 ??0 ?

?x?y?x?y?x?x ○3如果

?2????,则????arctgv; uv,与○2的讨论一样,可得到f?z?是常数。 u25) au?bv?c,其中a,b与c为不全为零的实常数。

如果???????,则?????arctg证明:因为au?bv?c,且a,b与c为不全为零,所以a和b不能同时为零。

假设a?0,则有u?1?c?bv?,于是?u??b?v,?u??b?v

a?ya?xa?x?y?u?v?u?v?u?u?v?v???????0, ,,??x?y?y?x?x?y?x?yf?z??u?iv在区域D内解析,

所以v为常数,故f?z?是常数。 11. 下列关系是否正确? 1) e?e

解:设z?x?iy,则e?e2) cosz?cosz 解:cosz?zx?iyzz?exeiy?exe?iy?ex?iy?ez

1iz11?e?e?iz??eiz?e?iz??e?iz?eiz??cosz 222??3) sinz?sinz 解:sinz?1iz1iz1?e?e?iz???e?e?iz???e?iz?eiz??sinz 2i2i2i??12. 找出下列方程的全部解:

1) sinz?0

解:?sinz?0,?e?eiz?iz?0 ? e2iz?1,即z?n??n?0,?1,?2,?3??

2) cosz?0

解:?cosz?0,?e?e3) 1?e?0

解:?1?e?0,?e??1,即z??2n?1??i?n?0,?1,?2,?3??

zziz?iz?0 ? e2iz??1,即z??2?n??n?0,?1,?2,?3??

z4) sinz?cosz?0 解:?sinz?cosz?0,?即z?1iz1?e?e?iz???eiz?e?iz??0 ? e2iz??i, 2i2?2?n??n?0,?1,?2,?3??

13. 证明:

1) cos?z1?z2??cosz1cosz2?sinz1sinz2,sin?z1?z2??sinz1cosz2?cosz1sinz2 证明:cosz1cosz2?sinz1sinz2?1iz11iz21iz11iz2e?e?iz1e?e?iz2?e?e?iz1e?e?iz2 222i2i11?ei?z1?z2??ei?z1?z2??e?i?z1?z2??e?i?z1?z2??ei?z1?z2??ei?z1?z2??e?i?z1?z2??e?i?z1?z2? 441?ei?z1?z2??e?i?z1?z2??cos?z1?z2? 21iz11iz211iz2sinz1cosz2?cosz1sinz2?e?e?iz1e?e?iz2?eiz1?e?iz1e?e?iz2

2i222i1i?z1?z2?1i?z1?z2??e?ei?z1?z2??e?i?z1?z2??e?i?z1?z2??e?ei?z1?z2??e?i?z1?z2??e?i?z1?z2? 4i4i1i?z1?z2??e?e?i?z1?z2??sin?z1?z2? 2i????????????????????????????2) sinz?cosz?1

22?1??1?证明:sin2z?cos2z???eiz?e?iz?????eiz?e?iz??

?2i??2?1i2z1?e?2?e?i2z???ei2z?2?e?i2z??1 443) sin2z?2sinzcosz

??证明:?sin?z1?z2??sinz1cosz2?cosz1sinz2 令z?z1?z2,则sin2z?2sinzcosz 4) tg2z?222tgz 21?tgz证明:?cos?z1?z2??cosz1cosz2?sinz1sinz2,sin?z1?z2??sinz1cosz2?cosz1sinz2 令z?z1?z2,则cos2z?cosz?sinz,sin2z?2sinzcosz

222sinzsin2z2sinzcoszcosz?2tgz

?tg2z???cos2zcos2z?sin2z1?sin2z1?tg2zcos2z5) sin?????z??cosz,cos?z?????cosz ?2??????????z??i??z???i?1?iz1?i2?iz???1?i?iz?2??2?2证明:sin??z???e?e???ee?ee??e?eiz?cosz ?2?2i???2??2i???cos?z????coszcos??sinzsin???cosz

6) cosz2?cos2x?sh2y,sinz?sin2x?sh2y

22证明:cosz?1??1?1?coszcosz?coszcosz???eiz?e?iz????eiz?e?iz????eiz?e?iz?eiz?e?iz

?2??2?4?? 令z?x?iy,则z?x?iy

1?iz?iz1ee?eize?iz?e?izeiz?eizeiz?e?iz?iz?eiz?iz?e?iziz?eiziz 441i2x11 ??e?e?2y?e2y?e?i2x???ei2x?e?i2x?2???e?2y?e2y?2?

444 ??????1??1????eix?e?ix?????e?y?ey???cos2x?sh2y ?2??2?同理可证:sinz222?sin2x?sh2y

2?1??1?sinz?sinzsinz?sinzsinz???eiz?e?iz???eiz?e?iz?

?2i??2i???1iziz1ee?e?izeiz?eize?iz?e?ize?iz??eiz?iz?e?iz?iz?eiz?iz?e?iz?iz 44111???ei2x?e2y?e?2y?e?i2x??2?ei2x?2?e?i2x???e2y?2?e?2y?

444i???????1??1????eix?e?ix?????ey?e?y???sin2x?sh2y ?2i??2?14. 说明:

1) 当y??时,sin?x?iy?和cos?x?iy?趋于无穷大; 解:?sin?x?iy??22sin2x?sh2y,而limsh2y???,?limsin?x?iy????

y??y?? 同理:?limcos?x?iy????

y??2) 当t为复数时,sint?1和cost?1不成立。

解:由于t为复数,可设t?iy?y?0?,则cost?cosiy?1?sh2y?1

ey?e?y????y???? sint?siniy?shy?2 故当t为复数时,sint?1和cost?1不成立。 15. 求Ln??i?,Ln??3?4i?和它们的主值。

解:Ln??i??ln?i?iArg??i??ln1?i?arg??i??2n???i?? 主值为ln??i???????2n?? n?0,?1,?2,? ?2??2i

??4??2n?? n?0,?1,?2,? 3? Ln??3?4i??ln?3?4i?iArg??3?4i??ln5?i???arctg 主值为ln??3?4i??ln5?i???arctg16. 证明对数的下列性质: 1) Ln?z1z2??Lnz1?Lnz2

??4?? 3?证明:Ln?z1z2??lnz1z2?iArg?z1z2??lnz1z2?iArg?z1z2??lnz1?lnz2?iArg?z1??iArg?z2? Lnz1?Lnz2?lnz1?iArgz1?lnz2?iArgz2 所以:Ln?z1z2??Lnz1?Lnz2 2) Ln???z1???Lnz1?Lnz2 ??z2??z1??z1?z1???ln?iArg??z???lnz1?lnz2?iArg?z1??iArg?z2??Lnz1?Lnz2 zz2?2??2??z1???Lnz1?Lnz2 ??z2?证明:Ln?? 所以:Ln??17. 说明下列等式是否正确: 1) Lnz?2Lnz 解:设z?rei?

2Lnz2?Ln?rei???Ln?r2ei2???lnr2?i?2??2n???2lnr?i?2??2n?? n?0,?1,?2,?

2 2Lnz?2Lnre?i???2lnr?i2???2m???2lnr?i?2??4m?? m?0,?1,?2,?