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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)
f(z)?(4) 解:因为
f(z)?x?yx?y?ix2?y2x2?y2.
x?y?i(x?y)x?iy?i(x?iy)(x?iy)(1?i)z(1?i)1?i????2x2?y2x2?y2x2?y2zz.所以f(z)除z=0外处处可导,且
f?(z)??(1?i)z2.
6. 试判断下列函数的可导性与解析性.
22f(z)?xy?ixy; (1)
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
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.
22(2) f(z)?x?iy.
22u(x,y)?x,v(x,y)?y解:在全平面上可微.
?u?2x,?x?u?0,?y?v?0,?x?v?2y?y
?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) ;
33解:u(x,y)?2x,v(x,y)?3y在全平面上可微.
?u?6x2,?x?u?0,?y?v?9y2,?x?v?0?y
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)2?x3?xy2?i(y3?x2y) u(x,y)?x3?xy2,v(x,y)?y3?x2y
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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)
?u?3x2?y2,?x?u?2xy,?y?v?2xy,?x?v?3y2?x2?y
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?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
?u?v???y?x
?u?u?,?x?y而f(z)为解析函数,所以
?v?v??,?x?x?v?v??,?y?y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
?u?u??0?x?y
?u?u??0?x?y即u=C2
?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
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
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
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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)
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 所以即u=C1,v=C2,于是f(z)为常数.
?v?0?x
(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)
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?x?x?y?y解得,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?u?2nxy,?3my2?nx2?x?y ?v?3x2?ly2,?x?u?v??n?l?x?y?v?2lxy?y
?u?v???n??3,l??3m?y?x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?3x2?3y2,?x?u??6xy,?y?v?6xy,?x?v?3x2?3y2?y 13 / 66
复变函数与积分变换(修订版)课后答案(复旦大学出版社)
f?(z)??u?vxx?i?3x2?3y2?6xyi?3(x2?y2?2xyi)?3z2f(z)?e(xcosy?ysiny)?ie(ycosy?xsiny). ?x?x.(2)
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?v?ex(ycosy?xsiny)?ex(siny)?ex(ycosy?xsiny?siny)?x?u?v?u?v?v????ex(cosy?y(?siny)?xcosy)?ex(cosy?ysiny?xcosy)?x ?y所以?x?y, ?y所以f(z)处处可导,处处解析.
f?(z)??u?x?i?v?x?ex(xcosy?ysiny?cosy)?i(ex(ycosy?xsiny?siny))?excosy?iexsiny?x(excosy?iexsiny)?iy(excosy?iexsiny)?ez?xez?iyez?ez(1?z)10. 设
?x3?y3?i?x3?y3?f?z????x2?y2,z?0.??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.
证明.(1)∵limz?0f(z)??x,ylim???0,0?u?x,y??iv?x,y?
而?x,ylim???0,0?u?x,y???x,ylimx3?y3???0,0?x2?y2
x3?y3?xy?∵
x2?y2??x?y????1?x2?y2??
0≤x3?y3≤3∴x2?y22x?y
x3?y3lim∴?x,y???0,0?x2?y2?0
limx3?y3?同理?x,y???0,0?x2?y20 ∴?x,ylim????0,0?f?z??0?f0?
∴f(z)在z=0处连续.
(2)考察极限limf(z)?f?0?z?0z
当z沿虚轴趋向于零时,z=iy,有
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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)
3?1?i?11?y???lim?fiy?f0?lim??1?i???y?0iyy?0iy?y2.
当z沿实轴趋向于零时,z=x,有 lim1?f?x??f?0???1?ix?0x
?u?v?i?,?x?x?u?v???y?x?v?u?i?y?y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.
?u?v?,?x?y?u?v???y?x所以u(x,y),v(x,y)在D内可微且满足C-R方程,即.
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
,得
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fz在D1内解析
13. 计算下列各值
(1) e2+i=e2?ei=e2?(cos1+isin1) (2)(3)
e2??i3?e?e23π?i32?1??π?3??π??3?e??cos????isin?????e???i??22? ?3????3?23 15 / 66