z2?213 (2) f?z??2?(1???)
z?z?3?zz Res(f,?)??C?1??1
??z?4f?z?dz?2?i??Res(f,?)??2?i
i?dzz2?17. 解: 设z?e 则d??, sin??
iz2izd?2dz2??dz ?05?3sin???z?13z2?10iz?3??z?1i3(z?3i)(z?)3i2 令f?z??,则f在z?1内只有一级权点, z??,依离数定理有
i33(z?3i)(z?)3
2?
?2?0d???2?iRes?f??z?,5?3si?n?i????2??i??3?i????
2?48. 解: (1) ?1?i?z?1 即 z?11. 故R?
22(n!)2 (2) Cn?
nn R?limn??cn1?1??lim?1????0 cn?1n???n?n?1n9.解 设u(x,y)?my3?nx2y,v(x,y)?x3?lxy2, 则
?u?u?2nxy,?3my2?nx2, ?x?y2nxy?2lxy??v?v?3x2?ly2,?2lxy,因f(z)解析,由C?R条件有?,解得2222?x?y3my?nx??3x?ly?l??3,m?1,n??3.
三 1. 证明 设f?u?iv,由f?H(D) 有
?u?v?u?v?,??,(1) ?x?y?y?x又f(z)?u?iv也在D也解析,有
?u?(?v)?u?(?v)?,??,(2) ?x?y?y?x由(1)与(2)得
?u?u?v?v?????0 ?x?y?y?x41
故f在D内为常数.
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