复变函数习题答案

本文格式为Word版，下载可任意编辑——复变函数习题答案习题一

1.用复数的代数形式a+ib表示以下复数

e?iπ/4;3?5i7i?1;(2?i)(4?3i);1i?31?i.

①解：e?π4i2?2?π??π??cos????isin????????22?4??4???22i???i?22?②解：

3?5i7i?1??3?5i??1?7i??1+7i??1?7i???1625?1325i

③解：?2?i??4?3i??8?3?4i?6i?5?10i④解：

1i?31?i=?i?3?1?i?2?32?52i

2.求以下各复数的实部和虚部(z=x+iy)

3???1?i3?n3??1?i(a?);z;??;??;i.z?a?2??2?z?a33

①解：∵设z=x+iy则

z?az?a??x?iy??a?x?iy??a2??x?a??iy?x?a??iy22????x?a??iy?????x?a??iy???x?a?2?y2

x?a?y?z?a??∴Re??22?z?a??x?a??y,Im??z?a????z?a?2xy?x?a?2?y2．

②解：设z=x+iy

∵z3??x?iy???x?iy??x?iy???x2?y2?2xyi??x?iy?

?x?x?y23232??2xy22222??y?x?y??2xy?i??3?x?3xy??3xy?y2?i3

∴Re?z3??x3?3xy2,

3Im?z??3x2y?y．

3??1?i3??③解：∵????2????1?i38?31???1?3???1???8???3?22???3???1??????3??3?3????

?18?8?0i??1

∴Re???1?i3??1,???2??3??1?i3?Im??0．???2???3???1???3??1?3??22??3???1??3??3?3?i④解：∵???1?i3???????2??8

?18?8?0i??1

∴Re??1?i3???1,Im?????1?i3??2?????0．?2???in???1?k⑤解：∵?,n?2k?n?2k?1k??．

?k???1??i,∴当n?2k时，Re?in????1?k，Im?in??0；

当n?2k?1时，Re?in??0，Im?in????1?k．

3.求以下复数的模和共轭复数

?2?i;?3;(2?i)(3?2i);

①解：?2?i?4?1?5．

?2?i??2?i

②解：?3?3

?3??3

③解：?2?i??3?2i??2?i3?2i?5?13?65．

?2?i??3?2i???2?i???3?2i???2?i???3?2i??4?7i④解：

1?i?1?i?2222

?1?i????1?i?1?i?2??2?2

4、证明：当且仅当z?z时，z才是实数．证明：若z?z，设z?x?iy，

则有x?iy?x?iy，从而有?2y?i?0，即y=0

??

1?i2.

∴z=x为实数．

若z=x，x∈?，则z?x?x．∴z?z．命题成立．

≤5、设z,w∈?，证明：z?w

2z?w

证明：∵z?w??z?w???z?w???z?w??z?w?

?z?z?z?w?w?z?w?w

?z?z≤2?zw?z?w?w?w22??2Re?z?w??2z?w?2z?w

?2

z2?w?w?w2

?z?22?z?2∴z?w≤z?w．

6、设z,w∈?，证明以下不等式．

z?wz?w2?z?z2?2Rez?w?w?2Rez?w?w2??2

22??2z?w2?z?w?2z?2?w2?

2并给出最终一个等式的几何解释．

证明：z?w?z?2Re?z?w??w在上面第五题的证明已经证明白．

22下面证z?w?z?2Re?z?w??w．

222

∵z?w??z?w???z?w???z?w??z?w?

2?z2?z?w?w?z?w2

2?z2?2Rez?w?w2??2．从而得证．

∴z?w?z?w?2z?w?22?

几何意义：平行四边形两对角线平方的和等于各边的平方的和．

7.将以下复数表示为指数形式或三角形式

3?5i7i?1;i;?1;?8π(1?3i);2π2π??cos?isin??.

99??3①解：

3?5i7i?1??3?5i??1?7i?

?1?7i??1?7i??175?ei???38?16i50?19?8i25其中??π?arctan819．

②解：i?ei??其中??

iπ2π2．

i?e

23③解：?1?eiπ?eπi

④解：?8π?1?3i??16π???π.

∴?8π?1?3i??16π?e3?23πi

2π2π???isin⑤解：?cos?99??2π2π???isin?cos?99??3解：∵?1．

i?π.32π2π??9?isin?1?e?e∴?cos?99??322π3i

8.计算：(1)i的三次根；(2)-1的三次根；(3)3?⑴i的三次根．

13i的平方根.

解：3i??cos??π2?isinπ?3??cos2?2kπ?3π2?isin2kπ?3π2?k56?0,1,2?

3?12i

∴z1?cos

z3?cosπ696?isinπ6?96132?12i．z2?cos3?12i

56π?isinπ??2

π?isinπ??2⑵-1的三次根

解：?1??cosπ?isinπ?3?cos32kπ+π33i

?isin2kπ?π3?k?0,1,2?

∴z1?cos

π3?isinπ3?12?2z2?cosπ?isinπ??1

z3?cos53π?isin53π??12?32i

⑶3?3i的平方根．解：3??223i=6????22??i????12π6?e4

i

∴3?13i??π6?e4i?ππ??2kπ?2kπ??4?isin4??64??cos??22?11π?k?0,1?

iππ??∴z1?6??cos?isin??64?e888??411

9

πi99??z2?6??cosπ?isinπ??64?e8．

88??49.设z?e

i2πnn?1?0,n?2.证明：1?z???z证明：∵z?e

i?2πn∴zn?1，即zn?1?0．

∴?z?1??1?z???zn?1??0

又∵n≥2．∴z≠1

从而1?z?z2+??zn?1?0

11.设?是圆周{z:z?c?r},r?0,a?c?rei?.令

???z?a?L???z:Im???0?,

?b???其中b?e.求出L?在a切于圆周?的关于?的充分必要条件.

解：如下图．

i?

由于L?={z:Im??z?a???b?=0}表示通过点a且方向与b同向的直线，要使得直线在a处与

圆相切，则CA⊥L?．过C作直线平行L?，则有∠BCD=β，∠ACB=90°

故α-β=90°

所以L?在α处切于圆周T的关于β的充要条件是α-β=90°．

12.指出以下各式中点z所确定的平面图形，并作出草图.

(1)argz?π;(2)z?1?z;(3)1?z?i|?2;(4)Rez?Imz;(5)Imz?1且z?2.

解：

(1)、argz=π．表示负实轴．

(2)、|z-1|=|z|．表示直线z=

12．

(3)、1Imz．

解：表示直线y=x的右下半平面

5、Imz>1，且|z|f?(z)?(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?7752223222222

?(3)f(z)?.

3(5z?7)?(3z?8)5(5z?7)2解：f(z)除z=(4)f(z)?外四处可导，且f?(z)??ix?yx?y22??61(5z?7)2.

x?yx?y22.

x?iy?i(x?iy)x?y22解：由于f(z)?x?y?i(x?y)x?y22??(x?iy)(1?i)x?y22?z(1?i)z2?1?iz.

所以f(z)除z=0外四处可导，且f?(z)??

6.试判断以下函数的可导性与解析性.(1)f(z)?xy2?ix2y;

(1?i)z2.

解：u(x,y)?xy2,v(x,y)?x2y在全平面上可微.

?y?x?y,2?u?y?u?x?2xy,??v?y?v?x?u?y?2xy,?v?x?v?y?x2

所以要使得，

??,

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)f(z)?x2?iy2.

解：u(x,y)?x2,v(x,y)?y2在全平面上可微.

?u?x?2x,?u?y?0,?v?x?0,??v?y?v?u?y?y?2y??.

只有当z=0时,即(0,0)处有

?u?x,

?v?y所以f(z)在z=0处可导，在全平面上不解析.(3)f(z)?2x3?3iy3;

解：u(x,y)?2x3,v(x,y)?3y3在全平面上可微.

?u?x?6x,2?u?y?0,?v?x?9y,2?v?y?0

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.(4)f(z)?z?z2.

解：设z?x?iy，则f(z)?(x?iy)?(x?iy)2?x3?xy2?i(y3?x2y)

3232u(x,y)?x?xy,v(x,y)?y?xy

?u?x?3x?y,22?u?y?2xy,?v?x?2xy,?v?y?3y?x22

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，四处不解析.

7.证明区域D内满足以下条件之一的解析函数必为常数.(1)f?(z)?0;

证明：由于f?(z)?0，所以

?u?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.

证明：设f(z)?u?iv在D内解析,则

?u?x?u?y?u?x??(?v)?y??u?x???v?y

???(?v)?x?v?y,???u?y?v?y?

?v?x??

?u?x?v?y??u?y,?u?y??u?y???v?x?v?x?而f(z)为解析函数，所以所以

?v?x???v?x,?v?y??

?v?y?0,即

?u?x?

从而v为常数，u为常数，即f(z)为常数.

(3)Ref(z)=常数.

证明：由于Ref(z)为常数，即u=C1,由于f(z)解析，C-R条件成立。故从而f(z)为常数.

(4)Imf(z)=常数.

?u?x?u?x???u?y?0?0

?u?y即u=C2

证明：与（3）类似，由v=C1得由于f(z)解析，由C-R方程得所以f(z)为常数.

5.|f(z)|=常数.

?u?x?v?x???u?y?v?y?0

?0,即u=C2

证明：由于|f(z)|=C，对C进行探讨.若C=0，则u=0,v=0,f(z)=0为常数.

若C?0，则f(z)?0,但f(z)?f(z)?C2，即u2+v2=C2则两边对x,y分别求偏导数，有

2u??u?x?2v??v?x?0,2u??u?y?2v??v?y?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所以?

?u?v?v??u??0??x??x所以

?u?x?0,?v?x?0

即u=C1,v=C2,于是f(z)为常数.

(6)argf(z)=常数.

证明：argf(z)=常数，即arctan???C,

?u??v?于是

??u?得??u??(v/u)?1?(v/u)???v?x?v?y?2u?(u??2?v?x?x222u(u?v)?v??u)?u(u22?v?y2?v?u?y2)?0

u(u?v)?u????u?????v?x?v?x?v??v??u?x?u?x?v??v??v?x?u?x?u?y??0?0C-R条件→

?0?v?y?0，即

?0解得

?u?x?u?y?u,v为常数，于是f(z)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：由于f(z)解析，从而满足C-R条件.

?u?x?v?x?u?x?2nxy,22?u?y?3my?nx?v?y22

?3x?ly,?v?y?n?l

?2lxy

??u?y???v?x?n??3,l??3m

所以n??3,l??3,m?1.

9.试证以下函数在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,22?u?y??6xy,?v?x?6xy,?v?y?3x?3y22

所以f(z)在全平面上满足C-R方程，四处可导，四处解析.

f?(z)??u?x?i?v?x?3x?3y?6xyi?3(x?y?2xyi)?3z22222.

(2)f(z)?ex(xcosy?ysiny)?iex(ycosy?xsiny).证明：u(x,y)?ex(xcosy?ysiny),?u?x?u?yxxxv(x,y)=e(ycosy?xsiny)四处可微，且

x?e(xcosy?ysiny)?e(cosy)?e(xcosy?ysiny?cosy)?e(?xsiny?siny?ycosy)?e(?xsiny?siny?ycosy)xxxxx

?v?x?v?y?e(ycosy?xsiny)?e(siny)?e(ycosy?xsiny?siny)?e(cosy?y(?siny)?xcosy)?e(cosy?ysiny?xcosy)?u?x??v?yxx

所以,

?u?y???v?x

所以f(z)四处可导，四处解析.

f?(z)??u?xxz?i?v?xz?e(xcosy?ysiny?cosy)?i(e(ycosy?xsiny?siny))xxxxxxx?ecosy?iesiny?x(ecosy?iesiny)?iy(ecosy?iesiny)?e?xe?iye?e(1?z)zz

?x3?y3?i?x3?y3?,?2210.设f?z???x?y?0.?z?0.z?0.

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．

z?0证明.(1)∵limf(z)?而

limu?x,y???x,y???0,0?limu?x,y??iv?x,y?

332?x,y???0,0??x,y???0,0?limx?yx?y2

∵

xy???x?y?1???2222??x?yx?y??x?yx?y3x?y33332∴0≤2≤32x?y

∴

?x,y???0,0?limx?yx?y32232?032

同理∴

?x,y???0,0?limx?yx?y?0?x,y???0,0?limf?z??0?f?0?

∴f(z)在z=0处连续．

(2)考察极限limz?0f(z)?f?0?z

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．

当z沿实轴趋向于零时，z=x，有

lim1xx?0?f?x??f?0???1?i

?u?x?i??v?x,?v?y?i?u?y它们分别为∴

?u?x??v?y,

?u?y???v?x

∴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?f?z?lim33x?1?i??x?1?i?32x?1?i?x?y?0?i1?i

∴lim

不存在．即f(z)在z=0处不可导．

z?0

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?在区域D1内解析．

证明：设f(z)=u(x,y)+iv(x,y)，由于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．

f?z??u?x,?y??iv?x,?y????x,y??i??x,y?，得

???x???x??u?x,?y??x

???y??u?x,?y??y???u?x,?y??y

???v?x,?y??x

???y???v?x,?y??y??v?x,?y??y

???y???y???x故φ(x,y),ψ(x,y)在D1内可微且满足C-R条件从而f?z?在D1内解析

13.计算以下各值

(1)e2+i=e2?ei=e2?(cos1+isin1)

2??i2???x?,??

(2)e3?e3?eπ?i3??π??π???e3??cos????isin?????e3?3??3???22?13????22?i??(3)Reex?x?iy2?y2?2

?yx?y22?Ree?xx?y2i?e??2x2x?y?Re?e??x?yy????????cos??2?isin?2??x2?y2????x?y????????ex?y22y???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，试探讨f(z)=z+e的极限．解：令z=reiθ，

对于?θ，z→∞时，r→∞．

??故lim?rei??ere??lim?rei??ercos??isin????．

i?z

r??r??所以limf?z???．

z??

15.计算以下各值．

(1)ln??2?3i?=ln13?iarg??2?3i??ln13?i?π?arctan??3??2?(2)ln?3?3i??ln23?iarg?3?3i??ln23?i????π?π??ln23?i6?6

(3)ln(ei)=ln1+iarg(ei)=ln1+i=i

(4)ln?ie??lne?iarg?ie??1?π2i

16.试探讨函数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除负实轴及原点外四处连续．设z=x+iy，g(z)?|z|?u?x,y???u?x?v?x22x?y?u?x,y??iv?x,y?

22x?y,v?x,y??0在复平面内可微．

?12?12?x2?y2??2x?xx?y22?u?y?yx?y22

?0?v?y?0

故g(z)=|z|在复平面上四处不可导．

从而f(x)=|z|+lnz在复平面上四处不可导．f(z)在复平面除原点及负实轴外四处连续．

17.计算以下各值．(1)?1?i??e?e?eln21?i?eπ4ln?1?i?1?i?e2i??2???1?i??ln?1?i??e?1?i????ln?2?π4?i?2kπi??

?π4π4i?ln?eπ4?2kπln2??2kπ?πi??ln?4ln2??2kπ??π??cos??ln?4???π??cos??ln?4?5??π2??isin??ln??4??π2??isin??ln??4??2??????2?????2?e2kπ?π4(2)??3??e?e?35?eln??3??e5?ln??3?

5??ln3?i?π?2kπi??e5ln3?5i?π?2kπ5i5?ln3?cos?2k?1?π5?isin?2k?1?π5?5?isin?2k?1?π5???5??cos?2k?1?π??i(3)1?i?eln1?e?iln1?e?i?ln1?i?0?2kπi

?e?i??2kπi??e2kπ1?i?1?i?(4)???2?1?i?e?1?i?ln???2??e?1?i?ln???1?i??2??e?e?π???1?i???ln1?i?????2kπi??4???π4π4π?e?2kππ??1?i???2kπi?i??4?π??i?2kπ???4?2kπi?i?2kπ??e4?e

π?e4π?2kπ?π?π????cos?isin????4?4????22????22?i???e4?2kπ

18.计算以下各值(1)cos?π?5i????e?5ei?π?5i??e2?5?i?π?5i??5e5iπ?5?e2?5?iπ?5

??ch55?e??1?2??e?e2??e?e2i?5(2)sin?1?5i????ei?1?5i??e2i2i?i?1?5i??e?e2i?i?5

5?5e?cos1?isin1??e??cos1?isin1?e?e25?5?sin1?i?e?e25?5cos1i?3?i?e?e?i?3?i?(3)tan?3?i??sin?3?i?cos?3?i??ei?3?i?sin6?isin22i????i3?i22?e2?ch1?sin3?2i

(4)sinz?212i2??e2?y?xi?ey?xi??sinx?chy?icosx?shy2

22?sinx?chy?cosx?shy222222?sinx??chy?shy???cosx?sinx??shy2?sinx?shy22(5)arcsini??iln?i?1?i2???iln?1?2???i?ln?2?1??i2kπ???????????i?ln?2?1??i?π?2kπ??k?0,?1,?

(6)arctan?1?2i???ln2?kπ?i1?i?1?2i?i?21????ln???i?1?i?1?2i?2?55?arctan2?i4?ln5

1219.求解以下方程

(1)sinz=2．

解：z?arcsin2?ln?2i?3i???ln???2?3?i??

i???i?ln?2??1???3???2k??πi??2??3?,k?0,?1,?11????2k??π?iln?2??2?(2)ez?1?3i?0

解：ez?1?3i即z?ln?1?3i??ln2?i1???ln2??2k??πi3??π3?2kπi

(3)lnz?π2i

π解：lnz?π2i即z?e2?i

i(4)z?ln?1?i??0解：z?ln?1?i??ln2?i?

20.若z=x+iy，求证(1)sinz=sinxchy+icosx?shy证明：sinz??e?e2i12i.?eiz?izπ4?2kπi?ln1??2??2k??πi．?4??ei?x?iy??e2i??x?yi??i

?y?xi?ey?xi??sinx?chy?icosx.shy(2)cosz=cosx?chy-isinx?shy证明：cosz????e?e21212iz?iz?12??ei?x?yi??e?i?x?yi??

?e?y?xi?ey?xi??e?y??cosx?isinx??ey.?cosx?isinx??y?y?yy??e?e?.cosx??isinx.??2?e?e2?cosx.chy?isinx.shy(3)|sinz|=sinx+shy证明：sinz?sinz2222

12i?e?y?xi?ey?xi??sinx?chy?icosx?shy

2222?sinxchy?cosx.shy222222?sinx?chy?shy???cosx?sinx?shy

?sinx?shy22(4)|cosz|2=cos2x+sh2y

证明：cosz?cosxchy?isinxshy

cosz2?cosx.chy?sinx.shy222222?cosx?chy?shy???cosx?sinx?.shy2222

?cosx?shy22

21.证明当y→∞时，|sin(x+iy)|和|cos(x+iy)|都趋于无穷大．证明：sinz?

∴sinz?e?y?xi12i?eiz?e?iz??1??e?y?xi?ey?xi?

2i12?e?y?y?xi?ey?xi

y?e≥ey?xi?e

而sinz12?e?y?xi-y

?ey?xi??1?e?y2y

y?e?

当y→+∞时，e-y→0，ey→+∞有|sinz|→∞．当y→-∞时，e→+∞，e→0有|sinz|→∞．

12e?y?xi同理得cos?x?iy???ey?xi≥12?e?y?ey?

所以当y→∞时有|cosz|→∞．

习题三

1.计算积分?(x?y?ix)dz,其中C为从原点到点1+i的直线段.

C2解设直线段的方程为y?x,则z?x?ix.0?x?1

??x?y?ix?dz???x?y?ix?d(x?ix)2201故

C??10ix(1?i)dx?i(1?i)?213x310?i3(1?i)?i?13

2.计算积分?(1?z)dz，其中积分路径C为

C(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

??1?z?dz???10C1?x??ix(d?x)ix?i(2)设z?x?ix2.0?x?1

??1?z?dz?C?10?1?x?ix2?d(x?ix)?22i3

3.计算积分?zdz，其中积分路径C为

C(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

?Czdz??1?1ydiy?i?ydy?i

?11(2)设z?ei?.?从

?i?3?2?到

?2

?Czdz??23?21de?i?32?de2i??2i

(3)设z?ei?.?从

?i?3?2到

?2

?Czdz???1de223?2i

6.计算积分??解

C?z?e?sinz?dz,其中C为

zz?a?0.

???zC?e?sinz?dz?z??Czdz???Ce?sinzdz

zz∵e?sinz在z?a所围的区域内解析

z∴??e?sinzdz?0

C从而

???zC?e?sinz?dz?z2?0??Czdz??2?0adaei?

?ai?2ed??0i?故??

C?z?e?sinz?dz?0

z7.计算积分??1z(z?1)2Cdz,其中积分路径C为

12（1）

C1:z?12（2）C212:z?32

（3）C13:z?i?（4）C4:z?i?32

解：（1）在

z?所围的区域内，

z(z?1)112只有一个奇点z?0.

??1z(z?1)2Cdz???C1(1z?12z?i?1?2z?i?)dz?2?i?0?0?2?i

（2）在C2所围的区域内包含三个奇点z?0,z??i.故

??1z(z?1)2Cdz???C2(1z?12z?i?1?12z?i?1)dz?2?i??i??i?0

（3）在C2所围的区域内包含一个奇点z??i,故

??1z(z?1)2Cdz???C3(1z?12z?i?1?12z?i?1)dz?0?0??i???i

（4）在C4所围的区域内包含两个奇点z?0,z?i,故

??

1z(z?1)2Cdz???C4(1z?12z?i?1?12z?i?1)dz?2?i??i??i

10.利用牛顿-莱布尼兹公式计算以下积分.(1)

???2i0zcosdz(2)

210??0?iedz(3)

?z?ii1(2?iz)dz

2(4)

?iln(z?1)z?1??2i01dz(5)

?z?sinzdz(6)

?1?tanzcosz21dz

解(1)?(2)?0??iz1zcosdz?sin222?z0??i??2i0?2ch1

edz??e?z??2

(3)

?i1(2?iz)dz?21i?ii1(2?iz)d(2?iz)?2113?(2?iz)i3i1??113?i3

(4)

?iln(z?1)z?11dz??1ln(z?1)dln(z?1)?12?ln(z?1)12i11?2??(?3ln2)

842(5)

?10z?sinzdz???zdcosz??zcosz0110?0coszdz?sin1?cos1

i1(6)

?i1?tanzcosz21dz??i1seczdz?2?i1secztanzdz?tanz2?12tanz2i111?????tan1?tan21?th21??ith1?22?e2z

11.计算积分??(1)

z?i?1Cz?1dz,其中C为

(2)z?i?1(3)e2zz?2

ezz?i解(1)

??e2Cz?1dz?dz???ezC(z?i)(z?i)ezdz?2?i?zz?i??ei

???e

?iz(2)(3)

??Cz?1??C(z?i)(z?i)dz?2?i?ez?iz??i??e2zCz?1dz???e2zC1z?1dz???e2zC2z?1dz??e??ei?i?2?isin1

16.求以下积分的值,其中积分路径C均为|z|=1.

ezz(1)

??Cdz(2)5??coszz3Cdz(3)

??C2dz,z?102(z?z0)2tanz解(1)

??ezzCdz?52?i4!(e)z(4)z?0??i12

(2)

??coszz3Cdz?2?i2!(cosz)(2)z?0???i

(3)

??C2dz?2?i(tanz)'2(z?z0)tanzz?z0??isec2z02

17.计算积分??(1)中心位于点z1(z?1)(z?1)33Cdz,其中积分路径C为

?1,半径为R?2的正向圆周

?2(2)中心位于点z??1,半径为R的正向圆周

解：(1)∴??(2)

C内包含了奇点z1dz?3?1

1(z?1)(2)z?12?i2!C(z?1)(z?1)3()3?3?i8

C内包含了奇点z1(z?1)(z?1)3??1,2?i2!1(z?1)3?i8∴??

Cdz?3()3(2)z??1??

19.验证以下函数为调和函数.(1)??x?6xy?3xy?2y;(2)??ecosy?1?i(esiny?1).xx3223

解(1)设w?u?i?,u∴?u?x2?x3?6xy?3xy22?2y3??0

2?3x?12xy?3y

2?u?y??6x?6xy?6y

2?u?x22?6x?12y

?u?y22??6x?12y

从而有?u?x22??u?y22?0,w满足拉普拉斯方程,从而是调和函数.

(2)?u?x设w?u?i?,u?ex?cosy?1??ex?siny?1

x∴

?e?cosy

?u?y?u?y22x??e?siny

?u?x22?e?cosy

x??e?cosy

x从而有?u?x???x22??u?yx22?0,u满足拉普拉斯方程,从而是调和函数.

?e?siny

???y2x?e?cosy

???x222?e?siny

x???y2??siny?e

x???x2????y22?0,?满足拉普拉斯方程,从而是调和函数.

20.证明:函数u?x?y,??22xx?y?u?x2222都是调和函数,但f(z)?u?i?不是解析函数

证明:

?u?x?2x

?u?y??2y?2

?u?y22??2

∴

?u?x22??u?y222?0,从而u是调和函数.

22???x???x22?y?x2(x?y)22

???y?(x???y22?2xy2

22?y)233??6xy?2x(x?y)2233?6xy?2x(x?y)22

∴

???x22????y22?0,从而?是调和函数.

但∵

?u?x????y

?u?y?????x

∴不满足C-R方程,从而f(z)?u?i?不是解析函数.

22.由以下各已知调和函数,求解析函数f(z)?u?i?(1)u?x2?y2?xy(2)u?yx?y22,f(1)?0

???x解(1)由于所以

?u?x?2x?y????y

?u?y??2y?x??

????(0,0)??x2(x,y)?u?ydx??u(x,y)x?xdx?y(2x?y)dy?Cdy?C??(2y?x)dx?(2x?y)dy?C??0?0(0,0)?x2?y222?2xy?Cf(z)?x?y?xy?i(?2x22?y22?2xy?C)

令y=0,上式变为f(x)?x?i(2x22?C)

从而f(z)?z?i?2z22?iC

?u?yx?y2222(2)

?u?x??2xy(x?y)222

?(x?y)2

用线积分法，取（x0,y0）为(1,0)，有???1x?(x,y)(1,0)(??u?yxdx??u?xdy)?C??xxx21dx?x?42yydy?C0(x2?y2)2?1?x?y22xy?2?1?C20x?y

f(z)?yx?y22?i(xx?y22?1?C)

由f(1)?0.，得C=0?1??f?z??i??1?

?z?

23.设p(z)?(z?a1)(z?a2)?(z?an),其中ai(i?1,2,?,n)各不一致，闭路C不通过

a1,a2,?,an,证明积分

12πi??Cp?(z)p(z)dz

等于位于C内的p(z)的零点的个数.

证明:不妨设闭路C内P(z)的零点的个数为k,其零点分别为a1,a2,...ak

nnk??2πi?11P?(z)P(z)1Cdz???2πi11?(z?ak?2C)?(z?a1)?(z?ak)?...(z?a1)...(z?an?1)k?3(z?a1)(z?a2)...(z?an)1z?a2Cdz??2πiCz?a1dz?12πi??2πi??Cdz?...?1??2πi??C11z?andz1Cdz?1?1?...?1??????k个1z?ak?1dz?...?2πiz?an?k

24.试证明下述定理(无界区域的柯西积分公式):设f(z)在闭路C及其外部区域D内解析，且limf(z)?A??，则

z????f(z)?A,d????A,2πiC??z?1f(?)z?D,z?G.

其中G为C所围内部区域.

证明：在D内任取一点Z，并取充分大的R，作圆CR:则f(z)在以C及f(z)?12πi[??CRz?R，将C与Z包含在内

为边界的区域内解析，依柯西积分公式，有

d?-??f(?)d?]

f(?)CR??zC??z由于

f(??z)??zf(?)在??R上解析，且

???lim????z?limf(?)????11?z?limf(?)?1????所以，当Z在C外部时，有f(z)?A???2πi1f(?)C??zd?

即

12πi??f(?)C??zd???f(z)?A

设Z在C内，则f(z)=0，即

12πif(?)d??f(?)d?]

0?[??1CR??z??C??z故有：2πi

??f(?)C??zd??A

习题四

????1.复级数?an与?bn都发散,则级数?(an?bn)和?anbn发散.这个命题是否成立?为

n?1n?1n?1n?1什么?

??答.不一定．反例:

??n?1?an??nn?11?i1n2??,?bn?n?1?n?1?1n?i1n2发散

但?(an?bn)?n?1??n?n?1i?2n2收敛

?(an?1?bn)??n?12n发散

???n?1anbn??[?(n?11n2?1n4)]收敛.

2.以下复数项级数是否收敛,是绝对收敛还是条件收敛?

iπ?(1)?n?1?1?i2n?1?nin(2)?(n?1?1?5i2cosin2n?)(3)

n?n?1enn

(4)

?lnn(5)?n?1?

n?nn?0解(1)

??n?11?i2n?1?n??n?1?1?(?1)?in2n?1??n?11n?(?1)n?i

由于?n?11n发散,所以?n?11?in发散

?(2)?n?11?5i2n???n?1n(262)发散1252n又由于lim(n???1?5i2n)?lim(n???i)?0

n所以?(n?11?5i2πi)发散

iπ?(3)

?n?1en?n??n?11n?发散,又由于?n?1enn?cosπn?isinnπn????n?1?n?11n(cosπn?isinπn)收

敛,所以不绝对收敛.

?(4)

?n?1in?lnn??lnn

n?111由于

lnn?1n?1

所以级数不绝对收敛.

?又由于当n=2k时,级数化为?k?1(?1)kln2k收敛

?当n=2k+1时,级数化为?k?1(?1)kln(2k?1)也收敛

所以原级数条件收敛

?(5)

?n?0cosin2n???n?012n?e?e2n?n?1??()?22n?02en1??n?0(12e)

n?其中?()发散,?(n?0e?n12e2)收敛

nn?0所以原级数发散.

??2?23.证明:若Re(an)?0,且?an和?an收敛,则级数?an绝对收敛.

n?1n?1n?12222证明:设an?xn?iyn,an?(xn?iyn)?xn?yn?2xnyni

??由于?an和?an2收敛

n?1n?1???2?所以?xn,?yn,?(xn?yn),?xnyn收敛

n?1n?1n?1n?1又由于Re(an)?0,

xn?limxn?0所以xn?0且limn??n??2当n充分大时,xn?xn

?2所以?xn收敛

n?12an2?xn?yn?2xn?(xn?yn)

?2n22222?22而?2x收敛，?(xn?yn)收敛n?1n?1?所以?ann?12?收敛，从而级数?an绝对收敛.

n?12

?4.探讨级数?(zn?0n?1?z)的敛散性

n

n解由于部分和sn??(zk?0k?1?z)?zkn?1?1，所以，当z?1时,sn??1

当z?1时,sn?0，当z??1时,sn不存在.

i?当z?e而??0时（即z?1,z?1），cosnθ和sinnθ都没有极限,所以也不收敛．

当z>1时,sn??.

?故当z?1和z?1时,

??(zn?0n?1?z)收敛.

n5.幂级数?Cn(z?2)能否在z=0处收敛而在z=3处发散.

n?0n解：设limn??Cn?1Cn??，则当z?2?1?时，级数收敛，z?2?1?时发散.

若在z=0处收敛,则

1?1?2

若在z=3处发散,则

??1

?显然矛盾,所以幂级数?Cn(z?2)不能在z=0处收敛而在z=3处发散

n?0n

6.以下说法是否正确?为什么?

(1)每一个幂级数在它的收敛圆周上四处收敛.

(2)每一个幂级数的和函数在它的收敛圆内可能有奇点.

答:(1)不正确,由于幂级数在它的收敛圆周上可能收敛,也可能发散.(2)不正确,由于收敛的幂级数的和函数在收敛圆周内是解析的.

?n?7.若?Cnz的收敛半径为R,求?n?0n?0Cnbnz的收敛半径。

nCn?1b解：由于limn??n?1Cnbn?limCn?1Cnn???1b?11Rb

所以R??R?b

?n8.证明：若幂级数?anz的系数满足limn??n?0nan??，则

(1)当0?????时,R?(2)当??0时,R???(3)当????时,R?0

?1?

证明：考虑正项级数?anzn?0n?a1z?a2zn2?...?anzn?...

由于limnn??anzn?limnn??an?nz???z，若0?????，由正项级数的根值判别法知，

n当??z?1时，即z?1??时，?anz收敛。当??z?1时，即z?n?01?时，anzn2不能

趋于零,limn??nanzn?1级数发散.故收敛半径R?1?.

当??0时,??z?1,级数收敛且R???.若????

9.求以下级数的收敛半径，并写出收敛圆周。

?n,对?z?0,当充分大时,必有anzn2不能趋于零,级数发散.且R?0

(1)(3)

?n?0(z?i)np?(2)

?2n?12n?nn?0p?zn

??n?0?(?i)n?1?z2n?1

(4)

?(n)n?0in?(z?1)n(n?1)

解:(１)n??(n?1)?R?1lim1p1np?lim(n??nn?1)?lim(1?n??p1n?1)?1p

收敛圆周

z?i?1

lim(n?1)npp(2)

n???1z?1

n?1R?1所以收敛圆周

(3)记fn(z)?(?i)由比值法,有

limfn?1(z)fn(z)?lim?2n?12n?z2n?1

(2n?1)?2?z(2n?1)?22n?1n2n?12n?1n??n???12z2?z

要级数收敛,则

z?2级数绝对收敛,收敛半径为R?2所以收敛圆周

z?2inn(n?1)(4)记fn(z)?()?(z?1)n

n(n?1)limnn??fn(z)?limn(z?1)nn??n?limz?1nn?1n?????,??????若????1若????1

所以

z?1?1时绝对收敛,收敛半径R?1

收敛圆周

z?1?1

10.求以下级数的和函数.

??(1)

?(?1)n?1n?1?nz(2)

n

?(?1)n?0n?z2n(2n)!

解:(1)limCn?1Cn?limn?1n?1n??n??

故收敛半径R=1,由逐项积分性质，有：

??0z??(?1)nznn-1dz??n?1(?1)z?nnz1?z

n?1所以

??n?1(?1)?nznn-1?(z1?z)??1(1?z)2,z?1

于是有：

???n?1(?1)n?1?nz??z?(?1)?nzn?1nnn?1??z(1?z)2z?1

?s(z)?(2)令：

?limCn?1Cn?(?1)n?0n?z2n(2n)!

1n???limn??(2n?1)(2n?2)?0.

故R=∞,由逐项求导性质

?s?(z)??(?1)n?1n?z2n?1(2n?1)!

?s??(z)??(?1)n?1n?z2n?2?(2n?2)!??(?1)m?0m+1?z2m?(2m)!(m?n?1)???(?1)?n?0nz2n(2n)!

由此得到s??(z)??s(z)即有微分方程s??(z)?s(z)?0

故有：s(z)?Acosz?Bsinz，A,B待定。

?由S(0)?A?[?(?1)?n?0nz2n(2n)!?]z?0?1?A?1

s?(0)??sinz?Bcosz?[?(?1)?n?1nz2n?1(2n?1)!]z?0?0?B?0

所以

??(?1)n?0n?z2n(2n)!?cosz.R???

???n11.设级数?Cn收敛，而?Cn发散，证明?Cnz的收敛半径为1

n?0n?0n?0?证明：由于级数?Cn收敛

n?0设

limCn?1ZCnZn?1nn????z.

若

??Cn?0nz的收敛半径为1

n则z?1?

??1现用反证法证明

limCn?1???1?若

0???1n??Cn则z?1，有

，即?Cn收敛，与条件矛盾。

n?0??n若

??1则z?1，从而?Cnz在单位圆上等于?Cn，是收敛的，这与收敛半径的概念

n?0n?0矛盾。

综上述可知，必有R?1?1??1，所以

?

?12.若?Cn?0nzn在

z0点处发散，证明级数对于所有满足

?z?z0点z都发散.

证明：不妨设当

z1?z0时，?Cn?0znn在

z1?处收敛

?则对

?z?z1，

??Cn?0nzn绝对收敛，则

?Cn?0nzn在

点

z0处收敛

nz?z0所以矛盾,从而?Cnz在处发散.

n?0

13.用直接法将函数ln(1?e解:由于ln(1?e)?ln(?z?z4)在z?0点处展开为泰勒级数,(到z项),并指出其收敛半径.

1?eezz)

奇点为zk?(2k?1)πi(k?0,?1,...)

所以R?π又

ln(1?e?z)z?0?ln2

[ln(1?e?z)]???e?z?zz?01?e??12122[ln(1?e?z)]????e?z?z(1?e?z)2z?0??

[ln(1?e?z)]?????e?e?z?2z(1?ee?z)3z?0?0

?2z[ln(1?e?z)](4)?(1?4e?z?z?e)4)z?0(1?e??12

3于是，有展开式

ln(1?e?z)?ln2?12z?12!22z?214!23z?...,R?π4

(z?1)4

114.用直接法将函数1?z2在解：又

f(z)?11?z2z?1?2点处展开为泰勒级数,(到

R?2项)

z??i1为1?z2的奇点，所以收敛半径

12

,f(1)?f?(z)??2z(1?z)22,f?(1)??1212

f??(z)??2?6z(1?z)223,f??(1)?f???(z)?24z?24z(1?z)243,f???(1)?0

24f(4)(z)?24?240z?120z(1?z)25,f(4)(1)?0

于是，f(z)在z?1处的泰勒级数为

11?z2?12?12(z?1)?14(z?1)?234!(z?1)?...,R?42

15.用间接法将以下函数展开为泰勒级数，并指出其收敛性.

1(1)2z?3分别在

arctanzz?0和

z?13sinz在z?0处处(2)

(3)在

z?0处(4)(z?1)(z?2)在z?2处

zz?0(5)ln(1?z)在处

1解（1）2z?312z?31??13?2z??13?1?123z??13???(n?023z),z?n32

?2z?2?1??12(z?1)?1??11?2(z?1)????2(z?1),z?1?n?0nn12

(2)

sinz??(2n?1)!zn?0?(?1)n2n?1?z?z33!?z55!?...

sinz?33?4(?1)?n32n?1n?0(2n?1)!1z2n?1,z??

?arctanz?(3)

?z01?z2dz

?z??i为奇点，?R?1z0arctanz??11?zdz?2??0z??(?1)zdz?n2n?n?0(?1)?n12n?1?z2n?1,z?1n?0

(4)

1(z?1)(z?2)1?n?1z?1?1z?21??1z?2?3?1z?2?4?13?1?1z?23?14?1?1z?24??(?1)?3n?0?n?(z?23?)?1n(?1)?4n?0nn?(z?24)n

?n?0(?1)?(13n?14n?1)(z?2),z?2?3(5)由于从z??1沿负实轴ln(1?z)不解析所以,收敛半径为R=1

[ln(1?z)]??11?zz?n???(?1)n?0nn?z

?nln(1?z)???(?1)0n?0?zdz??(?1)n?0n?1n?zn?1,z?1

16.为什么区域z?R内解析且在区间(?R,R)取实数值的函数f(z)展开成z的幂级数时，展开式的系数都是实数？

答：由于当z取实数值时，f(z)与f(x)的泰勒级数展开式是完全一致的，而在x?R内，f(x)的展开式系数都是实数。所以在z?R内，f(z)的幂级数展开式的系数是实数.

17.求

f(z)?2z?12z?z?2的以

z?0为中心的各个圆环域内的罗朗级数.

z?0z?1z2??2解:函数f(z)有奇点1与,有三个以为中心的圆环域,其罗朗级数.分别为：

在z?1内，f(z)?2z?1z?z?22=1z?1?1z?2????z?n?0n1?n(?1)?2n?0zn()2

???n?0((?1)?n12n?1?1)zn

119.在解:令

1?z???1?z内将f(z)?e展开成罗朗级数.

t?,1?z则

1f(z)?e?1?t?t12!?t?213!?t?...3

而

t?1?z???内展开式为在1?z111?z??1z?11?1z??1z?(1?1z?1z2?...)

所以,代入可得

f(z)?1??1?1z?1z12z2?(1??1z3??1z2?...)?1?112!z195?(1??...1z?1z2?...)?...216z

24z4120z

20.有人做以下运算,并根据运算做出如下结果z1?zzz?1?z?z?z?...23

?1?1z?1z2?...

z由于1?z1z3?zz?11z?0,所以有结果

...??1z2??1?1?z?z?z?...?023

你认为正确吗?为什么?

z答:不正确,由于1?zz1z1z2?z?z?z?...23要求z?1

而1?z?1???...要求z?1

zz?11z6z所以,在不同区域内1?z

??