解：令
z?1t,则z??,t?0.
1t2lim?lim?0z??1?z2t?01?t2于是.
Re(z)(2) z?0z;
limRe(z)x?x?iy有 解：设z=x+yi，则z
?x3y,?f(z)??x4?y2?0,?z?0,z?0.
解：因为
x3yxx3y0?4??x?y22x2y2,
x3ylim?0?f(0)(x,y)?(0,0)x4?y2所以
limRe(z)x1?lim?z?0x?0zx?ikx1?iky?kx?0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.
n?1(1) f(z)?(z?1) (n为正整数)；
显然当取不同的值时f(z)的极限不同 所以极限不存在. z?ilimz?iz(1?z2)（3） ；
解：因为n为正整数，所以f(z)在整个z平面上可导.
f?(z)?n(z?1)n?1.
：
f(z)?z?2(z?1)(z2?1).
解l(2) 1z?iz?iz?i11lim?lim??i2. z(?z2=z?iz(i?z)(z?i)z?iz(i?z)m2(z?1)(z?1))?0解：因为f(z)为有理函数，所以f(z)在
zz?2z?z?2limz?1z2?1（4） .
处不可导.
从而f(z)除z??1,z??i外可导.
zz?2z?z?2(z?2)(z?1)z?2??,2z?1(z?1)(z?1)z?1解：因为
limzz?2z?z?2z?23?lim?z?1z?1z2?12.
(z?2)?(z?1)(z2?1)?(z?1)[(z?1)(z2?1)]?f?(z)?(z?1)2(z2?1)2?2z3?5z2?4z?3?(z?1)2(z2?1)2f(z)?3z?85z?7.
z=75
所以
z?1
4. 讨论下列函数的连续性： (1)
?xy,?f(z)??x2?y2?0,?z?0,z?0;lim(3)
解：f(z)除f?(z)?外处处可导，且
3(5z?7)?(3z?8)561??(5z?7)2(5z?7)2x?yx?y?ix2?y2x2?y2.
f(z)?limf(z)?解：因为
z?0xy(x,y)?(0,0)x2?y2(4)
,
解：因为
f(z)?.
若令y=kx,则
xyklim?22(x,y)?(0,0)x?y1?k2x?y?i(x?y)x?iy?i(x?iy)(x?iy)(1?i)z(1?i)1?i????2x2?y2x2?y2x2?y2zz,
.
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.
22f(z)?xy?ixy; (1)
f?(z)??(1?i)z2.
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22u(x,y)?xy,v(x,y)?xy在全平面上可微. 解：
?y?y2,?x?u?2xy,?y?v?2xy,?x?v?x2?y
?u?u?v?v??0??0?(z)?0?x?yf?x?y证明：因为，所以,.
所以u,v为常数，于是f(z)为常数. (2) f(z)解析.
证明：设f(z)?u?iv在D内解析,则 ?u?(?v)?u?v?????x?y?x?y ?u??(?v)?v????y?x?y ?u?v??,?x?y?u?v??y?x
所以要使得
?u?v?u?v????x?y， ?y?x,
只有当z=0时,
从而f(z)在z=0处可导，在全平面上不解析.
22f(z)?x?iy(2) .
22解：u(x,y)?x,v(x,y)?y在全平面上可微.
?u?2x,?x?u?0,?y?v?0,?x?v?2y?y
?u?u?,?x?y而f(z)为解析函数，所以
?v?v??,?x?x?v?v??,?y?y?u?v???y?x
?u?v?u?v????y. 只有当z=0时,即(0,0)处有?x?y,?y所以f(z)在z=0处可导，在全平面上不解析.
33f(z)?2x?3iy(3) ;
所以即
从而v为常数，u为常数，即f(z)为常数. (3) Ref(z)=常数.
证明：因为Ref(z)为常数，即u=C1, 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.
?u?u?v?v????0?x?y?x?y
33u(x,y)?2x,v(x,y)?3y解：在全平面上可微.
?u?u??0?x?y
?u?6x2,?x?u?0,?y?v?9y2,?x?v?0?y?u?u??0?x?y即u=C2
所以只有当2x??3y时，才满足C-R方程. 从而f(z)在2x?3y?0处可导，在全平面不解析.
2f(z)?z?z(4) .
?v?v??0?x?y证明：与（3）类似，由v=C1得
因为f(z)解析，由C-R方程得
?u?u??0?x?y,即u=C2
解：设z?x?iy，则
f(z)?(x?iy)?(x?iy)2?x3?xy2?i(y3?x2y) u(x,y)?x3?xy2,v(x,y)?y3?x2y
?u?3x2?y2,?x?u?2xy,?y?v?2xy,?x?v?3y2?x2?y所以f(z)为常数. 5. |f(z)|=常数.
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.
2f(z)?f(z)?C??若C0，则f(z) 0,但，即u2+v2=C2
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;
则两边对x,y分别求偏导数，有
?u?v?u?v2u??2v??0,2u??2v??0?x?x?y?y
利用C-R条件，由于f(z)在D内解析，有 ?u?v?u?v????x?y?y?x ?v??uu??v??0???x?x??u?v??u?u??v?0?0,??x?x??x所以 所以
?v?0?x
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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,
(v/u)??21?(v/u)于是
u2?(u??v?u2?v?u?v?)u(u?v)?y?y?x?x??0222222u(u?v)u(u?v)
证明：
u(x,y)?ex(xcosy?ysiny),v(x,y)=ex(ycosy?xsiny)处处可微，且
?u?ex(xcosy?ysiny)?ex(cosy)?ex(xcosy?ysiny?cosy)?x
?u?ex(?xsiny?siny?ycosy)?ex(?xsiny?siny?ycosy)?y得
?u??vu??v??0??x??x??v?u??v??u?0??y??y C-R条件→ ?u??vu??v??0???x?x??u??v?v??u?0??x??x
?u?v?u?v????0解得?x?x?y?y，即u,v为常数，于是f(z)
?v?ex(ycosy?xsiny)?ex(siny)?ex(ycosy?xsiny?siny)?x?v?ex(cosy?y(?siny)?xcosy)?ex(cosy?ysiny?xcosy)?y?u?v?u?v????x 所以?x?y, ?y所以f(z)处处可导，处处解析.
f?(z)??u?v?i?ex(xcosy?ysiny?cosy)?i(ex(ycosy?xsiny?siny))?x?xx?ecosy?iexsiny?x(excosy?iexsiny)?iy(excosy?iexsiny)?ez?xez?iyez?ez(1?z)10. 设
?x3?y3?i?x3?y3?,z?0.?f?z???x2?y2?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?u?2nxy,?3my2?nx2?x?y ?v?3x2?ly2,?x?u?v??n?l?x?y?v?2lxy?y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． 证明.(1)∵
limf(z)?z?0?x,y???0,0?limu?x,y??iv?x,y?
x3?y3limu?x,y??lim?x,y???0,0??x,y???0,0?x2?y2而
∵
?u?v???n??3,l??3m?y?xx3?y3xy???x?y?1????x2?y2?x2?y2??0≤x3?y3x2?y23x?y2
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
所以f(z)在全平面上满足C-R方程，处处可导，处处解析.
?u?vf?(z)??i?3x2?3y2?6xyi?3(x2?y2?2xyi)?3z2?x?xxxf(z)?e(xcosy?ysiny)?ie(ycosy?xsiny). .(2)
≤∴
∴
x3?y3lim?0?x,y???0,0?x2?y2
?u?3x2?3y2,?x?u??6xy,?y?v?6xy,?x?v?3x2?3y2?y同理
x3?y3lim?0?x,y???0,0?x2?y2
∴?x,y???0,0?limf?z??0?f?0?∴f(z)在z=0处连续．
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f(z)?f?0?limz(2)考察极限z?0
当z沿虚轴趋向于零时，z=iy，有
11?y3?1?i???lim?f?iy??f0???1?i??limy?0iy?y?0iyy2．
e2??i3?e?e23π?i32?1??π?3??π??3?e??cos????isin?????e???i??22? ?3????3?23(3)
?Re?e?Re?e?ex?iyx2?y2xx2?y2x?x2?2?Re?ey??x?yx2?y2i?当z沿实轴趋向于零时，z=x，有 1lim?f?x??f?0???1?ix?0x
?u?v?i?,?x?x?u?v???y?x?v?u?i?y?y??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它们分别为?u?v?,?x?y
∴
∴满足C-R条件．
(3)当z沿y=x趋向于零时，有
f?x?ix??f?0,0?x3?1?i??x3?1?i?ilim?lim?x?y?0x?y?0x?ix2x3?1?i?1?i
?flim∴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
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?
????称区域，若f(z)在区域D内解析，求证Fz?fz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?v?,?x?y?u?v???y?x(2)
π?π?ln?3?3i??ln23?iarg?3?3i??ln23?i????ln23?i6 ?6?(3)ln(ei)=ln1+iarg(ei)=ln1+i=i
(4)
πln?ie??lne?iarg?ie??1?i2
．
，得
f?z??u?x,?y??iv?x,?y????x,y??i??x,y????u?x,?y?????u?x,?y????u?x,?y???y?y?x?x ?y ?v?x,?y??v?x,?y???????v?x,?y??????y?y?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，
g(z)?|z|?x2?y2?u?x,y??iv?x,y?在复平面内可微．
故φ(x,y),ψ(x,y)在D1内可微且满足C-R条件?????????,???x?y?y?x 从而
u?x,y??x2?y2,v?x,y??01?u122?2??x?y??2x??x2xx2?y2?u??yyx2?y2
f?z?在D1内解析
13. 计算下列各值
(1) e2+i=e2?ei=e2?(cos1+isin1) (2)
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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?v?0?x?v?0?y
17. 计算下列各值． (1)
?1?i?1?i?eln?1?i?1?i?e?1?i??ln?1??1?i?????ln2?π4i?2kπi?i??e???eln2?π4i?ln2i?π4?2kπln2?π??e4?2kπ?ei?π?4?ln2????eln2?π4?2kπ????cos??π?4?ln2???π????isin??4?ln2????π?2?e2kπ?4?? ??cos??π?4?ln2???π????isin??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?i?1ln??1?i??(4)?2??1?i?ln??1?i???i??2???e?e?2????e1?i?????π???ln1?i???4???2kπi???1?i???π??e?2kπi?4i???e2kπi?πππ4i?2kπ?4?e4?2kπ?ei??π??2kπ?4??π?e4?2kπ???π?π?cos4?isin????4?????π?e4?2kπ???2?2?2?2i??
18. 计算下列各值
(1)
ei?π?5i??e?i?π?5i?eiπ?5?e?iπ?5cos?π?5i??2?2?e?5?e5??1??e?5?e522??e5?e?5??2??ch5
(2)
????sin?1?5i??ei1?5i?e?i1?5iei?5?e?i?52i?2i?e5?cos1?isin1??e?5??cos1?isin1?2ie5?e?5e5?e?5?2?sin1?i?2cos1 (3)
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?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???i2ln1?i?1?2i?1?i?1?2i???i2?ln??21???5?5i???kπ?1i2arctan2?4?ln5
19. 求解下列方程
(1) sinz=2． 解：
z?arcsin2?1iln?2i?3i???ln???2?3?i????i???ln?2?3????1???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???1??2k?3??πi
(3) lnz?π2i
ππi解：
lnz?2i 即z?e2?i
(4)z?ln?1?i??0 解
：
z?l??????π?kπi?????k?1?44??πi．n
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