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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?