复变函数与积分变换（修订版）课后答案（复旦大学出版社）
1t解：令
z?,则z??,t?0.
2于是
lim11?zz???limt?0t22
?0?x3y,?f(z)??x4?y2?0,?z?0,z?0.3
x31?t.
解：因为
0?xyx?yxyx?y42342?y2(2)
limRe(z)z2xy?x2,
z?0;
Re(z)z?xx?iy所以
有
(x,y)?(0,0)lim?0?f(0)
解：设z=x+yi，则
limRe(z)z?limx所以f(z)在整个z平面连续.
5. 下列函数在何处求导？并求其导数.
z?0x?0y?kx?0x?ikx?11?ik
(1)
显然当取不同的值时f(z)的极限不同 所以极限不存在.
limz?iz(1?z)2f(z)?(z?1)n?1 (n为正整数)；
解：因为n为正整数，所以f(z)在整个z平面上可导.
n?1f?(z)?n(z?1)（3） 解
lz?i.
；
：
(2)
f(z)?z?2(z?1)(z?1)2.
m2(z?1)(z?1))?0z?iz(?z2z?ilimz?iz(i?z)(z?i)=
z?i?limz?i1z(i?z)??12i.
1解：因为f(z)为有理函数，所以f(z)在处不可导.
limzz?2z?z?2z?12（4）
z?1.
?(z?2)(z?1)(z?1)(z?1)?z?2z?1,从而f(z)除z??1,z??i外可导.
(z?2)?(z?1)(z?1)?(z?1)[(z?1)(z?1)]?(z?1)(z?1)?2z?5z?4z?3(z?1)(z?1)3z?85z?72223222222zz?2z?z?2解：因为
limz?12f?(z)?
?zz?2z?z?2z?12所以
z?1?limz?2z?1z?1?32
.
f(z)?
4. 讨论下列函数的连续性： (1)
?xy,?22f(z)??x?y?0,?z?0,z?0;(3) .
z=75解：f(z)除
f?(z)?2外处处可导，且
??61(5z?7)23(5z?7)?(3z?8)5(5z?7).
xyx?y22f(z)?x?yx?y22?ix?yx?y22limf(z)?解：因为
z?0(x,y)?(0,0)lim(4)
,
解：因为
f(z)?2.
若令y=kx,则
(x,y)?(0,0)limxyx?y22?k1?kx?y?i(x?y)x?y22?x?iy?i(x?iy)x?y22?(x?iy)(1?i)x?y22?z(1?i)z2?1?iz,
f?(z)??.
(1?i)z2因为当k取不同值时，f(z)的取值不同，所以f(z)在z=0处极限不存在.
从而f(z)在z=0处不连续，除z=0外连续. (2)
所以f(z)除z=0外处处可导，且
6. 试判断下列函数的可导性与解析性. (1)
f(z)?xy?ixy22.
;
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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）
解：
?y?xu(x,y)?xy,v(x,y)?xy222在全平面上可微.
?2xy,?v?y?x2?u?y,?u?y?2xy,?v?x证明：因为f?(z)?0，所以?x
??u?y?0?v,
?x??v?y?0.
所以u,v为常数，于是f(z)为常数. (2) f(z)解析.
所以要使得
?u?x??v?y?u???v?x， ?y,
证明：设f(z)?u?iv在D内解析,则
?u?x?u??(?v)?y??u?x???v?y只有当z=0时,
从而f(z)在z=0处可导，在全平面上不解析. (2) 解：
?u?x
f(z)?x?iy222.
2?y?u???(?v)?x?v?y,???u?y?v?y?
?v?xu(x,y)?x,v(x,y)?y?u?y在全平面上可微.
?0,?v?y?v?u?y?y?2y?x??
?u??u?y,?u?y?u?y?v?x?2x,?0,?v?x
???v?y而f(z)为解析函数，所以?x?v???v?x
?u只有当z=0时,即(0,0)处有
?x?,.
所以f(z)在z=0处可导，在全平面上不解析. (3) 解：
?u?x所以即
从而v为常数，u为常数，即f(z)为常数. (3) Ref(z)=常数.
?u?x???v?x,?v?y???v?y,?u?x????v?y?0f(z)?2x?3iy333;
3u(x,y)?2x,v(x,y)?3y2在全平面上可微.
2证明：因为Ref(z)为常数，即u=C1,
?u?x?u?y??u?y?0
?6x,?u?y?0,?v?x?9y,?v?y?0
所以只有当2x??3y时，才满足C-R方程. 从而f(z)在2x?2因为f(z)解析，C-R条件成立。故从而f(z)为常数. (4) Imf(z)=常数.
证明：与（3）类似，由v=C1得
?x??0即u=C2
?v?x??v?y?03y?0处可导，在全平面不解析.
?u
(4) f(z)?z?z. 解：设
z?x?iy因为f(z)解析，由C-R方程得所以f(z)为常数. 5. |f(z)|=常数.
23232?x??u?y?0,即u=C2
，则
f(z)?(x?iy)?(x?iy)?x?xy?i(y?xy)u(x,y)?x?xy,v(x,y)?y?xy?u?x?3x?y,22证明：因为|f(z)|=C，对C进行讨论. 若C=0，则u=0,v=0,f(z)=0为常数. 若C??u?x3232
?v?y?3y?x220，则f(z)
?v?x?0,但
f(z)?f(z)?C2，即u2+v2=C2
?u?y?2xy,?v?x?2xy,则两边对x,y分别求偏导数，有
2u??2v??0,2u??u?y?2v??v?y?0所以只有当z=0时才满足C-R方程.
从而f(z)在z=0处可导，处处不解析.
7. 证明区域D内满足下列条件之一的解析函数必为常数.
?(1) f(z)?0;
利用C-R条件，由于f(z)在D内解析，有
?u?x??v?y?u?y???v?x
?u?v?u??v??0???x?x??v??u?u??v?0?x??x所以??u
所以?x?0,?v?x?0
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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）
即u=C1,v=C2,于是f(z)为常数.
(6) argf(z)=常数.
?v?arctan???C?u?证明：argf(z)=常数，即
2证明：
u(x,y)?e(xcosy?ysiny),xv(x,y)=e(ycosy?xsiny)x处处可微，且
,
?v?u?y2?u?x?e(xcosy?ysiny)?e(cosy)?e(xcosy?ysiny?cosy)xxx
(v/u)?u?(u?2?v于是1?(v/u)得
??u???u???u????u????x?x22u(u?v)2?v??u)?u(u22?v?y2)?0?u?y?e(?xsiny?siny?ycosy)?e(?xsiny?siny?ycosy)xxu(u?v)
?v?x?e(ycosy?xsiny)?e(siny)?e(ycosy?xsiny?siny)xxx???v?x?v?y?v?x?v?x?v??v??u?x?u?y?u?x?u?x?v?0?0?y?e(cosy?y(?siny)?xcosy)?e(cosy?ysiny?xcosy)xx?u C-R条件→
?0所以?x??v?y?u, ?y???v?x
所以f(z)处处可导，处处解析.
f?(z)??u?xxz???v??v??i?v?xz?e(xcosy?ysiny?cosy)?i(e(ycosy?xsiny?siny))xxxxxxx?0?ecosy?iesiny?x(ecosy?iesiny)?iy(ecosy?iesiny)
?u?y??v?y?0?e?xe?iye?e(1?z)zz?u解得
?x??v?x10. 设
，即u,v为常数，于是f(z)
?x3?y3?i?x3?y3?,?22f?z???x?y?0.?z?0.z?0.?为常数.
8. 设f(z)=my3+nx2y+i(x3+lxy2)在z平面上解析，求
m,n,l的值.
解：因为f(z)解析，从而满足C-R条件.
?u?x?v?x?u?x?u?y?2nxy,?3x?ly,?v?y22求证：(1) f(z)在z=0处连续． (2)f(z)在z=0处满足柯西—黎曼方程．
(3)f′(0)不存在．
limf(z)?z?0
?u?y?3my?nx?v?y?2lxy22
证明.(1)∵
lim?x,y???0,0?limu?x,y??iv?x,y?
而
?x,y???0,0?u?x,y???x,y???0,0?limx?yx?y2332
??n?l
?n??3,l??3m???v?x
xy???x?y?1???2222?x?yx?y??? ∵
x?yx?y3x?y33所以n??3,l??3,m?1.
9. 试证下列函数在z平面上解析，并求其导数. (1) f(z)=x3+3x2yi-3xy2-y3i
证明：u(x,y)=x3-3xy2, v(x,y)=3x2y-y3在全平面可微,且
?u?x?3x?3y,223320≤2≤32x?y∴
∴
?u?y??6xy,?v?x?6xy,?v?y?3x?3y22?x,y???0,0?limx?yx?y232?0
32
所以f(z)在全平面上满足C-R方程，处处可导，处处解析.
f?(z)??u?x?i?v?xx同理∴
?x,y???0,0?limx?yx?y23?0
?x,y???0,0?limf?z??0?f?0??3x?3y?6xyi?3(x?y?2xyi)?3z22222∴f(z)在z=0处连续． .
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.(2)
f(z)?e(xcosy?ysiny)?ie(ycosy?xsiny)x
复变函数与积分变换（修订版）课后答案（复旦大学出版社）
2??i2(2)考察极限z?0limf(z)?f?0?ze3?e?e3π?i3
2?1?3??π??π??3?e??cos????isin?????e???i??22??3??3???32
(3)
当z沿虚轴趋向于零时，z=iy，有
lim1iyy?0????lim??f?iy??f0?y?01iy?3?y?1?i?y2?1?i．
?Re?e?Re?e?ex?iy2x?y2xx?y22?yx?y22i?当z沿实轴趋向于零时，z=x，有
lim1xx?0?2x2x?y?Re?e??x?yy????????cos??2?isin??2?2?2???x?yx?y????????f?x??f?0???1?i?u?v?x?ex?y22
?i?u?yy???cos?22??x?y?
它们分别为
?u??v?y,?x?i?,?v?y(4)
ei?2?x?iy??2x?e?e?2iyi?2?x?iy??e?e?e?2x?u?y∴?x???v?x
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→∞．
i1?i∴满足C-R条件．
(3)当z沿y=x趋向于零时，有
x?y?0limf?x?ix??f?0,0?x?ix?lim33x?1?i??x?1?i?32x?1?i?x?y?0?
(1)
故
lim?rer??i??erei???lim?rei?r???er?cos??isin?????．
∴z?0?z不存在．即f(z)在z=0处不可导．
11. 设区域D位于上半平面，D1是D关于x轴的对称区域，若f(z)在区域D内解析，求证
F?z??f?z?lim?f所以z??limf?z???．
15. 计算下列各值．
3??ln??2?3i?=ln13?iarg??2?3i??ln13?i?π?arctan??2?在
区域D1内解析．
证明：设f(z)=u(x,y)+iv(x,y)，因为f(z)在区域D内解析．
所以u(x,y),v(x,y)在D内可微且满足C-R方程，即
?u?x??v?y,?u?y???v?x(2)
ln?3?3i??ln23?iarg?3?π?π?3i??ln23?i????ln23?i6?6?
(3)ln(ei)=ln1+iarg(ei)=ln1+i=i (4)
π2．
，得
ln?ie??lne?iarg?ie??1?if?z??u?x,?y??iv?x,?y????x,y??i??x,y????x???x
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，
u?x,y???u12??u?x,?y????y??u?x,?y??y?????u?x,?y??y??y?y ?y
故φ(x,y),ψ(x,y)在D1内可微且满足C-R条件
?
??v?x,?y??x?x
???v?x,?y??v?x,?y?g(z)?|z|?22x?y?u?x,y??iv?x,y?22
x?y,v?x,y??012在复平面内可微．
?u2???x????y,???y?????x
?x?v?x??x2?y2???2x?xx?y2?y?yx?y22从而
f?z?
在D1内解析
13. 计算下列各值
(1) e2+i=e2?ei=e2?(cos1+isin1) (2)
?0?v?y?0
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．
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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）
17. 计算下列各值． (1)
?1?i?1?i?eln?1?i?1?i?e?1?i??ln?1?i???e1?i?????ln2?π4i?2kπi????eln2?π4i?ln2i?π4?2kπln2?π?e4?2kπ?ei??π?4?ln2???ln2?π?e4?2kπ????cos??π?4?ln2???π??isin??4?ln2???????2?e2kπ?π4???π ?2??cos??4?ln???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?i?1?i1?iln?1?i?1?i?ln?1?i(4)?2??e?????e????2???2???1?i?????ln1?i????π?4????2kπi???e??e1?i???π??2kπi?4i???e2kπi?πi?2kπ?ππ44?e4?2kπi???2kπ?π??e4??π?e4?2kπ??π??π???cos4?isin???4????π?e4?2kπ???22?2?2i???
18. 计算下列各值 (1)
?π?5i??i?π?5i?iπ?5iπ?5cos?π?5i??ei?e2?e?e?2?e??1??555?5??e?552??e?e2??e?e2??ch5
(2)
1?5i??e?i?1?5i?i?5?i?5sin?1?5i??ei?2i?e?e2i?e5?cos1?isin1??e?5??cos1?isin1?2i?e5?e?55?52?sin1?i?e?e2cos1
(3)
ei?3?i??e?i?3?i?tan?3?i??sin?3?i?2isin6?isin2cos?3?i??ei?3?i??e?i?3?i??2?ch21?sin23?2i(4)
2sinz2?12i??e?y?xi?ey?xi??sinx?chy?icosx?shy2?sin2x?ch2y?cos2x?sh2y?sin2x??ch2y?sh2y???cos2x?sin2x??sh2y?sin2x?sh2y(5)
arcsini??iln?i?1?i2???iln?1?2?????i?ln?2?1??i2kπ????k?0,?1,????i??ln?2?1??i?π?2kπ???
(6)
arctan?1?2i???i1?i?1?2i?2ln1?i?1?2i???i?21?2?ln???5?5i???kπ?12arctan2?i4?ln5
19. 求解下列方程 (1) sinz=2． 解：
z?arcsin2?1iln?2i?3i???ln???2?3?i????i?3????1?ln?2????2k??2??πi??????2k?1?2??π?iln?2?3?,k?0,?1,?
(2)ez?1?3i?0
解：ez?1?3i 即
z?ln?1?3i??ln2?iπ3?2kπi?ln2????2k?1?3??πi
(3)
lnz?π2i
lnz?π2iπi解：
即z?e2?i
(4)z?ln?1?i??0 解
：
z?l??????π4?kπi????1??k?4??πin．10 / 34