复变函数与积分变换 复旦大学出版社 习题二 答案

习题二

1. 求映射

w?z?1z下圆周|z|?2的像.

w?u?iv解：设z?x?iy,

u?iv?x?iy?1则

2x?iy?x?iy?x?iyx?y2?x?xx?y2?i(y?2yx?y2)

2 因为所以

x?y?45422,所以

34yu?iv?54x?34yi

u?u54xv??,

x?,y?v34

?2u5222u所以??542?v??342即???v3222???1，表示椭圆.

i?22. 在映射w?z下，下列z平面上的图形映射为w平面上的什么图形，设w??e或

w?u?iv.

4； （2） （1）

(3) x=a, y=b.(a, b为实数)

0?r?2,??π0?r?2,0???π4;

解：设

w?u?iv?(x?iy)?x?y?2xyi222

22所以u?x?y,v?2xy.

(1) 记w??e，则

0???4,??π2.i?0?r?2,??π4映射成w平面内虚轴上从O到4i的一段，即

(2) 记

w??ei?0???π4,0?r?20???4,0???π2.，则映成了w平面上扇形域，即

(3)

记w?u?iv，则将直线

x=a映成了

u?a?y,v?2ay.22即

v?4a(a?u).222是以原点为焦

22点，张口向左的抛物线将y=b映成了u?x?b,v?2xb.

222 即v?4b(b?u)是以原点为焦点，张口向右抛物线如图所示.

3. 求下列极限.

lim11?z1t2 (1) 解：令

z??;

z?,则z??,t?0.

2lim11?z于是

z???limt?0t221?t?0.

limRe(z)z(2)

z?0;

Re(z)z?xx?iy解：设z=x+yi，则

limRe(z)z?limx有

z?0x?0y?kx?0x?ikx?11?ik

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)2（3）

z?i；

limz?iz(1?z)2解：

z?ilimz?iz(i?z)(z?i)=

z?i?limz?i1z(i?z)??12.

limzz?2z?z?2z?12（4）

z?1.

?(z?2)(z?1)(z?1)(z?1)?z?2,z?1

zz?2z?z?2解：因为

limz?12zz?2z?z?2z?12所以

z?1?limz?2z?1z?1?32.

4. 讨论下列函数的连续性： (1)

?xy,?22f(z)??x?y?0,?z?0,z?0;

xyx?y22limf(z)?解：因为

z?0(x,y)?(0,0)lim,

若令y=kx,则

(x,y)?(0,0)limxyx?y22?k1?k2,

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)

?x3y,?f(z)??x4?y2?0,?z?0,z?0.3

x30?xyx?yxyx?y42342?y2解：因为

lim2xy?x2,

所以

(x,y)?(0,0)?0?f(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. (1)

f(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).

f(z)?z?2(z?1)(z?1)2(2) .

2解：因为f(z)为有理函数，所以f(z)在(z?1)(z?1)?0处不可导.

从而f(z)除z??1,z??i外可导.

(z?2)?(z?1)(z2?1)?(z?1)[(z?1)(z2f?(z)??1)]?(z?1)2(z2?1)2??2z3?5z2?4z?3(z?1)2(z2?1)2

3z?8(3)

f(z)?5z?7.

z=7f?(z)?3(5z?7)?(3z?8)5解：f(z)除

5外处处可导，且

(5z?7)2??61(5z?7)2.

f(z)?x?y?ix?y(4) x2?y2x2?y2.

解：因为

f(z)?x?y?i(x?y)x?iy)iy)(1?i)z(1?i)1?ix2?y2?x?iy?i(x2?y2?(x?x2?y2?z2?z.所以f(z)除z=0外处处可导，f?(z)??(1?i)z2.

6. 试判断下列函数的可导性与解析性. (1) f(z)?xy2?ix2y;

解：

u(x,y)?xy2,v(x,y)?x2y在全平面上可微.

?y?v?x?y2,?u?y?2xy,?x?2xy,?v?y?x2

所以要使得

?u?v?uv?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. (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?2y

且

?u只有当z=0时,即(0,0)处有?x??v?u?y?y???v?y,.

所以f(z)在z=0处可导，在全平面上不解析. (3)

f(z)?2x?3iy33;

33解：u(x,y)?2x,v(x,y)?3y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.

2(4) f(z)?z?z.

解：设z?x?iy，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,2223232

3232

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

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;

?u?u?y?v?v?y证明：因为

f?(z)?0，所以

?x??0,

?x??0.

所以u,v为常数，于是f(z)为常数. (2)

f(z)解析.

f(z)?u?iv??u?x??证明：设

?u?x?u?y?u?x???y?x?v?y在D内解析,则

?(?v)?v?y

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

?v?x??,

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

?u?x????v?y?0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证明：因为Ref(z)为常数，即u=C1,

?u?x???u?y?0?0 即u=C2

?u?y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.

?x?v证明：与（3）类似，由v=C1得

?u?x??v?y?0

因为f(z)解析，由C-R方程得所以f(z)为常数.

?x??u?y?0,即u=C2

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. 若C??u?x0，则f(z)

?v?x?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??2v??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??v??u?u??v?0?x??x所以?

?u

即u=C1,v=C2,于是f(z)为常数.

(6) argf(z)=常数.

所以?x?0,?v?x?0

?v?arctan???C?u?证明：argf(z)=常数，即

u?(u?22,

?v?u?y2?v(v/u)?于是得

1?(v/u)??x?x22u(u?v)2?v??u)?u(u22?v?y2)?0u(u?v)

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

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

?u?v?uv?x??y???y?0解得

?x?，即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：因为f(z)解析，从而满足C-R条件.

?u?x?2nxy,?u?y?3my2?nx2

?v?3x2?ly2?x,?v?y?2lxy

?u?x??v?y?n?l

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

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?3x2?3y2,?u?y??6xy,?v?x?6xy,?v?y?3x2?3y2

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)??u2222?x?i?v?x?3x?3y?6xyi?3(x?y?2xyi)?3z2f(z)?ex(xcosy?ysiny)?iex(ycosy?xsiny).

证明：

u(x,y)?ex(xcosy?ysiny),v(x,y)=ex(ycosy?xsiny)处处可微，且

?u?x?ex(xcosy?ysiny)?ex(cosy)?ex(xcosy?ysiny?cosy)

?ux?y?e(?xsiny?siny?ycosy)?ex(?xsiny?siny?ycosy)?v?x?ex(ycosy?xsiny)?ex(siny)?ex(ycosy?xsiny?siny).(2)

?v?y?e(cosy?y(?siny)?xcosy)?e(cosy?ysiny?xcosy)xx?u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?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)zz10. 设

z?0.z?0.?x3?y3?i?x3?y3?,?22f?z???x?y?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．

limf(z)?z?0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

xy???x?y?1???2222?x?yx?y??? ∵

x?yx?y3x?y333320≤2≤32x?y∴

∴

?x,y???0,0?limx?yx?y3232?0

32同理∴

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

?x,y???0,0?limf?z??0?f?0?∴f(z)在z=0处连续．

limf(z)?f?0?z(2)考察极限

z?0

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它们分别为?x?u??v?y,?i??v?x,?v?y?i?u?y

?u?y∴?x???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?lim33x?1?i??x?1?i?x?y?02x?1?i?3?i1?i

∴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．

?u??v?y,?u?y???v?x所以u(x,y),v(x,y)在D内可微且满足C-R方程，即?xf?z??u?x,?y??iv?x,?y????x,y??i??x,y????x???x．

，得

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

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

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

?y故φ(x,y),ψ(x,y)在D1内可微且满足C-R条件?x

??从而fz在D1内解析

13. 计算下列各值

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

2??i2π?i3(2)(3)

e3?e?e3??π??π???e??cos????isin?????e3?3??3???322?13????22?i??

?Re?e?Re?e?ex?iy2x?y2xx?y22?yx?y22i??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??2x?e?e?2iyi?2?x?iy??e?e?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， (1)

3??ln??2?3i?=ln13?iarg??2?3i??ln13?i?π?arctan??2?

对于?θ，z→∞时，r→∞． 故

lim?rer??i??erei???lim?rei?r???er?cos??isin?????．

所以z??limf?z???．

15. 计算下列各值．

(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)

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???u?x?v?xln?ie??lne?iarg?ie??1?π2ig(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?12?x2?y2???2x?xx?y2?y?yx?y22

?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． 17. 计算下列各值． (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???π2??isin??ln??4??π2??isin??ln??4??2??????2????

?2?e2kπ?π4 (2)

??3??e?e?35?eln??3?5?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??i??ln1?i?0?2kπi?5??cos?2k?1?π?

1?i?e?eln1?i?e?iln1?e(3)

?i??2kπi??e2kπ

?e?1?i?ln???1?i??2?1?i?1?i?(4)???2?1?i?e?1?i?ln???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?eπ?e4π?2kπ?π?π????cos?isin????4?4????22????22?i???e4?2kπ

18. 计算下列各值 (1)

cos?π?5i????e?5ei?π?5i??e2?5?i?π?5i??5e5iπ?5?e2?5?iπ?55?e??1?2??e?e2??e?e2??ch5

?i?5(2)

sin?1?5i????ei?1?5i??e2i2i?i?1?5i??ei?5?e2i5?5e?cos1?isin1??e??cos1?isin1?e?e25?5?sin1?i?e?e25?5cos1

etan?3?i??sin?3?i?cos?3?i??ei?3?i??e?i?3?i?i?3?i?sin6?isin22i??i?3?i?22?e2?ch1?sin3?2i(3)

sinz2(4)

?12i2??e2?y?xi?ey?xi??sinx?chy?icosx?shy222?sinx?chy?cosx?shy222222?sinx??chy?shy???cosx?sinx??shy2?sinx?shyarcsini??iln?i?1?i222(5)

???iln?1?2?k?0,?1,???i?ln?2?1??i2kπ???????????i?ln?2?1??i?π?2kπ??arctan?1?2i???i2ln

1?i?1?2i?i?21????ln???i?1?i?1?2i?2?55?arctan2?i4?ln5?kπ?12(6)

19. 求解下列方程 (1) sinz=2． 解：

z?arcsin2????i?ln?2??1iln?2i?

3i???ln???2?3?i??1???3???2k??πi??2??3?,k?0,?1,?1????2k??π?iln?2??2?z(2)e?1?3i?0

z解：e?1?3i 即

z?ln?1?3i??ln2?iπ3?2kπi1???ln2??2k??πi3??

(3)

lnz?π2i

π2iπlnz?解：

2即z?e?ii

(4)z?ln?1?i??0

z?ln?1?i??ln2?i?π4?2kπi?ln1??2??2k??πi?4?．

解：

20. 若z=x+iy，求证 (1) sinz=sinxchy+icosx?shy 证明：

sinz??e?e2i12i.?eiz?iz?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|2=sin2x+sh2y

证明：

sinz?12i?e?y?xi?ey?xi??sinx?chy?icosx?shy2

sinz?sinxchy?cosx.shy222222?sinx?chy?shy???cosx?sinx?shy2222

?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?12i?eiz?e?iz??1??e?y?xi?ey?xi?2i

?ey?xisinz?12?e?y?y?xi ∴

e?y?xi?eey?xi?ey

22而

当y→+∞时，e-y→0，ey→+∞有|sinz|→∞． 当y→-∞时，e-y→+∞，ey→0有|sinz|→∞． cos?x?iy??1e?y?xisinz≥1?e?y?xi?ey?xi??1?e?yy?e?2同理得

所以当y→∞时有|cosz|→∞．

?ey?xi≥12?e?y?ey?

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