复变函数论习题集解答(5)

?12?i12?[?f(?)|?|?r??zd????c0||?r??zd?]|

???r|f(?)?c0||??z|||?r|d?|

??r?|z|12?i??

故limr?????f(?)||?r??zd??c0

①z?D，则?12?if(?)|?|?r??zd???cf(?)???zd??0

故

?cf(?)???zd??0?c0

②z?D，亦可选取充分大的r0?r,使得

?2?ic从而

1f(?)???zd???2?i?1f(?)r0??z||?d??f(z)

?2?ic??z?1f(?)d??f(z)?c0

第六章

一．1．C 2．B 3．D 4．C 5．A 6．D 7．D 8．A 9．C 10．B 11．B 12．A 13．D 14．C 15．D

二．１．ＤＥ ２．ＢＣ ３．ＡＢ ４．ＡＢＣＥ ５．ＡＢＤＥ 三．1．

?(n?1)(a)(n?1)!,?(z)?(z?a)f(z)

n2．

?(z)?'(a)1,?(a)?0,?(a)?0,?'(a)?0

3．

2?i?4．??i

n?f(z)dz,?|z|???r,f(z)在0?r?|z|???

5．?Resf(z),Resf(z)

k?1z?az??6．（1）(n?m)?2，（2）在实轴上Q(z)?0，2?i7．limg(z)?0，0，m?0

R????Imak?0Resf(z)

z?ak8．高，在实轴上Q(z)?0，m?0，2?i9．limf(z)??，i(?2??1)?

R????Imak?0Res[g(z)ez?akimz]

10．lim(z?a)f(z)??，i(?2??1)?

r?0四．1．解：?z?1?0，则 zk?ei2k?1mm?，k?0,1,...,m?1 z2mm故f(z)?1?z的孤立奇点为zk(k?0,1,...,m?1)及? (z?zk)z1?zz2m2mResf(z)?limz?zkz?zk?lim1mzm?1z?zk?1mzm?1k?zkmzmk??zkm

当m?1时，f(z)?Resf(z)?1，

z?11?z，此时f(z)的孤立奇点为z??1及?

?Resf(z)??1，

z??m?1当m?1时，Resf(z)???(?z??zkm)?k?0?m1m?1zk?1mei?m2?i??mm1?(e)????k?01?ei2?m?0

2．解：f(z)?e2z4z(z??i)的孤立奇点为z?0,z??i,?，

Resf(z)?z?0??4i??16?55，Resf(z)?z??i2(???18)6?224?i(4??)2?5

Resf(z)?z????18?6??????6?i? ?３．解：?|a|?1，|b|?1且a?b

dz(z?a)(z?a)??n?1??|z|?1nn?2?i[Resf(z)?Resf(z)]

z?az?b?2?i?(?1)n(n?1)...(2n?2)(n?1)!dziz(1(b?a)2n?1??)?0 2n?1?(a?b)?11z)

４．解：设z?ei?，则d??,cos???122(z??2?0d?a?cos???dziz22a?z?2|z|?11z??2idzz?2az?1|z|?1

?a?1，故奇点为z0?a?1?a

2?a?12?2?0d?a?cos??4??Resf(z)??z?z0

５．解：设g(z)?ix1(z?2)(z?9)dx?2?i2222，则在实轴上(z?2)(z?9)?0

于是?????e?x2?1??x?9?2?Imak?0Res[g(z)e]

z?akiziz??e?1e?3??1???1e?2?i??(?) ?2?i?Res?Res?2????32z?iz?3i82e3e16i48i(z?2)(z?9)??????????????2cos?xdx?x?1??x?9?2??82e(1?13e3)

６．解：设f(z)?z222(z?1)(z?4)，则

?????f(x)dx?2?i?Imak?0Resf(z)?2?iResf(z)?Resf(z)

z?ak?z?iz?2i??2?i1???1 ???6i3i??3?????x0??x222?1??x?4?eizdx?1??x2????x222?1??x?4?dx??6

７．设f(z)?z?z?1?2，选取r?0充分小，R?0充分大

C?:|z|?r的上半部，CR:|z|?R的上半部，从而

?Rrf(x)dx??CRf(z)dz???r?Rf(x)dx??Cr?f(z)dz?2?iResf(z)

z?i由于在CR上lim因此?于是?1z(z?1)2R????0在C?上limzf(z)?1

r?0CRf(z)dz?0，?Cr?f(z)dz???i

Rrf(x)dx???r?Rf(x)dx??i?2?i?????1?2ie

令r?0，R?0，即得???02?4?? f(x)dx??i?1???ee??故?sin?x?x?1?2dx?12?????sin?x?x?1?2dx???2?1??? 2?e?五．１．证明：???1?在|z|?1上

|e|?ez|z|?e?|?e|?|z|?|?ez|

?n???n因而根据儒歇定理知?ez与e?ez在|z|?1内有相同个数的根， 故e?ez在|z|?1内有n个根 2．证明： 12?izz?n?n?zecnz?nn!??d???znn!??e?z??n?n!?z?|??0???

n!??n23．解：?1?在|z|?1内，10?1?8，故在|z|?1内z4?8z?10?0无根。

?2?在1?|z|?3内，由于在|z|?3上|10?8z|?10?8|z|?34?|z4|

故方程在|z|?3内有4个根，再结合?1?的讨论知z4?8z?10?0在1?|z|?3内有4个根 4．解：?z?c,?|z|?1

?1?|z|?1时，当z?0时?d?c?2?0当z?0时

?????z?cd???1??1?1?1?11?Res?Res???0????????0 ??z2?i2?i?????z?????z????z?z??2?当|z|?1时，

?????z?cd??12?i??1z?i2?z

5．证明：设f(z)?zlnz?1?z?c2，c为如图的围线，则支点0,?均不在C内，选在x轴上方取

2??i2i正值的那个分支，则?f?z?dz?2?iResf(z)?2?iz??1?2???i

2但?f?z?dz???c??cR???l2????crl1?f(z)dz ???0

在CR上，limzf?z??limR???zzlnz2R???(z?1)z在Cr上，limzf?z??limr???zlnz2r???(z?1)?0

在l1上，z?x

?l1f?z?dz=?lf1?x?dx=?rf(x)dx

，dz?exe?i2?iR在l2上，z?xe2?idx

?l2f?z?dz??Rr?lnx?2?i?2(1?x)dx=?Rr?xlnx2(1?x)dx??Rr??2?xi2?(1?x)dx

让r?0,R?0得

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