复变函数论 - 期末复习题 2(9)

(2) f?z??z2?2z2?z?3??1z(1?3z??)

Res(f,?)??C?1??1

??z?4f?z?dz?2?i??Res(f,?)??2?i

i?7. 解: 设z?e 则d??dziz, sin??z2?12iz

?2?d?5?3sin?20???2dz3z?10iz?32z?1???2i3(z?3i)(z?)3z?1dz

令f?z??i,则f在z?1内只有一级权点, z??,依离数定理有 i33(z?3i)(z?)3d?i????2??i?3??i????

2?4

?2?0??2?iRes?f??z?,5?3si?n?8. 解: (1) ?1?i?z?1 即 z?12. 故R?12

(2) Cn?(n!)2nncn

n R?limn??1?1??lim?1????0 cn?1nn?1?n???32329.解 设u(x,y)?my?nxy,v(x,y)?x?lxy, 则

?u?x?2nxy,?u?y?3my2?nx2,

2nxy?2lxy?,解得?3x?ly,?2lxy,因f(z)解析,由C?R条件有?22223my?nx??3x?ly?x?y??v22?v. l??3,m?1,n??3三 1. 证明 设f?u?iv,由f?H(D) 有

?u?x??v?u?v,??,(1) ?y?y?x又f(z)?u?iv也在D也解析,有

?u?x?v?x??(?v)?u?(?v),(2) ,???y?y?x由(1)与(2)得

?u?x??u?y??v?y???0

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故f在D内为常数.

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