9.本性 10. ?? 三、计算题.
1argz?2k?51.解：wk?ze55i k?0,1,2,3 ,??2k?5i 由?1??1 得?1?e 从而有k?2
w2(1?i)?2?e
2.解：（1）f(z)?110??4?45?i?2(cos1103?4?isin3?4)?1?i54
Lnzz?12的各解析分支为fk(z)?lnz?2k?z?12，(k?0,?1,?).
z?1为f0(z)的可去奇点，为fk(z)的一阶极点(k?0,?1,?)。
Res(f0(z),1)?0 Res(fk(z),?1)?k i (k??1,?2?, )（2）Resz?0ezzn?1??1zn?1?Res?n?1???? z?0n!n?0n!??z3.计算下列积分 解：（1）f(z)?z7(z?1)(z?2)232?112z(1?2)3(1?2)zz
Res(f,?)??C?1??1
?z?2f(z)dz?2?i[?Res(f,?)]?2?i
z2(z?a)z2(z?ai)2222（2）设f(z)??z2(z?ai)(z?ai)22
令?(z)?， ??(z)?2aiz(z?ai)314ai
则Res(f,ai)???(ai)1!?2(ai2)(2ai)3??
?Imz?0??f(z)dz?2?x2dx?iRes(f,?ai) 2a??四、证明题
?2a??(x2?a2)2
1证明：.设z?x?iy 有 f(z)?e?e
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zx(coys?isyin )u(x,y)?excosy,v(x,y)??exsiny
?u?x?excosy,?u?y??exsiny,?v?x??exsiny,?v?y??excosy
易知u(x,y)，v(x,y)在任意点都不满足C?R条件，故f在复平面上处处不解析。
2.证明：于高阶导数公式得 (e)z?(n)??0?2?i???1?n?1n!ez?d?
即z?n2?i??n!ez??1?n?1ez?d?
2?zn?1znez??d? ?d? 从而???故?C:??1n!?n?1n!2?in!2?i???1?n?1??zn1
《复变函数》考试试题（十三）参考答案
一、填空题．（每题２分） 1.
1re?i? 2. limu(x,y)?u0及limv(x,y)?v0 3. 0 4. ? x?xx?xoy?yooy?yo5. 2 6. 1?z?z?z????(?1)z246n2n???? 7.椭圆
8. ?12(1?2i) 9.
22(1??4)?1 10. ?1
二、计算题．
１．计算下列各题．（９分） 解: (1) cosi?12(e?e?1)
(2) ln(?2?3i)?ln?2?3i?iarg(?2?3i) ?(3) 33?i13ln13?i(??arctan) 22?e(3?i)ln3?e(3?i)(ln3?i?2k?)?e3ln3?2k??i(6k??ln3)
2k? ?27e3[cos(ln3)?isin(ln3)]
32. 解: z?8?0?z??8?8e3i??2ei??2k?3 (k?0,1,2)
3 故z?8?0共有三个根: z0?1?3, z1??2, z2?1?3
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3. 解: u?x?y?xy?ux?2x?y,uy??2y?x
22??2u?x2??2u?y2?2?2?0?u是调和函数.
(x,y) v(x,y)? ??(x,y)(0,0)?(uydx)?uxdy?c??y0(0,0)y(?2xdx?)x?(2ydy?)c
?x0(?x)dx??(2x?y)dy?c
x22?2xy?y22?c
x22y21?) 22 ?? ?f(z)?u?iv?(x?y?xy)?i(? ?4. 解 (1) (2)
22?2xy?12(2?i)z2?112i
15?i 66?c(x2?iy)dz??(x2?ix2)d(x?ix2)??0?1?i0[(x?y)?ix2]dz?i?(?y)dy??[(x?1)?ix2]dx
0011 ??i2?i3?11??(3?i) 261z?2?1z?1??1()??z?22n?on?0?5. 解: 0?z?1时f(z)?1(z?1)(z?2)??zn?n
??(1?n?01zn)z n?11?z?2时f(z)?1(z?1)(z?2)???1z?21zn?1z?1??1z2(1?)2?11z(1?)z
???n?ozn2n?1??n?0??
6. 解: (1)
????5z?2z(z?1)2sin2zz2(z?1)c?z?2dz?2?i[?Res(f,?)]??4?i
(2)
z?4dz?2?i[?Res(f,?)]?0
7.解: 设f(z)?z21?z4 z1?22(1?i)和z2?22(?1?i)为上半平面内的两个一级极点,
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且Res[f(z),z1]?limz2[z?2(?1?i)](z2?i)2z22(?1?i)z]2(?i2?1?i42i)?z?z1?1?i42i
Res[fz(z)2,?]z?z2lim[z??)1?i42i
???x21?x??dx?2?i(41?i42i?2
8. (1) R?1 (2) R??
9. 解: 设z?x?iy,则f(z)?z?x?y ux?2x,uy?2y,vx?222vy? 0当且仅当x?y?0时,满足C?R条件,故f(z)仅在z?0可导,在z平面内处处不解析. 三、
221. 证明: 设f?u?iv,因为f(z)为常数,不妨设u?v?C (C为常数)
则u?ux?v?vy?0 u?uy?v?vy?0
由于f(z)在D内解析,从而有ux?vy, uy??vx 将此代入上述两式可得ux?uy?vx?vy?0 于是u?C1,v?C2 因此f(z)在D内为常数.
《复变函数》考试试题（十四）参考答案
一、 1、 rn?cosn??isinn?? 2、limu?x,y??u0且limv?x,y??v0
x?x0y?y0x?x0y?y03、0 4、有限值 5、4 6、1?z?z???z7、椭圆 8、cos?二、计算题。
1、解（1）ln??3?4i?
242n??
1??????????isin???? 9、ie1?i 10、?
6?2??2? 39
4???ln5?i???argtan?2n??3??4???ln5?i??argtan??2n?1????n?0,?1,?2,??3???1?
?i6 （2）ie1?????cos?isine?66?1?3i??? ????e22??????????2?i???2k???4??? （3）?1?i?1?i?e?1?i?ln?1?i???e?1?i??ln?e???????ln2??2k???i???2k??ln2?4???4?
=?2e4?2k?????????cos??ln2?isin??ln2????? ?44??????32、解：z??2?2e z?33i?2ei???2n?3?n?0,1,? 23 故：方程z?2?0共有三个根，分别为：3、解：ux?2y,uy?2?x?1?
22?1?3i,?32 ??2u?x2?0??2u?y2
故u是调和函数。
v?x,y???4. 解: (1)
?x,y??0,0??uydx?uxdy?c
1?1?i022??(x?y)?ixdz?ix??0?d(x?ix) ??x311??i?1? ?i(1?i)?303 (2)
?1?i012??(x?y)?ixdz??i?1? ??31n11??z?1??????5. 解: f?z?? nz?1z?23n?03??3?1??3?? = ??n?0??z?1?3n?1n
6. 解: (1)
??z?2?2dz?2?i?cos(2)?0
(z?)240
sinz?