复变函数与积分变换(修订版-复旦大学)课后的习题答案
复变函数与积分变换(修订版)课后答案(复旦大学出版社)
复变函数与积分变换
(修订版)
主编:马柏林
(复旦大学出版社)
——课后习题答案
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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)
习题一
1. 用复数的代数形式a+ib表示下列复数
3513,i,iπ/4. eii;;(2)(43);,,,711iii,,
π,,,iππ2222,,,,?解 4ecosisinii,,,,,,,,,,,,,,,,,442222,,,,,,
35i17i,,,,,,35i1613,?解: ,,,,i7i11+7i17i2525,,,,,,
?解: 2i43i834i6i510i,,,,,,,,,,,,
31i,,,1335?解: ,,,,,=iii1i222,
2.求下列各复数的实部和虚部(z=x+iy)
33za,,,,,,,,,1313ii3n(a,); zi;;;.,,,,za,,,,,22? :?设z=x+iy
222zaxay,,,,,xayxay,,,,ii,,,,,,,,xyaxay,,,,ii,,,,za,,,,,Re,则 ?,,,,,,2222za,,,zaxyaxay,,,,,ii,,,,xay,,,,xay,,,,
zaxy,2,, ( Im,,,22za,,,xay,,,,
?解: 设z=x+iy
33232332Re3zxxy,,Im3zxyy,, ? ?, ( 322,,,,zxyxyxyxyxyxy,,,,,,,,,iii2ii,,,,,,,,,,
222222,,,,,,,,xxyxyyxyxy22i,,,,,,
3223,,,,xxyxyy33i,,
33,,1i323,,,,?解: ? ,,1i312,,,,,,,,,,,,,,,,13133133,,,,,,,,,,,,,,,,,,,,,,288,,
1 ,,,80i1,,8
,,,,,,1i3,,1i3Re1,Im0, ?, ( ,,,,,,,,22,,,,
2332,,3,,,,,,,,,,,13133133i,,,,,,1,,,,,,,,,,1i3,,?解: ? ,,,80i1,,,,,,,828,,
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,,,,,,1i3,,1i3 ?, ( Im0,Re1,,,,,,,,,22,,,,
k,1,,,,nk,2,n?解: ?( i,,k ,knk,,21,,1i,,,,,
knnnk,2 ?当时,,; Rei1,,Imi0,,,,,,,
knnnk,,21 当时,,( Rei0,Imi1,,,,,,,,3.求下列复数的模和共轭复数
1,i ,,,,,2;3;(2)(32);.iii2
?解:( ,,,,,2i415
,,,,,2i2i
?解: ,,33,,,33
?解:( 2i32i2i32i51365,,,,,,,,,,,,
2i32i2i32i2i32i47i,,,,,,,,,,,,,,,,,,,,,,,,
1i,1i2,,,?解: 222
1i,,,11i,,i,, ,, ,,222,,
4、证明:当且仅当zz,时,z才是实数(
zz, 证明:若,设zxy,,i,
2i0y, 则有 xyxy,,,ii,从而有,即y=0 ,,
?z=x为实数(
zxx,, 若z=x,x? ,则(
zz, ?(
命题成立(
zwzw,,?5、设z,w? ,证明:
2zwzwzwzwzw,,,,,,,, 证明? ,,,,,,,,
,,,,,,,,zzzwwzww
22,,,,,zzwzww ,,
22,,,,zwzw2Re,,
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22?zwzw,,,2
22 ,,,,zwzw2
2,,zw,,
?( zwzw,,?
6、设z,w? ,证明下列不等式(
222 zwzzww,,,,,2Re,,
222 zwzzww,,,,,2Re,,
2222zwzwzw,,,,,2 ,,
并给出最后一个等式的几何解释(
222证明:在上面第五题的证明已经证明了( zwzzww,,,,,2Re,,
222下面证( zwzzww,,,,,2Re,,
2zwzwzwzwzw,,,,,,,, ? ,,,,,,,,
22,,,,,,zzwwzw
22 (从而得证( ,,,,zzww2Re,,
2222zwzwzw,,,,,2 ? ,,
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.将下列复数表示为指数形式或三角形式
335,i2π2π,, ;;1;8ii,,,π(13);.cossin,i,,71i,99,,
35i17i,,,,,,35i,?解: ,7i117i17i,,,,,,,
3816i198i17,,8i,,,,,,e其中( ,,,πarctan5025519
πi,,i,e?解:其中( ,,2
πi2i,e
iiππ,,,1ee?解:
2?解:. ,,,,,,8π13i16ππ,,3
2,πi3 ? ,,,,8π13i16πe,,
32π2π,,cosisin,?解: ,,99,,
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32π2π,,解:?( cosisin1,,,,99,,
322π,iπ.3i2π2π,,93 ? cosisin1ee,,,,,,99,,
8.计算:(1)i的三次根;(2)-1的三次根;(3) 的平方根. 33,i
?i的三次根(
解:
ππ12kkπ,,2π3ππ,,322 icossincosisin0,1,2,,,,,ik,,,,2233,,
ππ315531?( z,,,,cosisiniz,,,,,cosπisinπi1266226622
9931 z,,,,,cosπisinπi36622
?-1的三次根
解:
12kkπ+π2ππ,3 3,,,,,,1cosπisinπcosisin0,1,2k,,,,33
ππ13 ? z,,,,cosisini13322
z,,,,cosπisinπ12
5513z,,,,,cosπisinπi 33322
33i,?的平方根(
πi,,224解: 33i=6i6e,,,,,,,,,22,,
ππ1,,2kkπ,,2ππ12,,i ? 4444,,,,33i6e6cosisin0,1,,,,,,,k,,,,22
π11iππ,,844z,,,,,6cosisin6e ? ,,188,,
911πi99,,844z,,,,,6cosπisinπ6e ( ,,288,,
2πin,1n10,,,,zz?9.设. 证明: zn,,e,2
2π,innnz,ez,,10z,1证明:? ?,即(
n,1,,,,zzz,,,,,110? ? 又?n?2( ?z?1
21n,1+0,,,,zzz? 从而
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i,11.设是圆周令 {:},0,e.zrracr,,,,,zc,
za,,,,,, Lz:Im0,,,,,,,b,,,,
i,其中.求出在a切于圆周的关于的充分必要条件. b,eL,,,
解:如图所示(
za,,,Im 因为={z: =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,,,b,,
(过C作直线平行,则有?BCD=β,?ACB=90? LL,,
故α-β=90?
所以在α处切于圆周T的关于β的充要条件是α-β=90?( L,
12.指出下列各式中点z所确定的平面图形,并作出草图.
(1)argz,π;
(2);,zz,1
(3)1|2;,,,zi
(4)ReIm;zz,
(5)Im12.z,,且z
解:
(1)、argz=π(表示负实轴(
1(2)、|z-1|=|z|(表示直线z=( 2
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(3)、1<|z+i|<2
解:表示以-i为圆心,以1和2为半径的周圆所组成的圆环域。
(4)、Re(z)>Imz(
解:表示直线y=x的右下半平面
5、Imz>1,且|z|<2(
解:表示圆盘内的一弓形域。
习题二
1wz,,||2z,z1. 求映射下圆周的像.
zxywuv,,,,i,i解:设则
1ixyxy,uvxyxyxy,,,,,,,,,,,iiii()222222xyxyxyxy,,,,i
53uivxy,,,i22xy,,444 因为,所以
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53ux,vy,,44所以 ,
uvxy,,,5344
22uvuv,,2,,122225353,,,,,,,,4422所以即,表示椭圆.
i,2w,,ewuv,,iwz,2. 在映射下,下列z平面上的图形映射为w平面上的什么图形,设或.
ππ,,,,,,,,,02,r02,0r44 (1); (2);
(3) x=a, y=b.(a, b为实数)
222wuvxiyxyxy,,,,,,,i()2i解:设
22uxyvxy,,,,2.所以
π,,,,02,ri,w,,e4(1) 记,则映射成w平面内虚轴上从O到4i的一段,即
π,,,,,04,.2
ππ,,,,,,,,,,,04,0.0,02ri,w,,e24(2) 记,则映成了w平面上扇形域,即
22222uayvay,,,,2.vaau,,4().wuiv,,(3) 记,则将直线x=a映成了即是以原点为焦点,张口向左的抛
22uxbvxb,,,,2.物线将y=b映成了
222vbbu,,4() 即是以原点为焦点,张口向右抛物线如图所示.
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3. 求下列极限.
1lim2z,,,z1 (1) ;
1z,zt,,,,0t解:令,则.
21t,,limlim022zt,,,0,,11zt于是.
Re()zlimz,0z(2) ;
Re()zx,zxy,i解:设z=x+yi,则有
Re()1zx,,limlimzx,,00,,zxkxki1iykx,,0
显然当取不同的值时f(z)的极限不同 所以极限不存在.
zi,lim2zi,(1)zz,(3) ;
zi,zi,11limlimlim,,,2zi,zizi,,(1)zz,zizziziz()()()2,,,解:=.
zzzz,,,22lim2z,1z,1(4) .
zzzzzzz,,,,,,22(2)(1)2,,,2zzzz,,,,1(1)(1)1解:因为
zzzzz,,,,2223limlim,,2zz,,11zz,,112所以.
4. 讨论下列函数的连续性:
(1)
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xy,,0,z,,22xy,fz(),,,0,0;z,,
xylim()lim,fz22zxy,,0(,)(0,0),xy解:因为,
xyklim,222(,)(0,0)xy,,,1xyk若令y=kx,则,
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)
3,xy,0,z,,42fz(),xy,,,0,0.z,,
33xyxxy0,,,422xy,22xy解:因为,
3xylim0(0),,f42xy,(,)(0,0),xy所以
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,1fzz()(1),,(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,1,fznz()(1),,.
z,2fz(),2(1)(1)zz,,(2) .
2(1)(1)0zz,,,解:因为f(z)为有理函数,所以f(z)在处不可导.
zz,,,,1,i从而f(z)除外可导.
22,,(2)(1)(1)(1)[(1)(1)]zzzzzz,,,,,,,,fz(),222(1)(1)zz,,
32,,,,2543zzz,222(1)(1)zz,,
38z,fz(),57z,(3) .
3(57)(38)561zz,,,7,fz(),,,z=22(57)(57)zz,,5解:f(z)除外处处可导,且.
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xyxy,,fz()i,,2222xyxy,,(4) .
解:因为
(1i),xyxyxyxyxyz,,,,,,,,,,i()ii(i)(i)(1i)(1i)1ifz(),,,,,,fz(),,22222222xyxyxyz,,,zz.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.
22fzxyxy()i,,(1) ;
22uxyxyvxyxy(,),(,),,解:在全平面上可微. ,,,,yuvv22,,,,yxyxyx,2,2,,,,,xyxy 所以要使得
,,uv,,uv,,,,,yx,,xy, ,
只有当z=0时,
从而f(z)在z=0处可导,在全平面上不解析.
22fzxy()i,,(2) .
22uxyxvxyy(,),(,),,解:在全平面上可微. ,,,,uuvv,,,,2,0,0,2xy,,,,xyxy
,,uv,,uv,,,,,yy,,xy只有当z=0时,即(0,0)处有,. 所以f(z)在z=0处可导,在全平面上不解析.
33fzxy()23i,,(3) ;
33uxyxvxyy(,)2,(,)3,,解:在全平面上可微. ,,,,uuvv22,,,,6,0,9,0xy,,,,xyxy
23xy,,所以只有当时,才满足C-R方程.
230xy,,从而f(z)在处可导,在全平面不解析.
2fzzz(),,(4) .
zxy,,i解:设,则
23232fzxyxyxxyyxy()(i)(i)i(),,,,,,,,
3232uxyxxyvxyyxy(,),(,),,,,
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,,,,uuvv2222,,,,,,3,2,2,3xyxyxyyx,,,,xyxy 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.
,fz()0,(1) ;
,,uu,,vv,,0,,0,,,xyfz()0,,,xy证明:因为,所以,. 所以u,v为常数,于是f(z)为常数.
fz()(2) 解析.
fzuv()i,,证明:设在D内解析,则
,,,,,uvuv(),,,,,,,,xyxy
,,,,,uvv(),,,,,,yxy
,,,,uvuv,,,,,,,,xyyx
,,,,uuuv,,,,,,,,xyyx而f(z)为解析函数,所以
,,,,vvvv,,,,uuvv,,,,,,,,,,0,,,,xyxy,,,,xxyy所以即 从而v为常数,u为常数,即f(z)为常数. (3) Ref(z)=常数.
,,uu,,0,,xy证明:因为Ref(z)为常数,即u=C1,
,,uu,,0,,xy因为f(z)解析,C-R条件成立。故即u=C2 从而f(z)为常数.
(4) Imf(z)=常数.
,,vv,,0,,xy证明:与(3)类似,由v=C1得
,,uu,,0,,xy因为f(z)解析,由C-R方程得,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.
2fzfzC()(),,,,若C0,则f(z) 0,但,即u2+v2=C2 则两边对x,y分别求偏导数,有
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,,,,uvuv220,220uvuv,,,,,,,,,,,,xxyy 利用C-R条件,由于f(z)在D内解析,有 ,,,,uvuv,,,,,,,xyyx
,,uv,uv,,,,0,,,,xx,,,uv,,uv,vu,,,,0,,0,0,,,xx,,,xx所以 所以 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,
,,vu2,,vu2uuv(),uuv,,,,(),(/)vu,,yy,,xx,,,022222221(/)()(),,,vuuuvuuv于是 得
,,vu,uv,,,,0,,,xx,,,,vu,uv,,,,0,,,yy, C-R条件?
,,vu,uv,,,,0,,,,xx,,,vu,uv,,,,0,,,xx,
,,,,uvuv,,,,0,,,,xxyy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.
,,uu22,,,2,3nxymynx,,xy
,,vv22,,,3,2xlylxy,,xy
,,uv,,,nl,,xy
,,uv,,,,,,,nlm3,3,,yx
nlm,,,,,3,3,1所以.
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
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,,,,uuvv2222,,,,,,,33,6,6,33xyxyxyxy,,,,xyxy 所以f(z)在全平面上满足C-R方程,处处可导,处处解析.
,,uv22222,xxfzxyxyxyxyz()i336i3(2i)3,,,,,,,,,fzxyyyyyxy()e(cossin)ie(cossin),,,,,,xx.(2) .
证明:
xxuxyxyyyvxyyyxy(,)e(cossin),(,)=e(cossin),,,处处可微,且 ,uxxxe(cossin)e(cos)e(cossincos),,,,,,xyyyyxyyyy,x ,uxx,vxxxe(sinsincos)e(sinsincos),,,,,,,,xyyyyxyyyye(cossin)e(sin)e(cossinsin),,,,,,yyxyyyyxyy,y,x
,,uv,,uv,vxx,,,e(cos(sin)cos)e(cossincos),,,,,,,yyyxyyyyxy,,yx,,xy,y所以,
所以f(z)处处可导,处处解析.
,,uvxx,()ie(cossincos)i(e(cossinsin))fzxyyyyyyxyy,,,,,,,,,,xxxxxxxxecosiesin(ecosiesin)i(ecosiesin),,,,,,yyxyyyyyzzzzeeiee(1),,,,,xyz10. 设
3333,xyxy,,,i,,,0.z,,22,,fz,xy,,,0.0.z,,
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(
lim()lim,i,fzuxyvxy,,,,,,,,zxy,,0,0,0,,证明.(1)?
33xy,lim,limuxy,,,22,,,,xyxy,,,,,,,0,0,0,0xy,而
33xyxy,,,,,,,xy1,,2222,,xyxy,,,,?
33xy,3??0xy,22xy,2?
33xy,lim0,22,,xy,,,,0,0xy,?
33xy,lim0,22,,xy,,,,0,0xy,同理
,,,,lim00fzf,,,,xy,0,0,,,?
?f(z)在z=0处连续(
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,,fzf()0,limz,0z(2)考察极限
当z沿虚轴趋向于零时,z=iy,有
3,,111i,,y,,,,limi0lim1ifyf,,,,,,,,,2yy,,00iiyyy( 当z沿实轴趋向于零时,z=x,有
1,,,,fxf,,,,,lim01ix,0x
,,,,uvvu,,,i,i,,,,xxyy它们分别为
,,,,uvuv,,,,,,,,xyyx?
?满足C-R条件(
(3)当z沿y=x趋向于零时,有
33,,,,,,,,fxxfxx,,,,,i0,01i1iilimlim,,3xyxy,,,,00,,xxx,,,i21i1i
,flimz,0,z?不存在(即f(z)在z=0处不可导(
,,,,Fzfz,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(
,,,,uvuv,,,,,,,,xyyx所以u(x,y),v(x,y)在D内可微且满足C-R方程,即(
,,fzuxyxyxyxy,,,,,,,iv,,i,,,,,,,,,,,,得
,,,,uxyuxy,,,,,,,,,,uxy,,,,,,,,,,,,yyy,,xx
,,,,vxyvxy,,,,,,,,,,,vxy,,,,,,,,,,,,yyy,,xx
,,,,,,,,,,,,xyyx,,,,故φ(x,y),ψ(x,y)在D1内可微且满足C-R条件
,,fz从而在D1内解析
13. 计算下列各值
(1) e2+i=e2?ei=e2?(cos1+isin1)
22,,iπ22,i,,,,ππ13,,,,33333eeeecosisinei,,,,,,,,,,,,,,,,,,3322,,,,,,,,(2) (3)
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xy,i22xy,,,Ree
xy,i2222xyxy,,,,,,Reee
x,,22,,yy,,,,xy,Reecosisin,,,,,,,2222,,,,,,,,,,xyxy,,,,,,,,x22y,,xy,ecos,,22,,xy,,,
(4)
i2i2i,,,,xyxy,,,,ieee,,
,,,22i2xyx,,,eee
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??(
i,,,,ieiisi,,,,rrcnos,,,,limeelimeerr,,,,,,,,,rr 故(
,,limfz,,z,, 所以(
15. 计算下列各值(
(1)
3,,,,,,ln23i=ln13iarg23iln13i,,,,,,,,πarctan,,,,2
ππ,,,,,,ln33iln23iarg33iln23iln23i,,,,,,,,,,,66,,(2) (3)ln(ei)=ln1+iarg(ei)=ln1+i=i
(4)
π,,,,lnielneiargie1i,,,,2
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(
22gzzxyuxyvxy()||,i,,,,,,,,,,设z=x+iy,
22uxyxyvxy,,,0,,,,,,,在复平面内可微(
1,,,uxuy1222,,,,,,,xyx22222,,xy2xyxy,, ,,vv,,00,,xy
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)
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π,,,,1iln2i2,,,,kπi1i,,,1i,,,,,,,ln1i1iln1i,,,,,,4,,1ieee,,,,
ππln2,,,,,eiln2i2πk44
ππ,,iln2,ln22,,kπ,,,,44,,ee
πln22,,kπ,,ππ,,,,4,,,,,ecosln2isinln2,,,,,,,,,,44,,
π2kπ,,,ππ,,,,4,,,,,,2ecosln2isinln2,,,,,,,,,,44,,
(2)
55,,,,,,,ln35ln3,,,,,3ee
,,kk5ln3i,,,,,,,π2πi5ln35iπ2π5i,,ee
5ln3,,,,,,,kk,,,,ecos21π5isin21π55,,,,,,kk,,,,,,3cos21π5isin21π5
,i,,,,,,,,,iln1iln1iln1i02kπi1eee,,,
,,,,i2kkπi2π,,ee(3)
1i,1i1i,,1i,,,,,,,ln1iln,,,,,1i,,,22,,,,(4),,ee,,2,,
,,π,,π,,,,,,1iln1i2,,,,,πik1i2,,πiik,,,,,,,,4,,,,4,,ee
ππππ,,i2π,k2πii2,,,,π2πkkk,,,,4444,,,eee
π,2πk,,ππ,,4,,,,ecosisin,,,,44,,,,
π,,,2πk224,,,ei,,,,22
18. 计算下列各值
(1)
,,,,iπ,,,,,,5iiπ5iiπ5iπ5eeee,,,,cosπ,,,5i22
,,,555555,,,,,,,,ee1eeee,,,,,,ch5222
(2)
,,,,i15ii15ii5i5,,,,,,eeee,,,,sin15i,,,2i2i55,,,,,ecos1isin1ecos1isin1,,,,,2i5555,,eeee,,,,,,sin1icos122
,,,,i3ii3i,,,ee,
,,sin3isin6isin2,,2i,,tan3i,,,,,,,,i3ii3i,,,22,,,,cos3i,ee,2ch1sin3,
2i(3)(4)
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2122,,,yxyxii,,zxyxy,,,,,,,sineesinchicossh2i
2222,,,,sinchcosshxyxy
222222,,,,,,,,,,xyyxxysinchshcossinsh
22,,xysinsh(5)
2,,,,arcsiniilni1iiln12,,,,,,,
,,,,,,,,iln21i2kπ,,,,,,k0,1,?,,,,,,,,,,,,iln21iπ2kπ,,,
,,i1i12ii21,,,,,,arctan12ilnlni,,,,,,,,,,,,21i12i255,,,,
1i,,,,kπarctan2ln524(6) 19. 求解下列方程
(1) sinz=2(
解:
1,,,,,,z,,,,,,arcsin2ln2i3iln23i,,i
,,1,,,,,,,,,iln232kπi,,,,,,2,,
1,,,,,,,,,,2kkπiln23,0,1,?,,,,2
ze13i0,,,(2)
ze13i,,解: 即
π,,zk,,,,,ln13iln2i2πi3
1,,,,,ln22kπi,,3,,
(3)
πlniz,2
ππilniz,2z,,ei2解: 即
,,z,,,ln1i0(4)
π1,,,,zkk,,,,,,,,,ln1iln2i2πiln22πi,,44,,解:( 20. 若z=x+iy,求证
(1) sinz=sinxchy+icosx?shy
证明:
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iiixyxyi,,,,,,,,iizz,eeee,,sinz,,2i2i
1,,,yxyxii,,,,.ee2i
xyxy,,,sinchicos.sh
(2)cosz=cosx?chy-isinx?shy
证明:
iizz,ee1,iiiixyxy,,,,,,,,,zcosee,,,,22
1,,,yxyxii,,,,ee2
1,yy,,,,,,xxxx,,,,,ecosisine.cosisin2
yyyy,,,,eeee,,,xx,,.cosisin.,,22,,
xyxy,,cos.chisin.sh (3)|sinz|2=sin2x+sh2y
证明:
1,,,yxyxii,,zxyxy,,,,,,sineesinchicossh2i
22222sinsinchcos.shzxyxy,,
222222,,,,,,,,sinchshcossinshxyyxxy
22,,sinshxy (4)|cosz|2=cos2x+sh2y
coscoschisinshzxyxy,,证明:
22222coscos.chsin.shzxyxy,,
222222,,,,,,,,coschshcossin.shxyyxxy
22,,cosshxy
21. 证明当y??时,|sin(x+iy)|和|cos(x+iy)|都趋于无穷大( 证明:
11iiiizzyxyx,,,,,,,,z,,,,,sineeee2i2i
1,,,yxyxiize,,,sine2
,,,,yxyyxyii,,eeee ?
11,,,,yxyxyyii,,,,z?,,,sineeee22 而
当y?+?时,e-y?0,ey?+?有|sinz|??(
当y?-?时,e-y?+?,ey?0有|sinz|??(
11,,,,yxyxyyii,,xy,,,,?,,cosieeee22同理得 所以当y??时有|cosz|??(
习题三
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2()dxyixz,,,C1. 计算积分,其中C为从原点到点1+i的直线段.
yx,01,,xzxix,,解 设直线段的方程为,则.
122xyixdzxyixdxix,,,,,,(),,,,,,0C
111ii,231,,,,,,,,ixidxiixi(1)(1)(1)0,0333故
(1)d,zz,C2. 计算积分,其中积分路径C为 (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.
01,,xzxix,,解 (1)设.
111(),,,,,,zdzxixdxixi,,,,,,0C
201,,xzxix,,(2)设.
12i2211(),,,,,,zdzxixdxix,,,,,,03C
dzz,C3. 计算积分,其中积分路径C为
(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.
,,,11yziy,解 (1)设.
11zdzydiyiydyi,,,,,,,,11C
3,,
i,,ze,22(2)设. 从到
,,ii,,22zdzdeidei,,,1233,,,,,22C
3,,
i,,ze,22(3) 设. 从到
,i,2zdzdei,,123,,,2C
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zzezdz,,sin,, ,za,,0CC6. 计算积分,其中为.
zzzezdzzdzezdz,,,,,sinsin,, ,,,CCC解 zza,ez,sin?在所围的区域内解析
zezdz,,sin0 ,C?
从而
2,zi,zezdzzdzadae,,,,sin,, ,,,CC0
2,2i,,,,aied0,0
zzezdz,,,sin0,, ,C故
1dz2 ,Czz(1),C7. 计算积分,其中积分路径为
113Czi:,,Cz:,Cz:,312222(1) (2) (3)
3Czi:,,42(4)
112z,zz(1),z,02解:(1)在所围的区域内,只有一个奇点.
111111dzdzii,,,,,,,,,()2002,,2 ,,CCzzi,,,0,C1zzzzizi(1)22,,,2(2)在所围的区域内包含三个奇点.故
111111dzdziii,,,,,,,,,,,,()202 ,,CCC2zzzzizi,,,(1)22zi,,2(3)在所围的区域内包含一个奇点,故
111111dzdzii,,,,,,,,,,,,()002 ,,CCzzi,,0,C3zzzzizi,,,(1)224(4)在所围的区域内包含两个奇点,故
111111dzdziii,,,,,,,,()2,,,2 ,,CC4zzzzizi(1)22,,, 10.利用牛顿-莱布尼兹
公式
小学单位换算公式大全免费下载公式下载行测公式大全下载excel公式下载逻辑回归公式下载
计算下列积分.
i0,,2iz2,zcosdz(2),izdzedz,,0,2,1,i (1) (2) (3)
i1tan,ziln(1)z,1dzdz2,,zzdz,sin11,coszz,10(4) (5) (6)
解 (1)
,,2izz1,2i,cossin21dzch,,0,0222
(2)
0,,zz0edze,,,,2,,i,,,i
ii11111i223i(2)(2)(2)(2),,,,,,,,,,izdzizdiziz1,,11ii333(3)
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2iiln(1)11z,,22idzzdzz,,,,,,,,ln(1)ln(1)ln(1)(3ln2)1,,11z,1284(4)
1111zzdzzdzzzzdz,,,,,,,,sincoscoscossin1cos10,,,000(5)
iii1tan1,z222iidzzdzzzdztanzz,,,,secsectantan112,,,111cos2zze11,,22dz,,,ith1tan1tan1t1,,h,,2 ,Cz,C1,,22(6) 11. 计算积分,其中为
zi,,1zi,,1z,2(1) (2) (3) 解 (1)
zzzeeei,,,,2,,dzdzie,zi2 ,,CC,,,,1()()zzizizi
zzzeee,i,,,,,2,,dzdzie,,zi2 ,,CC,,,,1()()zzizizi(2)
zzzeeeii,,,,,,2sin1dzdzdzeei,,,222 ,,,CCC12zzz,,,111(3)
16. 求下列积分的值,其中积分路径C均为|z|=1.
ztanz1cosz2edzz,,dz02dz ,3C ,5 ,C()2zz,Cz0z (1) (2) (3) 解 (1)
zeii2,,(4)z,,dze(),0z5 ,Cz4!12
(2)
cos2zi,(2),,,dzzi(cos),z,03 ,Cz2!
(3)
ztanz'202,,,,2(tan)secdzizizz,2 ,0C,()2zz0
1dz33 ,C(1)(1)zz,,C17. 计算积分,其中积分路径为
z,1R,2(1)中心位于点,半径为的正向圆周
z,,1R,2(2) 中心位于点,半径为的正向圆周
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Cz,1解:(1) 内包含了奇点
1213,,ii(2)dz,,()1z,333 ,C(1)(1)2!(1)8zzz,,,? Cz,,1(2) 内包含了奇点,
1213,,ii(2)dz,,,()1z,,333 ,C(1)(1)2!(1)8zzz,,,? 19. 验证下列函数为调和函数.
3223(1)632;,,,,,xxyxyy
xx(2)ecos1(esin1).,,,,yiy,
3223uxxyxyy,,,,632,,0w,ui,,解(1) 设,
?
,u,u2222,,,,666xxyy,,,3123xxyy,y,x
22,u,u,,,612xy,,612xy22,y,x
从而有
22,,uu,,022,,xyw,满足拉普拉斯方程,从而是调和函数.
xx,,,,eysin1uey,,,cos1wui,,,(2) 设,
,u,uxxsin,,,ey,,eycos,y,x?
22,u,uxxcos,,,eycos,,ey22,y,x
从而有
22,,uu,,022u,,xy,满足拉普拉斯方程,从而是调和函数.
,,,,xxeycos,,,,eysiny,,x
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22,,,,xxsinye,,,eysin,,22y,x,
22,,,,,,022,,,xy,满足拉普拉斯方程,从而是调和函数.
x,,2222fzui(),,,uxy,,,xy20.证明:函数,都是调和函数,但不是解析函数
证明:
22,u,u,u,u,,2y,,2,2x,222,y,y,x,x 22,,uu,,022,,xyu?,从而是调和函数.
22,,,2xyyx,,,,,222222,yxxy()()xy,,,
223223,,,,62xyx,,,62xyx,,22232223,,yxy(),,xxy() 22,,,,,,022xy,,,?,从而是调和函数.
,,u,,,u,,,,,,xy,,yx但?
fzui(),,,?不满足C-R方程,从而不是解析函数.
fzui(),,,22.由下列各已知调和函数,求解析函数
yuf,,,(1)02222uxyxy,,,xy,(1) (2)
u,,,,,u,2yx,,,,,2xy,,,xyyx,,,, 解(1)因为 所以
,,uu(,)(,)xyxyyx,dxdyCyxdxxydyCxdxxydyC(2)(2)(2),,,,,,,,,,,,,,,,,,00(0,0)(0,0),,yx
22xy2xyC,,,,,22
22xy22fzxyxyxyC()i(2),,,,,,,,22 令y=0,上式变为
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2x2fxxC()i(),,,2
从而
2z2fzzC()ii,,,,2
22,uxy2,,uxy,,,222222,,xxy(),,yxy()(2)
用线积分法,取(x0,y0)为(1,0),有
2xyx(,),,uuxy2y,,,,,,,,()dxdyCdxxdyC,4222,,0(1,0)1,,,yxxxy()yx1xxyfzC()i(1),,,,,,,,,,11C222222220xyxy,,xxyxy,, f(1)0.,由,得C=0
1,,,,?,fi,1z,,,,z
pzzazaza()()()(),,,,?ain(1,2,,),?aaa,,,?12ni12n23.设,其中各不相同,闭路C不通过,证明积
分
,1()pzdz ,2πipz()C
等于位于C内的p(z)的零点的个数.
aaa,,...Pz()12k证明: 不妨设闭路C内的零点的个数为k, 其零点分别为
nn()()()...()...()zazazazaza,,,,,,,,,kkn111,,1()1Pzkk,,23dzdz, ,,CC2πi()2Pzzazazaπi()()...(),,,n12111111,,,,dzdzdz... ,,,CCC2πi2zazaza,,,πi2πin121111,,,,,,,11...1...dzdz,,,,, ,,CC2πi2zaza,,πikn,1k个,k24.试证明下述定理(无界区域的柯西积分公式): 设f(z)在
lim()fzA,,,z,,闭路C及其外部区域D内解析,且,则
,,,fzAzD(),,,,1()fd,,,,AzG,.,2πiz,,,C
其中G为C所围内部区域.
,Rz证明:在D内任取一点Z,并取充分大的R,作圆CR: ,将C与Z包含在内
CR则f(z)在以C及为边界的区域内解析,依柯西积分公式,有
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1()()ff,,,fzdd()[-],, ,,CCR,,2πizz,,
fz(),,
,,R,z,因为 在上解析,且
f()1, ,,,,limlim()lim()1ff,,,,,,,,,,,,z,z,,1,
所以,当Z在C外部时,有
1()f,,,fzAd(), ,C,2πiz,
1()f,,,,dfzA(), ,C,2πiz,即
设Z在C内,则f(z)=0,即
1()()ff,,,,0[]dd,, ,,CCR,,2πizz,,
1()f,,dA, ,C,2πiz,故有:
习题四
,,,,1. 复级数与都发散,则级数和发散.这个命题是否成立?为什么? ab()ab,,,abnn,nn,nn,n1n11n,,n1,答.不一定(反例: ,,,,1111 ab,,,,,i,i发散,,,,nn22nnnn1111nnnn,,,,
,,2但收敛 ab,,,()i,,nn2nnn11,,,,2发散 ab,,(),,nnn11nn,,,,11收敛. ab,,,[()],,nn24nnnn11,,
2.下列复数项级数是否收敛,是绝对收敛还是条件收敛?
iπ21n,,,,1i,15i,enn()(1) (2) (3) ,,,n2n1n,1n,
,n1n,,icosin,(4) (5) ,nlnn2n,1n0,
21nnn,,,,1i1(1)i1(1),,,,,,,,,i解 (1) ,,,nnnnnnn,,,111
21n,,,11i,
,,因为发散,所以发散 nn,n1n,1
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n,,15i26,n,()(2)发散 ,,2211nn,,
15i15,nnlim()lim(i)0,,,又因为 nn,,,,222
n,15i,()所以发散 ,21n,
ππiππi,cosisin,,,n,,e1ππnnne1(3),,,(cosisin) 发散,又因为收敛,所以不绝对收敛. ,,,,,,nnnnn,,,nnn111nnnn,,11
n,,i1,(4) ,,lnlnnnnn,,11
11,因为 ln1nn,
所以级数不绝对收敛.
k,又因为当n=2k时, 级数化为收敛 (1),,k1ln2,k
k,(1),当n=2k+1时, 级数化为也收敛 ,kln(21),1k,所以原级数条件收敛
nn,,,,,cosi1ee1e11n,nn,,,,()() (5) ,,,,nn2222222e0000nnnn,,,,
,,1enn()()其中 发散,收敛 ,,2e2n,0n,0
所以原级数发散.
,,,22Re()0a,aa3.证明:若,且和收敛,则级数绝对收敛. na,,nn,n,n1n,1n,1
证明:设
2222axyaxyxyxy,,,,,,,i,(i)2i nnnnnnnnnn
,,2因为和收敛 aa,,nnn,1n,1
,,,,2xyxyxy,,(),,所以收敛 ,,,,nnnnnnnnnn1111,,,,
Re()0a,n又因为,
2limlim0xx,,x,0nnn所以且 nn,,,,
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2xx,当n充分大时, nn
,2x所以收敛 ,n,n1
222222axyxxy,,,,,2()nnnnnn
,,222而2x()xy,收敛,收敛 ,,nnnn,1n1,
,,22aa所以收敛,从而级数绝对收敛. ,,nn,n1n1,
,nn,1()zz,4.讨论级数的敛散性 ,n,0
n,,11kkn当时zs,,,1,1szzz,,,,()1解 因为部分和,所以, n,n,0k
当时zs,,1,0当时zs,,1,,不存在. nn
i,,,0z,e,,1,1zz当而时(即),cosnθ和sinnθ都没有极限,所以也不收敛(
当zs>1时,,,. n
,nn1,z,1z,1()zz,故当和时, 收敛. ,n0,
,nCz(2),5.幂级数,能否在z=0处收敛而在z=3处发散. n0n,
C11n,1,,z2lim,,,,z2解: 设,则当时,级数收敛,时发散. n,,,,Cn
1,2若在z=0处收敛,则 ,
1,1若在z=3处发散, 则 ,
,nCz(2),,显然矛盾,所以幂级数不能在z=0处收敛而在z=3处发散 n0n,
6.下列说法是否正确?为什么?
(1)每一个幂级数在它的收敛圆周上处处收敛. (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,因为幂级数在它的收敛圆周上可能收敛,也可能发散.
(2) 不正确,因为收敛的幂级数的和函数在收敛圆周内是解析的.
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,,CnnnCzz7.若的收敛半径为R,求的收敛半径。 ,,nnbn0n0,,
Cn,1n,1111Cn,1b,,,limlim解: 因为 Cnn,,,,nCbRbnnb
,RRb,,所以
,nnlima,,azn8.证明:若幂级数的 系数满足,则 ,n,,nn0,
10,,,,,R,(1)当时, ,
R,,,,,0(2) 当时,
R,0,,,,(3) 当时,
证明:考虑正项级数
,nn2azazazaz,,,,,...... ,nn12n0,
nnnnn0,,,,,,,,z1时limlimazazz,,,,,由于,若,由正项级数的根值判别法知,当,即nn,,,,nn
,211nnnnaz,时,,,z1时,时zzazlim1az,,收敛。当,即,不能趋于零,级数发散.故收,nnn,,n,,n0,
1敛半径R,,.
R,,,,,0,,,z1级数收敛且. 当时, ,
2nR,0,,,,,,z0,az若不能趋于零,级数发散.且 n,对当充分大时,必有
9.求下列级数的收敛半径,并写出收敛圆周。
,n,pn(i)z,nz,(1) (2),,pn,n00,n ,21n,nn121,,(i),,,z(3) ,2nn0,
,innn(1),,,z()(1),(4) nn0,
解: (,)
n111pp,,,,limlim()lim(1)1pp,,,,,,nnn,(1)nn,,nn11收敛圆周
?,R1
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z,,i1
(2)
p(1)n,lim1,p,,nn
R,1
所以收敛圆周
z,1
21n,nn,,121fzz()(i),,,,n(3) 记 n2由比值法,有
21n,n(21)2nz,,,fz()12n,1limlim,,z21n,21n,nn,,,,fz()2 (21)2nz,,,n
要级数收敛,则
z,2
级数绝对收敛,收敛半径为 R,2
以收敛圆周 所
z,2
innn(1),fzz,,,()()(1)(4) 记 nn
n,1nn(1),z,1(1)z,若,~ ,,,,1nnlim()limlimfz,,,,n若,,,,,,1nnnn,,,,,, nn
R,1z,,11时绝对收敛,收敛半径 所以
z,,11 收敛圆周
10.求下列级数的和函数.
2n,,znnn1,(1),,(1),,nz(1) (2),,(2)!nn1,,n0
(1) 解:
Cn,1n,1limlim1,,nn,,,,Cnn
故收敛半径R=1,由逐项积分性质,有:
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,,zz-1nnnn(1)(1),,,,nzdzz,,, 01,z11nn,,
所以
,z1nn-1,,,,,,(1)(),1nzz,2,,1(1)zzn1, 于是有:
,,znnnn11,,(1)(1)1,,,,,,,,,nzznzz,,2(1),znn11,,
令: (2)
2n,zn()(1),,,sz, (2)!nn,0
C1n,1?,,limlim0.nn,,,, ,,Cnn(21)(22)n
故R=?, 由逐项求导性质
21n,,zn,()(1),,,sz,(21)!,nn,1
2222nmn,,,,,,szsz()(),,zzznmn+1由此得到,,()(1)(1)(1)(1),,,,,,,,,,,,szmn,,,(22)!(2)!(2)!,nmnnmn,,,100
,,szsz()()0,,即有微分方程
szAzBz()cossin,,故有:, A, B待定。
2n,zn由S(0)[(1)]11,,,,,,,AA ,z,0(2)!n,n0
21n,,zn,(0)sincos[(1)]00,,,,,,,,,szBzB ,z,0(21)!n,n,1
所以
2n,zn(1)cos.,,,,,,zR,(2)!n,n0
,,,nCCCz11.设级数收敛,而发散,证明的收敛半径为1 ,n,,nnn0,n0,n0,
,
C,证明:因为级数n收敛 n0,
设
n,1CZn,1lim.,,znn,,CZ n
若
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,nCz的收敛半径为1 ,nn0,
1z,则 ,
,,1现用反证法证明
C,n,1lim1,,,01,,,z,1Cn,,,若则,有,即收敛,与条件矛盾。 nCn,n0
,,n,,1z,1CzC,,nn若则,从而在单位圆上等于,是收敛的,这与收敛半径的概念矛盾。 n0n0,,
,,1综上述可知,必有,所以
1R,,1 ,
,nzzz,zCz,00n12.若在点处发散,证明级数对于所有满足点都发散. n0,
,nzzz,Cz,110n证明:不妨设当时,在处收敛 n0,
,,nn,,zzCzCz1,n,则对nn0,,n0,绝对收敛,则在 z0点处收敛
,nzz,Cz,0n所以矛盾,从而在处发散. n0,
4,zz,0zln(1e),13.用直接法将函数在点处展开为泰勒级数,(到项),并指出其收敛半径.
z1e,,zln(1e)ln(),,:因为解ze
zkk,,,,(21)πi(0,1,...)奇点为k
R,π所以
又
,zln(1e)ln2,,0z,
,ze1,z,,,,,,[ln(1e)]0z, ,z,1e2
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,ze1,z,,,,,,,[ln(1e)] 0z,22,z,(1e)2
,,zz2,,ee,z,,,[ln(1e)]0,,, z,0,z3(1e),
,,,zzz2e(14ee)1,,,z(4)[ln(1e)],,,,z,0,z43(1e)2, 于是,有展开式
111,z24,,,,,,,zzzRln(1e)ln2...,π2322!24!2
14z,,12(1)z,214.用直接法将函数在点处展开为泰勒级数,(到项) 1,z
1z,,iR,22解:为的奇点,所以收敛半径 1,z
又
11fzf(),(1),,2 12,z
,21z,,fzf(),(1),,,22 (1)2,z
2,,261z,,,,fzf(),(1),,23 (1)2,z
32424zz,,,,,,,fzf(),(1)0,,24 (1),z
2424240120,,zz(4)(4)fzf(),(1)0,,25 (1),z
z,1fz()于是,在处的泰勒级数为
1111324,,,,,,,,,(1)(1)(1)...,2zzzR212244!,z 15.用间接法将下列函数展开为泰勒级数,并指出其收敛性.
1z,0z,1(1) 分别在和处 23z,
3z,0sinz(2) 在处
arctanzz,0(3) 在处
zz,2(4) 在处 (1)(2)zz,,
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z,0ln(1),z(5) 在处
解 (1)
,1111123n,,,,,,,,,zz(), ,2zz,,233233320n,,z13
,11111nn,,,,,,,,,zz 2(1),1,zzzz,,,,,,,232212(1)112(1)20n,
n35,(1),zz21n,sin...zzz,,,,,,(2) (21)!3!5!n,,n0
2n,331,321nn,zzzsin(1),,,,,,, n4(21)!,,n0
z1?zdz,arctan2,0,z1(3)
?,,?,zR为奇点,i1
,,zz11221nnnn,zdzzdzzz,,,,,,,,arctan(1)(1),1 ,,2,,00,,zn12100nn,,
(4)
111111111,,,,,,,,zz,,22zzzzzz,,,,,,,,(1)(2)12232434,,1134 ,,zz,,1212nnnn,,,,,,(1)()(1)(),,334400nn,,,11nnzz,,,,,,,(1)()(2),23,11nn,,340n,
ln(1),zz,,1(5)因为从沿负实轴不解析
所以,收敛半径为R=1
,1nn,,,,,,zz[ln(1)](1), z,10n,
,,z11nnnn,,,,,,,,,,zzdzzzln(1)(1)(1),1,,,0n00nn,,
fz()(,),RRzR,z为什么区域16.内解析且在区间取实数值的函数展开成的幂级数时,展开式的系数都
是实数,
fz()fx()fx()xR,z答:因为当取实数值时,与的泰勒级数展开式是完全一致的,而在内,的展开式
fz()zR,系数都是实数。所以在内,的幂级数展开式的系数是实数.
21z,fz(),z,0217.求的以为中心的各个圆环域内的罗朗级数. zz,,2
z,0fz()z,1z,,212解:函数有奇点与,有三个以为中心的圆环域,其罗朗级数.分别为:
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,,21111z,znnn在内,z1()=,,,,,,fzz(1)(), ,,2zzzz,,,,2122200nn,,,1nn,,,,((1)1)z,1n,20n,
1,z11z,,,,fz()e,19.在内将展开成罗朗级数.
1t,,解:令则 1,z
11t23fzttt()e1...,,,,,,,, 2!3!
1t,1z,,,,而在内展开式为 1,z
111111,,,,,,,,,(1...)2 11,zzzzz1,z
所以,代入可得
11111112fz()1(1...)(1...)...,,,,,,,,,,,,22zzzzzz2! 111119,,,,,,,1...2345zzzzz2624120
20.有人做下列运算,并根据运算做出如下结果
z23,,,,zzz... 1,z
z11,,,,1...2 zzz,1
zz,,0因为,所以有结果 11,,zz
11123...11...0,,,,,,,,,,zzz32 zzz
你认为正确吗?为什么?
z23,,,,zzz...z1,答:不正确,因为要求 1,z
z11,,,,1...z1,2而要求 1,zzz
所以,在不同区域内
zz11123,,,,,,,,,,,,...11...0zzz62 11,,zzzzz
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1fzz()cos(),,21.证明: 用z的幂表示的罗朗级数展开式中的系数为 z
2π1Cndn,,,cos(2cos)cos.0,1,...,,,n, 02π
11cos()z,cos()z,z,00,,,zz,,证明:是内,因为和的奇点,所以在的罗朗级数为 zz
n,,1nzCz,,cos(),nzn,,,
1cos(),,1,,,,,,0,1,2,...Ccdn,n其中n,1, 2πi,
0,,,z其中C为内任一条绕原点的简单曲线.
1cos()z,1i,zCdzz,,,,,(e,02,π)n1n, ,1z,2πiziiii,,,,,,2π2π1cos(ee)1cos(ee),,i,,,iedd,,i(1)inn,,,,,002πie2πe
2π1ii, ,,,,,,cos(ee)(cosisin)nnd,,,,02π
2π1,,,cos(2cos)cos.0,1,...ndn,,,,02π
1z,0fz,()22.是函数的孤立奇点吗?为什么? 1cos()z
1z,0fz,()解: 因为的奇点有 1cos()z
1π1,,,,,,,kzkπ(0,1,2,...) πz2kπ,2
1z,0k,,z,所以在的任意去心邻域,总包括奇点,当时,z=0。 πkπ,2
1z,0从而不是的孤立奇点. 1cos()z
336z,023. 用级数展开法指出函数在处零点的级. 6sin(6)zzz,,
解:
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336393fzzzzzzz()6sin(6)6sin6,,,,,,
11 391593,,,,,,6(...)6zzzzz3!5!
故为f(z)的15级零点 z=0
z,024. 判断是否为下列函数的孤立奇点,并确定奇点的类型:
1cos,z1/z? ; ? e2z
1zz,0e是的孤立奇点 解:
因为
1111ze1......,,,,,,2nzznz2!!
1zz,0e是的本性奇点. 所以
(2)因为
112411...,,,,zz1cos11,z22!4!,,,,z...22 zz2!4!
1cos,zz,02所以是的可去奇点. z
25. 下列函数有些什么奇点,如果是极点,指出其点:
1sinz1? ? ? 232zsinzzz,(e1)
1135zzz,,,...sin111z23!5!,,,,,z...332解: (1) zzz3!5!
z,0是奇点,是二级极点. 所以
zkk,,,2πi(0,1,...): (2) 解
z,02kπi是奇点,是一级极点,0是二级极点. 解: (3)
z,0 2sin0,z,z,0
22,(sin)cos20.zzz,,,0z, 2222,,(sin)4sin2cos20zzzz,,,,,,0z,
2z,0sinz是二级零点 的
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22sinzsinzzk,,πizk,,π而是是一级零点, 一级零点 的的所以
11z,0,,kkπi,π22是是二级极点,一级极点. 的 的sinzsinz26. 判定下列各函数的什么奇点, z,,
22z1/zcossinzz,? ? ? e23,z
12ze1,z,,解: (1)当时,
12zez,,所以,是的可去奇点.
(2)因为
11112435cossin1......zzzzzzz,,,,,,,,,2!4!3!5! 11112345,,,,,,,1...zzzzz2!3!4!5!
cossinzz,z,,所以,是的本性奇点.
2z,0z,,2(3) 当时, 3,z
2zz,,2所以,是的可去奇点. 3,z
1z,127. 函数在处有一个二级极点,但根据下面罗朗展开式: fz(),2zz(1),
1111 . ,,,,,,?, 11z2543zzzzz(1)(1)(1)(1),,,,
z,1我们得到“又是的本性奇点”,这两个结果哪一个是正确的,为什么? fz()
011,,,z解: 不对, z=1是f(z)的二级极点,不是本性奇点.所给罗朗展开式不是在内得到的
011,,,z在内的罗朗展开式为
1111112,,,,,,,,,,,1(1)(1)...zz 222zzzzzzz(1)1(1)(1)1,,,,,
fzdz()z,3 ,28.如果C为正向圆周,求积分的值 C
z1fz(),fz(),(1) (2) (1)(2)zz,,zz(2),
解:(1)先将展开为罗朗级数,得
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1111,,[]2zzz(2)2,z(1),z 1248,,,,,,,,(...),2z2342zzz
C,02,,,,zz,1而 =3在内,,故 fzdzC()2,,,πi0,1 , C
z2,,,,z(2)在内处处解析,罗朗展开式为 (1)(2)zz,,
z1111,,,,z[]12(1)(2)12zzzz,,,,11,,zz 137,,,,,,,,...,2z23zzz
C,12,,,,zz,1而=3在内,,故 fzdzC()2,,,πi2πi,1 , C
习题五
1. 求下列函数的留数(
z,e1,,(1)在z=0处( fz,5z
ze1,解:在0<|z|<+?的罗朗展开式为 5z
234zzzz11,,,,,,z?,,,e11111111112!3!4!,,,Res,01? ,,,,,,,,?,,55432,,z4!24zzzzz2!3!4!
1z,1,,(2)fz,e在z=1处(
1z,1e解:在0<| <+?的罗朗展开式为 z,1
11111111z,1 e1,,,,,,,,,,??n23,,,,,,zn,12!3!!zzz,,,111
1,,z,1?( Rese,11,,,
2. 利用各种方法计算fz)(在有限孤立奇点处的留数(
32z,,,(1) fz,2,,zz,2
32z,,,解:的有限孤立奇点处有z=0,z=-2(其中z=0为二级极点z=-2为一级极点( fz,2,,zz,2
1,,13232324zzz,,,,,,,,? Res,0limlim1fz,,,,,,,,,2zz,,00,,1!24,,,zz,2
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32z,,, ,,Res,2lim1fz,,,,2z,,2z
12,,3. 利用罗朗展开式求函数在?处的留数( z,,1sinz
1122,,,,解: zzz,,,,,,1sin21sinzz
11111,,2,,,,,,,,,,,zz21?,,35zzz3!5!,,
1,, ? ,,Res,01fz,,3!
1,,从而 ,,Res,1fz,,,,3!5. 计算下列积分(
(1),n为正整数,c为|z|n=取正向( tanπzzd ,c
sinπz解:( tanπzzzdd, ,,cccosπz为在c内tanπz有
1 (k=0,?1,?2„?n-(1))一级极点 zk,,k2
sinπz1,, 由于,,,,,Res,fzzk,,π,,,cosπz,2zk
1,,,,? zzfzznn,,,,,,,,,,tanπd2πiRes,2πi24i,,,k,, ,c,,πk
dz(2) c:|z|=2取正向( 10 ,c,,,,,,zzz,,,i13
1解:因为在c内有z=1,z=-i两个奇点( 10,,,,,,zzz,,,i13
所以
dz,,,, ,,,,2πiRes,iRes,1fzfz,,,,,,10 ,c,,,,,,zzz,,,i13
,,,,,,,,,2πiRes,3Res,fzfz,,,,,,
πi,,10,,3i,6. 计算下列积分(
πcosm,(1) d,,0,54cos,
π1cosm,因被积函数为θ的偶函数,所以 ,Id,,,π,254cos,
π1sinm,令则有 ,Id,1,,π,254cos,
im,π1e II,,,id1,,π,,254cos
2z1,1i,z,e设 则 cos,d,,dz,iz2z
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m1dzz II,,i1 ,2z,12iz,,1,z54,,,,,2z
m1z,dz ,2z,1,,2i,,521z
mz1,,被积函数fz,在|z|=1内只有一个简单极点 z,2,,521zz,,2
m11z,,,,但 ,,Res,limfzm,,1,,,232,,2z,,,,2,,521zz
11π所以 II,,,,,i2πi1mm,,2i3232
π1sinm,又因为 ,,Id0,1,,π,254sco,
πcosm,π? ,d,m,0,,54cos32,
2πcos3,(2) ,|a|>1( d,,20,,12cosaa,
解:令
2πsin3,2πcos3, ,Id,,Id,21,2,200,,,,12cosaa12cosaa,,
3,i2πe ,,,IIid12,20,,,12cosaa
3z1iθ令z=e(,则 ,,,,cosddz2zzi
3z1,,, IIzid122 ,z,1,z1iz2,,,12aa2z
31z,dz ,22z,1,,i,,,,azaza1
,112π,,,,,,,,2πiRes,fz32,,,,ia,,,aa1
2π得 I,132,,aa,1
,,dx(3),a>0,b>0( ,2222,,,,,,,,xaxb
1,,Rz,解:令,被积函数R(z)在上半平面有一级极点z=ia和ib(故 2222,,,,zazb,,
2,,xdx(4). ,a>0( 2,022,,,,,,,,,,IRzaRzb,,2πiRes,iRes,i,,,xa
11,,,,,,,,,,2πilimilimizazb22222222,,zazbii,,,,,,,,,,zazbzazb,,,,,,
11,,,,2πi2222,,,,,,2i2iababab,,,,
π,,,abab,
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22,,,,xx1解: ,ddxx22,,,,022222,,,,,,xaxa
2z,,令,则z=?ai分别为R(z)的二级极点 Rz,222,,za,
,,x,sin,x2故(5) ,β>0,b>0( dx0x12,022,,,,,,,,d2xRzaRza,,,,,πiRes,iRes,i,,,,xb,2,,,222,,,xa
,,,,22,,,,zz,,,,πilimlim,,,,22,,zaza,,,ii,,,,,,zazaii,,,,,,
π,2a
解:
,,,,,,xxx,,cossin,,xxi,x ,,,eddidxxx222,,,,,,,,,222222,,,,,,xbxbxb,,,
z,,而考知,则R(z)在上半平面有z=bi一个二级极点( Rz,222,,zb,
,,xxzii,,,,,,,,,,ed2πiRese,ixRzb2,,,22,,,xb ,zi,,,,eπzb,,,,,,,2πilimei,,zb,i,,,i2zbb,,
,,xx,sin,,π,b, xde,,2,,,22b2,,xb,
,,x,sin,,,xππ,b,从而 dex,,,2b,,02244ebb,,xb,
ix,,e(6) ,a>0 xd22,,,,xa
1,,解:令,在上半平面有z=ai一个一级极点 Rz,22za,
iixza,,,eeeπiz7. 计算下列积分 ,,,,d2xRza,,,,,,,,πiRese,i2πilim2πi22a,,,,zaixazaaa,,i2ie
,,sin2x(1) dx,20,,,xx1
1,,:令解,则R(z)在实轴上有孤立奇点z=0,作以原点为圆心、r为半径的上半圆周c,使C,[-R, Rz,rR2,,zz1,
-r], C,[r, R]构成封闭曲线,此时闭曲线内只有一个奇点i, r
2izedz2i2xiz,,,,1e1elim,,,πi于是:而( ,,,,IxRzz,,,,ImdIm2πiRes,ilimd,,,2,,c,,r0,22r,,,,c,0rzr,,,,1,z22zz1,xx1,,,
故:
22iz,z,,,,,,1e1eπ1a,2,,,,,,,,,,,IIm2πilimπiIm2πiπi1e((2),其中T为直线zRe=c, c>0, 0
0. 00, 00. Im)>0. (若=+i, 则 wwwuv
uv yx,,,2222uvuv,,
u1122因为00, 00,Im(w)>0, (以(,0)为圆心、为半径的圆) w,12222
223. 在将经过点求w=zz=i处的伸缩率和旋转角,问wz=z=i且平行于实轴正向的曲线的切线方向映成w平面
上哪一个方向,并作图.
π,,,,wwww解:因为=2z,所以(i)=2i, ||=2, 旋转角arg=. 2于是, 经过点i且平行实轴正向的向量映成平面上过点-1,且方向垂直向上的向量.如图所示. w
?
24. 一个解析函数,所构成的映射在什么条件下具有伸缩率和旋转角的不变性,映w=z在射z平面上每一点
都具有这个性质吗,
2答:一个解析函数所构成的映射在导数不为零的条件下具有伸缩率和旋转不变性映射w=z在z=0处导数为零,
所以在z=0处不具备这个性质.
wz,,,,,5. 求将区域00. Re()0z,
解:(1) Re(z)=0是虚轴,即z=yi代入得.
22i1(1i)12yyyy,,,,,w,,,,,i 222i1111yyyy,,,,
22y,,1yv,u,写成参数方程为, , . ,,,,,,y221,y1,y
消去y得,像曲线方程为单位圆,即
22u+v=1.
i,2e1,i,z,,,2e,02,π(2) |z|=2.是一圆围,令.代入得化为参数方程. w,i,2e1,
34sin,02,,,π ,,uu,,,54cos54cos,
,消去得,像曲线方程为一阿波罗斯圆.即
54222 ()()uv,,,33
ww,,11 (3) 当Izm()>0时,即, ,,,,zIm()0ww,,11
令w=u+iv得
wuvv,,,,1(1)i2Im()Im()0,,,. 22wuvuv,,,,,1(1)i(1)
即v>0,故Im(z)>0的像为Im(w)>0. 9. 求出一个将右半平面z)>0Re(映射成单位圆w||<1的分式线性变换.
w,,解:设映射将右半平面z映射成w=0,则z关于轴对称点的像为, z000
zz,0所以所求分式线性变换形式为wk,,其中k为常数. zz,0
zz,0wk,,又因为,而虚轴上的点z对应w|=1,|不妨设z=0,则 zz,0
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zz,i,0 wkkk||1e(),,,,,,,,Rzz,0
zz,i,0故. e(Re()0)wz,,,0zz,0
z,,i,10. 映射将映射成,实数的几何意义显什么, ||1z,||1w,ew,,,1,,z,
解:因为
2(1)()()1||,,,,,,,,,zz,,ii, wz()ee,,,,22(1)(1),,,zz,,
21||1,,,,ii,w()ee,,,,从而 ,222(1||)1||,,,,
i2,,所以 arg()argearg(1||)w,,,,,,,,
z,,i,,故表示在单位圆内处的旋转角. arg()w,we,,,,1z,,
11. 求将上半平面Im(z)>0,映射成w|<1|单位圆的分式线性变换w=f(z),并满足条件
1,(1) f(i)=0, =0; (2) f(1)=1, f(i)= . arg(i)f5
z,,,:将上半平面Im()>0, 映为单位圆|<1|的一般分式线性映射为=(Im()>0). 解zwwk,z,,
2ii,,,(1) 由f(i)=0得=i,又由arg,即, ,,,f(i)0,fz()e2,(i)z
π,,i()1π2,,,,,f(i)e0,得,所以 22
z,iw,,i. z,i
1,,i,,1(2) 由f(1)=1,得k=;由f(i)= ,得k=联立解得 51,,5(i),,
3+(52i)z,. w,
(52i)3,,z
12. 求将|z|<1映射成w||<1的分式线性变换w=f(z),并满足条件:
1(1) f()=0, f(-1)=1. 2
π11,arg()f, (2) f()=0, , 222
,(3) arg()fa,f(a)=a, ,.
解:将单位圆|z|<1映成单位圆|w|<1的分式线性映射,为
z,,i,we,, , ||<1. 1z,,,
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11(1) 由f()=0,知.又由f(-1)=1,知 ,,22
1,,1,,,iii2. ,ee(1)1e1,,,,,,,,,π11,2
1z,21z,2故. w,,,,1z12,,z2
54,z1i,1,(2) 由f()=0,知,又 w,,e,,222(2),z
4πi,11,,, ,,,,,ff()earg()2232
1πz,21z,i22于是 . w,,,e()iz12,,z2
,(3) 先求,使z=a,,且|z|<1映成||<1. ,,=()zarg(),,a,,,,0,
za,i,,,=()=ez,则可知 1az,,
,=(),使=0=, ,且||<1映成||<1. 再求wgwaw,,,arg(0)0g,,先求其反函数,它使|w|<1映为||<1,w=a映为=0,且 ,,=()w,,
,,,则 arg()arg(1/(0))0,wg,,
,wa,,=()=w . 1,,aw
因此,所求w由等式给出.
waza,,i,=e,. 11awaz,,,,
13. 求将顶点在0,1,i的三角形式的内部映射为顶点依次为0,2,1+i的三角形的内部的分式线性映射.
解:直接用交比不变性公式即可求得
w,01i0,,z,0i0,?=? z,2i1,1i2,,w,2
w1i2,,i1,z.=. w,2z,11i,i
,4zw,. (i1)(1i),,,z
14. 求出将圆环域2<|z|<5映射为圆环域4<|w|<10且使f(5)=-4的分式线性映射.
解:因为z=5,-5,-2,2映为w=-4,4,10,-10,由交比不变性,有
25,,,25,,104104,?=? 25,,,25104,,,104
故w=f(z)应为
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z,5,,25w,4104,?=? z,5,,25105,w,4
w,4z,520即 =. ,,,,wz,5w,4z
讨论求得映射是否合乎要求,由于w=f(z)将|z|=2映为|w|=10,且将z=5映为w=-4.所以|z|>2映为w|<10|.
又w=f(z)将|z|=5映为|w|=4,将z=2映为w=-10,所以将z||<5映为w|>4|,由此确认,此函数合乎要求.
211,,2215.映射将z平面上的曲线映射到w平面上的什么曲线, wz,xy,,,,,24,,
解:略.
z16. 映射w=e将下列区域映为什么图形.
(1) 直线网Re(z)=C,Im(z)=C; 12
(2) 带形区域; ,,,,,,,,,Im(),02zπ
(3) 半带形区域
. Re()0,0Im(),02zz,,,,,,,π
解:(1) 令z=x+iy, Re(zC)=, 1
Ciy1z=C+iy,,w=ee, Im(z)=C,则 12
iCx2z=x+iC,,w=ee 2
zC1w=e将直线Re()映成圆周;直线Im()=. 故zzC映为射线,,C,,e22
zxyxy,ii(2) 令z=x+iy,,则 ,,,,ywy=eeee,,,,,,,,
zw=e故将带形区域映为的张角为的角形区域. ,,,,Im()z,,,,arg()w,,,
02,,,π(3) 令z=x+iy,x>0,00,01, (). 0arg,,w,
17. 求将单位圆的外部z|>|1保形映射为全平面除去线段-11映为w||<1,再用分式线性映射. 11z
w,121将w||<1映为上半平面Im(w)>0, 然后用幂函数映为有割痕为正实轴的全平面,最ww,w,,,i12232w,11
w,13w,后用分式线性映射将区域映为有割痕[-1,1]的全平面. w,13
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221,,,,1,w1,z11,i1,,,,,,,211,w1,w1,w111,,,,,1z故. 32wz,,,,,,()2221wwz112,,,,,,32w11,,1zi1,,,1,,,,,1w1,1,1,,z,,
ππ18. 求出将割去负实轴,Im(z)=0的带形区域映射为半带形区域,,,Im()z,,,,Re()0z22
,Re()>0的映射. w,,,πIm()wπ
w,1z1解:用将区域映为有割痕(0,1)的右半平面Rew)>0(;再用将半平面映为有割痕w,ew,ln121w,11(-,-1]的单位圆外域;又用将区域映为去上半单位圆内部的上半平面;再用将区域ww,iww,ln,4332
映为半带形00;最后用映为所求区域,故 ww,,2iππ444
ze1,. w,lnz,e1
19. 求将Im(z)<1去掉单位圆z||<1保形映射为上半平面Im(w)>0的映射. 解:略.
将半带形区域00保形映射为平面上的什么区域. 20. 映射zzwz,cosπ,解:
1iziz,wzee,,,因为 cos()2
可以分解为
11w1w,ew=iz ,, ww,,()12322w2
由于在所给区域单叶解析,所以 wz,cos
π(1)w =iz将半带域旋转,映为00. w,e222
11(3) 将区域映为下半平面Im(w)<0. ww,,()22w2
习题 七 1.证明:如果f(t)满足傅里叶变换的条件,当f(t)为奇函数时,则有
,,f(t),b(,),sin,td, ,0
,,2,,b(,),ft,sin,tdt其中 ,0π
,,f(t),a(w),cos,td,当f(t)为偶函数时,则有 ,0
,,2a(,),f(t),cos,tdt其中 ,,0
证明:
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,,1i,t因为其中为()的傅里叶变换 ftf(t),G(,)ed,G(,),,,2π
,,,,it,, Gftedtfttitdt()()()(cossin),,,,,,,,,,,,,
,,,, ,,,,fttdtifttdt()cos()sin,,,,,,,,
,,当()为奇函数时,为奇函数,从而 ftf(t),cos,tdt,0f(t),cos,t,,,
为偶函数,从而 f(t),sin,t
,,,, f(t),sin,tdt,2f(t),sin,tdt.,,0,,
,,故 有 G(,),,2if(t),sin,tdt.,0
为奇数。 G(,,),,G(,)
,,,,1i,,,,11i,t = ,,,,,,,,,GitGt()sind()sindf(t),G(,),ed,,G(,),(cos,t,isin,t)d,,,,,0,,,,,,2ππ22,,所以,当f(t)为奇函数时,有
,,,,2同理,当f(t)为偶函数时,有 ,,,,,,,,其中fttftt()b()sind.b()=()sindt.,,00π
,,ftat()()cosd,,,,,.其中 ,0
,,2,,,,afttdt()()cos ,0π
2,,t1t,,2.在上一题中,设.计算的值. ft(),a(),,0,t,1,,解:
,,,,12222aftttttttt()()cosdcosd0cosd,,,,,,,,,,,,,001πππ
1122122,,,,tttttcosdd(sin),,,,00ππ, 11222,,,,,tttttsinsin2d,,,00ππ,,
12sin4,,,,,tdt(cos)2,,0π,,,12sin41,,,,,,,tttdtcoscos2,0,,0,,,,π,,,s4sin,,2sin4co,,,,23,,,,,,
,,sin,6ttπ,3.计算函数. ft(),的傅里叶变换,0,6t,π,,
解:
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,,6π,,itit,,Ffftettet()()dsind,,,,,,,,,,,,6π
6π,,,sin(cossin)dttitt,,,,6π 6π,,,2sinsindittt,,0
isin6π,,2π(1),,
4.求下列函数的傅里叶变换
,t (1)()fte,
解:
,,,,,,ittittit,,,||(||),,,,,Ff()()ddd,,,,,fteteetet,,,,,,,,,,, 0,,2titi(1)(1),,,,,,,,etetdd2,,,,01,,
2,tftte(),,(2)
解:因为
2,,2222tttt/,,,,4 F[]eeeette,,,,,,,,π.()(2)2.而
2,,2π,t,4所以根据傅里叶变换的微分性质可得 Gtee,,,,()F(),i2
sinπt(3)ft(), 21,t
解:
,,sinπt,it,Gfet()F()()d,,,,,2,,,1,t
,,sinπt,,,(cossin)dtitt,,2,,,,1t 1,,,,[cos(π)cos(ttπ)],,,,,,sinπtt,sin2,,,,,ititd2d22,,,,011,,tt
,,,,cos(π+)cos(ttπ-),,利用留数定理,,ititdd()22,,0011,,tt
i,当,,sin,||π,,,,2,,当|,π.0,|,,
1ft(),(4) 41,t
解:
,,,,,,1cossin,,tt,it,Gettit()ddd,,,,1444,,,,,,,,,111,,,tttR(z)=令,则R(z)在上半平面有两个一级极点41z,,,,,coscostt,,,,2ddtt44,,,,011,,tt
22(1),(1),,,ii. 22
,,22itiziz,,, R()d2tetiReszeiiReszei,,,,,,,,,,π[R(),(1)]2π[R(),(1)],,,22
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it,,,,,cos1||||te,,,||/2,,故. ,,,dRe[d](cossin)tte44,,,,,,,,1122tt22
t(5) ft(),41,t
解:
,,tit,,Get()d,,,4,,,1,t
,,,,,sinttt, ,,,cosddttit,44,,,,,,11,,tt
,,,sintt,,,itd4,,,1,t
z同(4).利用留数在积分中的应用,令R()=z 41,z则
it,,,,,ttte,,sin,,,,ititd()Im(d)44,,,,,,11,,tt.
i,||/2,,,,,,esin22
5.设函数F(t)是解析函数,而且在带形区域Im()t,,内有界.定义函数为 G(),L
L/2,,it GFtt()()ed.,,L,,L/2
L,,证明当时,有
,1it, GFt,,,p.v.()ed()L,2π,,
对所有的实数t成立.
(
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上有推理过程)
,,1,0t,t6.求符号函数 的傅里叶变换. sgnt,,,1,0t,||t,
解:
1,,,,,F(())utπ().因为把函数sgn()t与u(t)作比较. i,
不难看出 sgn()()().tutut,,,
故:
11F[sgn()]F(())F(())tututπ()[π()],,,,,,,,,,,,,,iiπ(),, 22,,,,,π()(),,,,,,ii,,
F()=,,,,,,,π()(),,,,ft()ft()7.已知函数的傅里叶变换求 ,,00解:
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,,1-1it,()F(F())=π()()dfte,,,,,,,,,,,,,,,00,,,2π
,,it,,F(cos)=cosdttet,而,,00,,,
itit,00,,,,ee, ,it0,,,etd,,,2
,,,,π[()()]00,,,,,,
()cosftt,所以0,
8.设函数f(t)的傅里叶变换,a为一常数. 证明 F(),
1,,, [()]().fatF,,,,aa,,
,,,,1itit,,,, 解F[()]()()d()d():fatfatetfateat,,,,,,,,,,,a当a>0时,令u=at.则
u,,,,i11,,,a fatfueuFF[()]()()d,,,,,,,,,aaa,,
1,,,<0时,令,则. 当au=atF[()]()F()fat,aa故原命题成立.
9.设证明 FF,,,,,;f,,,,
. F,,,F,,f,,,,,,,,t
证明:
,,,,itiu,,, Fftfu,,,,,,,,,,,ededf,,,,,,tu,t,,,,,,
,,,,,,iiu,,,,,(),,,,u,,,,,,,fufu,,eded,,uu,,,,,,
,,,it,,,,,,,,,,ftFed.,,,,t,,,,,
FF,,,,,10.设,证明: f,,,,
1以及 ,,,,,,,,,,,,,,,,FftFFcos,,,,,,,t000,,2
1 ,,,,,,,,,,,,,,,,FftFFsin.,,,,,,,t000,,2
证明:
itit,,,00,,e+e,,cosFftF,,,,,,tf,,t0,,,,2
itit,00,,,,1,,,,ee,,FFff,,,,,,,,tt,,,,2,,,,22,,
1,,,,FF,,,,,,00,,,,,,2
同理:
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itit,,,00,,ee,FftF,,,,sin,,,f,,,tt0,,2i,,
1itit,00,,,,FF,,,,,,,,ff,,ee,,tt,,,,2i
1,,,,FF,,,,,,00,,,,,,2i
11.设
π,0,tt,0sin,,,,0t, ,,,,fg,,2tt,,,tt,,00e,,,其他,计算. fg*,,t
,,解: fg*(d,,fygy),t,y,,t,,,
当时,若则故 tyo,,f,0,t,0,y,,
=0. fg*,,t
,0,0,,,,,tyt若则 2
tt,y fg*,,,fygyey()dsind,,t,,yty,,,,t,,00
,,,ttytyt,,,,,,,,,0..若 222
t,y则 fg*,,,ey,sindt,y,,t,,,t2
0,0t,,
,,1,t,,0,,t,,故 sincosett,,,,fg*,t,22,1,,,t,e.t,,,21e,,22
12.设为单位阶跃函数,求下列函数的傅里叶变换. u,,t
,at ,,ftu,,,,,,esin,tt10
,,,,atit,GFtu解:,,,,,,,,esine,,,,tdf,,t,,0,,,
,,,,atit,,esine,,ttd,0,0
itit,00,,,,ee,,,atit,,ee,,td,0i2
,,,,11,,,,,,,,,,aitait,,,,00,,,,,,,,,,ettded,,00ii22
0,,22 ,,,ai,0,,
习题八
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1.求下列函数的拉普拉斯变换.
,4t2(1)(2)(3)fttt()sincos,,ft()e,ftt()sin,,,
2(4)(5)ftbt()sinh,ftt(),,
1解: (1) ftttt()sincossin2,,,2 1121LftLt(())(sin2),,,,222244ss,,
11,4t(2) LftL,,(())(e)s,24
1cos2,t2(3)ftt()sin,,2 1cos21111122,tLftLLt(())()(1)(cos2),,,,,,,,22222224(4)ssss,,
22(4)Lt(),3s
btbt,ee111111,bbtbt,(5)LftLLL(())()(e)(e),,,,,,,,2222222sbsbsb,,, 2.求下列函数的拉普拉斯变换.
2,01,,t,,1)(ftt()1,12,,,,
, 0,2t,,
cos,0tt,,π,(2) ft(),,0,t,π,
解: (1)
,,121stststss,,,,,2(())()e2ee(2ee)Lftftdtdtdt,,,,,,,,,,,001s
,πs,,π11e,,,,ststsπ(())()ecose(1e),,,,,,,Lftftdttdt(2)2,,00,1ss .设函数3,其中函数为阶跃函数, 求的拉普拉斯变换. fttttut()cos()sin(),,,,,ut()ft()
解:
,,,,,,ststst,,,Lftftdtttdttutdt(())()ecos()esin()e,,,,,,,,,,,,000
,,,,stst,,,,,,,cos()esinettdttdt,,,,,0
211s,st,,,,,,cose1t,t0222sss,,,111 4.求图8.5所表示的周期函数的拉普拉斯变换
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解:
T,stftdt()e,T1,asa,0Lft(()),,,T2,,asas1e(1e),,ss 5. 求下列函数的拉普拉斯变换.
t,2t(1)ftlt()sin,,(2) ftt()esin5,,2l
t,4t(3)(4)ftt()1e,,,ftt()ecos4,,
(5ftut()(24),,
(6 fttt()5sin23cos2,,
12,t2(7) (8) fttt()32,,,ftt()e,,
解:(1)
t1ftlttlt()sin[()sin],,,,,,22ll
t1FsLftLltLtlt()(())(sin)[()sin],,,,,,,22ll
112llss,,,,,,,,()2222222222()()lsllslsl,,,
5,2t(2)FsLftLt,,,,()(())(esin5)2s,,(2)25
1ttt(3)()(())(1e)(1)(e)(e)FsLftLtLLtLt,,,,,,,,,,,s
1111 ,(),,,,21(1)ssss,,
s,4,4t(4) FsLftLt()(())(ecos4),,,,2(4)16s,,
1,2t,,(5) ut(24),,,0,其他,
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,st,FsLftLututdt()(())((24))=(24)e,,,,,,0 ,12sts,,=e=edt,2s
(6)
FsLftLttLtLt()(())(5sin23cos2)5(sin2)3(cos2),,,,,
2103ss,,,,,,53222sss,,,444
13,,,(1)()1,t222(7) FsLftLt,,,,,()(())(e)3322,,ss,,()()
1222FsLftLttLtLtLss()(())(32)()3()2(1)(232),,,,,,,,,,(8)s 6.记,对常数,若 sLfsFs[]()(),0
st0,证明 Re()ss,,,LfsFss[e]()(),,,000证明:
,ststst,00Lfsftdt[e]()e()e,,,,,()nn0FsLfts()[(t)()](),,,7 记,证明: LfsFs[]()(),,,()()sstsst,,,00,,,,,,ftdtftdtFss()e()e()0,,00
证明:当n=1时,
,,st,Fsftdt()()e,, ,0
,,st,,,Fsftdt()[()e],,,0 st,,,,,,,[()e]ftst,,,,,,,,,dttftdtLtft()e(()),,00,s
()nn所以,当n=1时, 显然成立。 FsLfts()[(t)()](),,,
假设,当n=k-1时, 有
(1)1kk,, FsLfts()[(t)()](),,,
现证当n=k时
,,kst1,,(1)k,dtftdt()()e,,,dFs(),()k0Fs(),,dsds
kst,,1,,,,,,,[()()e]tftkst,,,,,,dttftdt()()e,,00,s
k,,,Lfts[(t)()]() 8. 记LfsFs[]()(),,如果a为常数,证明:
1sLfatsF[()]()(), aa
LfsFs[]()(),证明:设,由定义
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,,udu,stLfatfatdtatutdt[()]()e.(,,),,,,,令,0aa
ss,,uu,,,,du1aa,,,,fufudu()e()e,,00aa
1sF,()aa
9. 记,证明: LfsFs[]()(),
,ft(),,,ft()st,,即 ,LFsds[](),,e()dtFsds,,,s0stt
证明:
,,,,,,stst,,Fsdsftdtdsftdsdt()[()e]()[e],,,,,,,,sss0
,,,,1()()ftftstst,,,,,,,,,ftdtdtL()[e]e[]s,,00ttt
10.计算下列函数的卷积
11,(1) (2)tt,
tsinsinatat,(3)t,e(4)
sinsinatat,(5)(6 ,,()()tft,,
t)解:(1 1111,,,,dt,,0 t13tttdt,,,,,,,,()(2) , 06
(3) ttttttt,,,,,,tddd,,,,,,,,,,eeeeee,,,,,,,,000
t ttt,,,,,,,,,,e[e]ee1dt,,0,0
(4)
tt1sinsinsinsin()[coscos(2)]atataatdataatd,,,,,,,,,,,,,,,002
1tsincos2,,atat22a
(5) tt
,,,,,,,,,,()()()()()()()tfttftdtftdt,,,,,,,,,,,,,,00t0 t0,,,,,,,,,()()()()fdfd,,,,,,,,,ftt(),0,,,,,,,t0
(6)
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tt1sincossincos()[sinsin(2)]tttdttd,,,,,,,,,,,,,,002
ttt,,,sinin(2)tstd,,,022
1ttsincos(2),,,tt0,24
1tt,,,,,sin[coscos()]sintttt242 11.设函数f, g, h均满足当t<0时恒为零,证明 fgtgft,,,()()以及
()()()()fghtfhtght,,,,,,
证明:
tt0,,,hth()d,,,fgfg,,,(),,,,,,t,,,令,=,tu,fgtfgftugu,,,,,,,,,,()()d()d,,,,,,0tu,, ,,0ttttt,,,,,,fhtgh()d()d,,,,,,,,t,,,,,,,,ftugugfgft()d()d(),,,,,,,,ut,,,,,,0000
,,,,fhtgft()()
12.利用卷积定理证明
tFs()Lftdt[()],,0s
t,gtftdt()(),gtftg()(),(0)0,,且,0证明:设,则
,LgtsLgtgsLgt[()][()](0)[()],,,,则
,Lgt[()]Lgt[()],,所以 s
tFs()Lftdt[()],,0ds
13. 求下列函数的拉普拉斯逆变换.
s(1)Fs(),(1)(2)ss,,
2s,8(2) Fs(),22(4)s,
1(3)Fs(),sss(1)(2),,
sFs(),(4)22(4)s,
s,1Fs()ln,(5) s,1
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2ss,,21(6Fs(),2ss(1),
s21解:(1)Fs(),,,(1)(2)21ssss,,,,
2111,,,1112ttLLL()2()()2ee,,,,,
,,,,ssss2121 (2)
22ss,,8321431,,11FsLLttt()()()sin2cos2,,,,,22222(4)442(4)42sss,,,
1111(3Fs(),,,,ssssss(1)(2)212(2),,,,
11,,,12tt故LFs,,,(())ee22
ss1412,,(4)Fs()(),,,,,,,222222(4)4(4)42sss,,, 因为
2,1,Lt()sin222,s2
所以
1st,,11,,,,LFsLt(())()sin222,4(4)4s
,sgt,111()FsduL()ln()(),,,,,(5),0suut,,,111 其中
11,,1ttgtL,,,,()()ee ss,,11
所以
,,tttteeee,,FsLL()()(),,, tt
,,tttteeee,,sht1,ftLFs()(())2,,,,,, ttt
2ss,,21122(6)Fs(),,,,,22sssss(1)1(1),,, 所以
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122,,,,1111LFsLLL,,,,(())()()()2sss,,1(1)
tttt,,,,,,,tt12e2e2e2e1 14.利用卷积定理证明
st,1Lat[]sin,,22()2saa,
证明:
ssa1,,11,,,LL[]()2222222,,,()sasasaa 又因为
sa LatLat(cos),(sin),,2222sasa,,
所以,根据卷积定理
sa11,1Latat()cossin,,,,2222sasaaa,,
tt111,,,,,,,cossin()[sinsin(2)]aatadataatd,,,,,,,00aa2
t,,sinat2a 15.利用卷积定理证明
t212,,ty1Ldy,[]ee,0ss,(1)π 证明:
111,,11,,LL[][],s1,sss(1)
t212,,ty1Ldy,[]ee,0ss,(1)π 因为
1,111t,,112 LtL,,(),()es,1sπ
所以,根据卷积定理有
111tt,,,1121,,,1()ttyty222Ltdydy[]eyeeye,,,,,,,00(1)ss,πππ
ttt22222,yu令2tytuty,,,,,,,,,,eeeeeedydudy,,,000πππ
16. 求下列函数的拉普拉斯逆变换.
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11(1)(2) Fs(),Fs(),4222ss,,54(4)s,
s,2(3)Fs(),22(45)ss,,
2233ss,,(4)Fs(),2 (1)(3)ss,,
解:(1)
22112(4)14ss,,Fs(),,,,,222222(4)16(4)8(4)sss,,, 21214s,,,,,2221648(4)ss,,
2121411s,111,,,故 LFsLLttt(())()()sin2cos2,,,,,2221648(4)168ss,,
(2):
1111Fs()(),,,4222ssss,,,,54314
1112 ,,()2223122ss,,
1112,,,111LFsLL(())()(),,2223162ss,,
11,,sinsin2)tt36
ss,,2211,Fs()(),,,,(3)22222(45)[(2)1]2(2)1ssss,,,,,,
1,,12tLFstt,,,(())esin故 2
(4)
2233ssABCD,,Fs(),,,,,223(1)(3)13(3)(3)ssssss,,,,,,
113 ,,,,,,ABCD,,,3442
故
113,3442Fs(),,,,23ssss,,,,13(3)(3) 且
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1111,,,(),()2,,,,23ssss,,,,3(3)3(3)
所以
113,,,,,13323ttttLFstt,,,,,,(())eee3e442
17.求下列微分方程的解
,t,,,,(1)yyyyy23e,(0)0,(0)1,,,,,
(2) ,,,,yyttyy,,,,,,,4sin5cos2,(0)1,(0)2
t(3) ,,,,yyytyy,,,,,,222ecos2,(0)(0)0
2t,,,,,,,(4)yyyyy,,,,,e,(0)(0)(0)0
(4)(5) ,,,,,,,,yyyyyyy,,,,,,,20,(0)(0)(0)0,(0)1
解: (1)设
,LytYsLytsYsysYs[()](),[(()]()(0)(),,,,, 22,,,LytsYssyysYs[(()]()(0)(0)()1,,,,,
得 方程两边取拉氏变换,
12 sYssYsYs,,,,,,()12()3()s,1
12s,2 (23)()1ssYs,,,,,ss,,11
ss,,22Ys(),, 2(1)(23)(1)(1)(3)ssssss,,,,,,
sss,,,,,1,1,3为Y(s)的三个一级极点,则 123
31,stytLYssYss()[()]Re[()e;],,,,k1k,stst(2)e(2)ess,,,,,,,Re[;1]Re[;1]ss (1)(1)(3)(1)(1)(3)ssssss,,,,,,
st(2)es,,,,Re[;3]s(1)(1)(3)sss,,,
1313ttt,,,,,,eee488
(2) 方程两边同时取拉氏变换,得
1s2sYssYs,,,,,,,,()2()45 222ss,,12
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1s2(1)()45(2)sYss,,,,,,,222ss,,12
452ss,Ys(),,,222222(1)(1)(1)(2)(1)sssss,,,,, 11112s,,,,,,,2()()s2222222ssssss,,,,,,111211
2s,,,222ss,,12
,1 ytLYstt()[()]2sincos2,,,,
(3)方程两边取拉氏变换,得
s,12 sYssYsYs,,,,,,()2()2()22(1)1s,,
2(1)s,2(22)()ssYs,,,2(1)1s,,
2(1)1s,, Ys()[],,,222[(1)1](1)1ss,,,,
因为由拉氏变换的微分性质知,若L[f(t)]=F(s),则
,LtftFs[()()](),,,
即
,,11,LFstfttLFs[()]()()()[()],,,,,,
1,1t因为 Lt,,[]esin2s,,(1)1
所以
2(1)1s,,,11,LL{}[()],,222[(1)1](1)1ss,,,,
1t,1,,,,,,()[]esintLtt2(1)1s,,
t故有 ytt,,,,,esint
(4)方程两边取拉氏变换,设L[y(t)]=Y(s),得
132,,,sYssysyysYsy,,,,,,,,,,()(0)(0)(0)()(0)s,2
13sYssYs,,,,()()s,2
111 Ys(),,,22ssssss,,,,2(1)(2)(1)
故
113,,,,,12323tttt ytLYs,,,,,,,()[()]eete3te442
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(5)设L[y(t)]=Y(s),则
,LytsYsysYs[(()]()(0)(),,,,
22,,,LytsYssyysYs[(()]()(0)(0)(),,,,,
323,,,,,,LytsYssysyysYs[(()]()(0)(0)(0)()1,,,,,,,, (4)4324,,,,,,LytsYssysysyysYss[(()]()(0)(0)(0)(0)(),,,,,,,,,,,
方程两边取拉氏变换,,得 42sYsssYsYs,,,,,,()2()()042(21)()ssYss,,,,
ss1211,Ys()(),,,,,,22222(1)2(1)21sss,,,
故
s111,,11,ytLLtt()[][()]sin,,,,,, 222(1)212ss,,
18.求下列微分方程组的解
t,,xxy,,,e,xy(0)(0)1,,(1),t ,yxy,,,,322e,,
,,xygt,,2(),,,xxyy(0)(0)(0)(0)0,,,,,(2) ,,,,xyy,,,0,
解:(1) 设
LxtXsLytYs[(()](),[(()](),,
,LxtsXsxsXs[(()]()(0)()1,,,,,,
,LytsYsysYs[(()]()(0)()1,,,,,,,微分方程组两式的两边同时取拉氏变换,得
1,sXsXsYs,,,,,()1()(),,s,1, 2,sYsXsYs,,,,,()13()2(),s,1,
得
s,YssXs()(1)()...(1),,,,,s,1, 21s,,3()(2)()1...(2)XssYs,,,,,,,ss,,11,
(2)代入(1),得
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ss,13()(2)[(1)()]XsssXs,,,,,,ss,,11
2sssss,,,,1(2)12(1)()ssXs,,,,, sss,,,111
1t故于是有Xsxt()()e...(3),,s,1
(3)代入(1),得
11st Yssyt()(1)()e,,,,,,,sss,,,111
(2)设
LxtXsLytYsLgtGs[(()](),[(()](),[(()](),,,
方程两边取拉氏变换,得,,LxtsXsLytsYs[(()](),[(()](),,,,
22,,,,LxtsXsLytsYs[(()](),[(()](),,,,,
sXssYsGs,,,,()2()()...(1),
,22,,sXssYsYs,,,,,()()()0...2,
(1)(2),,,s得
sYsGs()()...(3),,,2s,1
t,1?,,,,,ytLYsgttgd()[()]()*coscos,,,,t,,,0 (3)代入(1):
ssXssGsGs,,,,,,()2[()]()2s,1
即:
22 21ss,sXsGsGs,,,,,()(1)()()22ss,,11
21,s12s,,XsGsGs()()(),,,,,,22,,sss1,,,s,1
所以
t,1?,,,,,,,,xtLXstgtgtd()[()](12cos)()(12cos)(),,,,0
故
txtgtd()(12cos)(),,,,,,,,0
tytgtd()()cos(),,,,,,,,0
19.求下列方程的解
t,xtxtdt()()e23,,,,,,,(1),0
tyttydt()()(),,,,,,,(2),0
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解:(1)设L[x(t)]=X(s), 方程两边取拉氏变换,得
123XsXs()(),,,,2sss,1
123,sXs()[1],,2 ss,1
2(23)(1)352352,,,,,ssssXs(),,,,,,3323sssss
2,,,,,xttt()35
(2)设L[y(t)]=Y(s), 方程两边取拉氏变换,得
1YsLtyt()(()),,,2s
11YsYs()(),,,22ss 1Ys(),2s,1
111,,,,,,ytLYsLsht()(())()2s,1
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