9.本性 10. ?? 三、计算题. 1.解：wk?ze515argz?2k?i5 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?3?1?i?isin)?5 444Lnzlnz?2k?f(z)?的各解析分支为，(k?0,?1,?). k22z?1z?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?1?zn?1ez（2）Resn?1?Res?n?1????
z?0zz?0n!n?0n!??z3.计算下列积分
z71?解：（1）f(z)?2 3212(z?1)(z?2)z(1?)3(1?)z2z2Res(f,?)??C?1??1
?z?2f(z)dz?2?i[?Res(f,?)]?2?i
z2z2（2）设f(z)?2 ?2222(z?a)(z?ai)(z?ai)2aizz2??(z)?令?(z)?， 32(z?ai)(z?ai)则Res(f,ai)???(ai)1!2(ai2)1???i (2ai)34a
?Imz?f(z)dz?2?iRes(f,?ai)
02a??四、证明题
????x2dx? ?(x2?a2)22a1证明：.设z?x?iy 有 f(z)?ez?ex(coys?i
36
)syinu(x,y)?excosy,v(x,y)??exsiny
?u?u?v?excosy,??exsiny,??exsiny,?x?y?x?v??excosy ?y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。
z?(n)2.证明：于高阶导数公式得 (e)??0n!ez??d? 2?i???1?n?1n!ez?即z?d? ???12?i?n?1n?zn?1znez?zn1ez??d? 故?d? 从而????C:??1n!?n?1n!2?in!2?i???1?n?1??
2《复变函数》考试试题（十三）参考答案
一、填空题．（每题２分） 1.
1?i?e 2. limv(x,y)?v0 3. 0 4. ? u(x,y)?u0及limx?xox?xory?yoy?yo246n2n5. 2 6. 1?z?z?z????(?1)z???? 7.椭圆
8. ?12?(1?2i) 9. (1?)?1 10. ?1 224二、计算题．
１．计算下列各题．（９分） 解: (1) cosi?1(e?e?1) 2(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?3?8?8e?2e3i?i??2k?3 (k?0,1,2)
故z?8?0共有三个根: z0?1?3, z1??2, z2?1?3
37
3. 解: u?x2?y2?xy?ux?2x?y,uy??2y?x
?2u?2u?2?2?2?2?0?u是调和函数. ?x?y v(x,y)? ???(x,y)(0,0)?(uydx)?uxdy?c??y0(x,y)(0,0)y(?2xdx?)x?(2ydy?)c
x0(?x)dx??(2x?y)dy?c
x2y2?2xy??c ??22x2y21?2xy??) ?f(z)?u?iv?(x?y?xy)?i(?22222 ?4. 解 (1) (2)
11(2?i)z2?i 221152222(x?iy)dz?(x?ix)d(x?ix)???i ?c?066?1?i0[(x?y)?ix2]dz?i?(?y)dy??[(x?1)?ix2]dx
0011 ??ii11????(3?i) 2326n?1111?z5. 解: 0?z?1时f(z)??????()??zn
(z?1)(z?2)z?2z?12n?o2n?0? ??(1?n?01zn )zn?11?z?2时f(z)?111?11 ????z1(z?1)(z?2)z?2z?12(1?)z(1?)2z????zn1 ?????n
n?o2n?1n?0z6. 解: (1)
5z?2??c?z?2z(z?1)2dz?2?i[?Res(f,?)]??4?i
sin2z (2) ??z?4z2(z?1)dz?2?i[?Res(f,?)]?0
z222z?(1?i)z?(?1?i)为上半平面内的两个一级极点,7.解: 设f(z)? 和121?z422 38
且Res[f(z),z1]?limz?z1z2[z?2(?1?i)](z2?i)2z2?1?i 42i Res[fz(z)2,?]1?i lim?z?z2242i[z?(?1?i)z]2(?i)2x21?i1?i? dx?2?i(?)????1?x4242i42i??8. (1) R?1 (2) R??
229. 解: 设z?x?iy,则f(z)?z?x?y ux?2x,uy?2y,vx?vy? 02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、
1. 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.
22《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?y0242n3、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?
??
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?? （2）ie?1??i61????1?3i???cos?isin?????? e?66?e?22???e?1?i?ln?1?i?? （3）?1?i?1?i?e?1?i??ln??????2?i???2k???4????e???????ln2??2k???i???2k??ln2?4???4?
=?2e4?2k?????????cos??ln2?isin??ln2???? ??4??4????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?
21?3i,?32 2???2u?2u ?0?2
?x2?y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)?033 (2)
?1?i012??(x?y)?ixdz??i?1? ??3n111??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?