内容发布更新时间 : 2024/11/6 0:43:03星期一 下面是文章的全部内容请认真阅读。
复变函数与积分变换（修订版）课后答案（复旦大学出版社）
?Re?e?Re?e?ex?iyx2?y2xx2?y2x?x2?2?Re?ey??x?yx2?y2i???y?y??????cos??2?isin?2??x2?y2?????????x?y???ex2?y2?y??cos?22??x?y?
(4)
ei?2?x?iy??ei?e?2?x?iy??e?2x?e?2iy?e?2x
14. 设z沿通过原点的放射线趋于∞点，试讨论f(z)=z+ez的极限． 解：令z=reiθ， 对于?θ，z→∞时，r→∞．
故r??lim?rei??erei???lim?rei??er?cos??isin?????r??．
所以z??limf?z???．
15. 计算下列各值． (1)
3??ln??2?3i?=ln13?iarg??2?3i??ln13?i?π?arctan??2?
π?π?ln?3?3i??ln23?iarg?3?3i??ln23?i????ln23?i6 ?6?(2)
(3)ln(ei)=ln1+iarg(ei)=ln1+i=i
(4)
πln?ie??lne?iarg?ie??1?i2
16. 试讨论函数f(z)=|z|+lnz的连续性与可导性．
解：显然g(z)=|z|在复平面上连续，lnz除负实轴及原点外处处连续． 设z=x+iy，
g(z)?|z|?x2?y2?u?x,y??iv?x,y?在复平面内可微．
u?x,y??x2?y2,v?x,y??01?u122?2??x?y??2x??x2xx2?y2?u??yyx2?y2
故g(z)=|z|在复平面上处处不可导．
从而f(x)=|z|+lnz在复平面上处处不可导． f(z)在复平面除原点及负实轴外处处连续． 17. 计算下列各值． (1)
?v?0?x?v?0?y 16 / 66
复变函数与积分变换（修订版）课后答案（复旦大学出版社）
?1?i?1?i?eln?1?i?1?i?e?1?i??ln?1?i??e?1?i????π??ln2?4i?2kπi???eln2?π4i?ln2i?π4?2kπ?eln2?π4?2kπ?ei??π?4?ln2????eln2?π4?2kπ????π??π?cos??4?ln2???isin????4?ln2?????2?e2kπ?π4?? ??cos??π?4?ln2????isin??π?4?ln2?????? (2)
??3?5?eln??3?5?e5?ln??3??e5??ln3?i?π?2kπi??e5ln3?5i?π?2kπ5i?e5?ln3?cos?2k?1?π5?isin?2k?1?π5??35??cos?2k?1?π?5?isin?2k?1?π5? 1?i?eln1?i?e?iln1?e?i??ln1?i?0?2kπi?(3)
?e?i??2kπi??e2kπ
1?i1?iln??1?i???2??1?i?ln?1?i?(4)??1?i??e??2???2???e??e1?i?????π???ln1?i???4???2kπi???1?i???π??e?2kπi?4i???e2kπi?πππ4i?2kπ?4?e4?2kπ?ei???2kπ?π?4??π?e4?2kπ????cosπ4?isin??π????4????π?e4?2kπ???22??2?2i??
18. 计算下列各值
(1)
?????eiπ?5i?e?iπ?5ieiπ?5?e?iπ?5cos?π?5i?2?2?5??e?e5??1??555?52??e?ee?e2??2??ch5
(2)
ei?1?5i??e?i?1?5i?ei?5?5i???e?i?5sin?12i?2i?e5?cos1?isin1??e?5??cos1?isin1?2ie5?e?5e5?e?5?2?sin1?i?2cos1
ei?3?i??e?i?3?i?tan?3?i??sin?3?i?cos?3?i??sin6?isin2ei?3?i?2i?e?i?3?i??2?ch21?sin23?(3)2i 17 / 66
(4)
复变函数与积分变换（修订版）课后答案（复旦大学出版社）
12sinz???e?y?xi?ey?xi??sinx?chy?icosx?shy2i?sin2x?ch2y?cos2x?sh2y22?sin2x??ch2y?sh2y???cos2x?sin2x??sh2y?sin2x?sh2yarcsini??iln?i?1?i2???iln?1?2???ln?2?1??i2kπ????i???k?0,?1,????ln?2?1??i?π?2kπ????i?
i1?i?1?2i?i?21?arctan?1?2i???ln???ln???i?21?i?1?2i?2?55?1i?kπ?arctan2??ln524(6)
(5)
19. 求解下列方程
(1) sinz=2． 解：
1z?arcsin2?ln?2i?3i???ln???2?3?i??i?1?????i?ln?2?3???2k??πi??2???1????2k??π?iln?2?3?,k?0,?1,??2?
z(2)e?1?3i?0
z解：e?1?3i 即
z?ln?1?3i??ln2?i1???ln2??2k??πi3??π?2kπi3
(3) lnz?πi2
ππii2 即z?e2?i
解：
lnz?(4)z?ln?1?i??0
π1??z?ln?1?i??ln2?i??2kπi?ln2??2k??πi4?4?． 解：
20. 若z=x+iy，求证
(1) sinz=sinxchy+icosx?shy 证明：
18 / 66
复变函数与积分变换（修订版）课后答案（复旦大学出版社）
eiz?e?izei?x?iy??e??x?yi??isinz??2i2i1?.?e?y?xi?ey?xi?2i?sinx?chy?icosx.shy (2)cosz=cosx?chy-isinx?shy
证明：
eiz?e?iz1?i?x?yi??i?x?yi??cosz???e?e221??e?y?xi?ey?xi?21??e?y??cosx?isinx??ey.?cosx?isinx??2?ey?e?y?e?y?ey??.cosx??isinx.?2?2??cosx.chy?isinx.shy (3)|sinz|2=sin2x+sh2y 证明： sinz?1??y?xie?ey?xi??sinx?chy?icosx?shy2i
2sinz?sin2xch2y?cos2x.sh2y?sin2x?ch2y?sh2y???cos2x?sin2x?sh2y
?sin2x?sh2y
(4)|cosz|2=cos2x+sh2y
证明：cosz?cosxchy?isinxshy
cosz?cos2x.ch2y?sin2x.sh2y?cos2x?ch2y?sh2y???cos2x?sin2x?.sh2y
2?cos2x?sh2y
21. 证明当y→∞时，|sin(x+iy)|和|cos(x+iy)|都趋于无穷大． 证明： sinz?1?iz?iz?1??y?xie?e??e?ey?xi?2i2i
1sinz??e?y?xi?ey?xi2?y?xi?yy?xiye?ee?e∴ sinz≥
而
当y→+∞时，e-y→0，ey→+∞有|sinz|→∞． 当y→-∞时，e-y→+∞，ey→0有|sinz|→∞．
11cos?x?iy??e?y?xi?ey?xi≥?e?y?ey?22同理得 所以当y→∞时有|cosz|→∞．
习题三
19 / 66
1?y?xi?e?ey?xi??1?e?y?ey?22
复变函数与积分变换（修订版）课后答案（复旦大学出版社）
1. 计算积分C?(x?y?ix2)dz,其中C为从原点到点1+i的直线段.
解 设直线段的方程为y?x,则z?x?ix. 0?x?1
22x?y?ixdz?x?y?ix????d(x?ix)??0101C11??ix2(1?i)dx?i(1?i)?x3ii?1?(1?i)?故 033(1?z)dz2. 计算积分C?，其中积分路径C为
(1) 从点0到点1+i的直线段;
(2) 沿抛物线y=x2，从点0到点1+i的弧段. 解 (1)设z?x?ix. 0?x?1
?dz?1x??ix(d?x)ix?i
C??1?z?0?1?
(2)设z?x?ix2. 0?x?1
?1?z?dz?C??1220?1?x?ix?d(x?ix)?2i3
3. 计算积分C?zdz，其中积分路径C为
(1) 从点-i到点i的直线段;
(2) 沿单位圆周|z|=1的左半圆周，从点-i到点i; (3) 沿单位圆周|z|=1的右半圆周，从点-i到点i. 解 (1)设z?iy. ?1?y?1
zdz?1C??1?1ydiy?i??1ydy?i
3??(2)设z?ei?. ?从2到2
??dz??1dei?32??i?i?32?de?2iC?z22
3??(3) 设z?ei?. ?从2到2
?C?zdz??2i?3?1de?2i2
3
20 / 66