收藏 分享(赏)

复变函数部分习题解答分析(复拉).pdf

上传人:weiwoduzun 文档编号:3294840 上传时间:2018-10-10 格式:PDF 页数:8 大小:503.69KB
下载 相关 举报
复变函数部分习题解答分析(复拉).pdf_第1页
第1页 / 共8页
复变函数部分习题解答分析(复拉).pdf_第2页
第2页 / 共8页
复变函数部分习题解答分析(复拉).pdf_第3页
第3页 / 共8页
复变函数部分习题解答分析(复拉).pdf_第4页
第4页 / 共8页
复变函数部分习题解答分析(复拉).pdf_第5页
第5页 / 共8页
点击查看更多>>
资源描述

1、Mf s5ssT 1+3i. , L H, V1 vl.2. zB ,5z 6= z.p. l,B yi; y 6= 0, z = yi,5z = yi.3.f w = arg(z)z = 3) .p.zV/Z!3 H, w = arg(z)K;zV Z!3 H, w = arg(z)K.4.f(z) = u+ivz0 = x0 +iy0 sA1Hq u(x;y); v(x;y)(x0; y0) .p. Th1.4.3.5. Zz = t2 +ti (t L ) V U wL Ly = x2. x = y2.=A b51. Ti(57i) = (x+i)(y i) ,5x= ,y = .s:

2、Ml. x = 6;y = 1,x = 1;y = 6.2.ZIm(i z) = 3V U wL .s: M, Im(i z) = Imi(xiy) = Imx+(1+y)i = 1+y = 3,#Ay = 2.3.Zz3 +27 = 0.s: z3 = 27ei;z = 271=3(cos(+2k3 )+sin(+2k3 );k = 0;1;2, z = 3; 32 32p3i.4.Mf w = z2z+1 Lu(x;y) = ;v(x;y) = .s:|z = x+iy ,s La,u(x;y) = x2x+y22(x+1)2+y2 ; v(x;y) = 3y(x+1)2+y2.5. !z

3、1 = 2i;z2 = 1i,5Arg(z1z2) = .s: arg(z1) = 2; arg(z2) = 4; Arg(z1z2) = 2 4 +2k = 4 +2k;(k = 0;1;2;)6. z = p122i V U T, V U T.s: 4cos(56)+isin(56); 4ei(56).9 a 51. p z = (1+p3i)4 .z = (1+p3i)4 = 24(cos 23 +isin 23 )4 = 16ei83 ;jzj = 16; Arg(z) = 23 +2k;k = 0;1;2;.2. !z = x+iy Re(z2 +3) = 4, pxy1“ T.Re

4、(z2 +4) = Re(x2 y2 +3+2xyi) = 4; x2 y2 = 1.3. pf(z) = 1z| Ly = 1 w wLZ.w = 1zz = 1w;x+iy = 1u+iv = uu2+v2 vu2+v2i.y = 1 vu2+v2 = 1; u2 +v2 +v = 0.4. p0 0).5.f f(z) = u(x;y)+iv(x;y)z0 = x0 +iy0 VNu(x;y)v(x;y)(x0; y0) V.f f(z) = u(x;y) + iv(x;y)z0 = x0 + iy0 VNu(x;y)v(x;y)(x0;y0) V O CRHq.Q u = x; v =

5、 y: du = dx+0dy; dv = 0dxdy; u; v Vf(z) = u+iv = xiy) V.6.f ez f .p. 2i .=A b51. !ez = 3+4i,5Re(iz) =sz = 3 + 4i H |1 z = Ln(3 + 4i) = lnj3 + 4ij+ iarg(3 + 4i) + 2ki,V7Re(iz) = iiarg(3+4i)+2ki = arctan 43 +(2k +1).( V“ )2. 3i =s3i = eiLn3 = eiln3+iarg(3)+2ki = eiln3+2ki = e2k(cosln3+isinln3).3. (1+i

6、)i =s(1+i)i = eiLn(1+i) = eilnj1+ij+iarg(1+i)+2ki = eilnp2+i4+2ki = e2k4(coslnp2+isinlnp2)4. cos2i =scos2i = ei2i+ei2i2 = e2+e22 = cosh2.( T V)5.Zeiz = eizz =s Heiz,e2iz = 1. H |1 2iz = Ln1 = lnj1j+iarg(1)+2ki = 2ki; z =k.6. !z = x+iy,5ei2z sjei2zj = jei2(x+iy)j = e2x.7.f f(z) = u+ivz0 = x0 +iy0 f(z

7、)Hq.sf(z),5f(z) B # = V bAA1.9 a 51.k | Hf(z) = kln(x2 +y2)+iarctan yxx 0 = f .s: 1Hq. ux = 2kxx2+y2;uy = 2kyx2+y2;vy = 1x1+yx2 =xx2+y2;vx =yx2+y2.ux = vy;uy = vx:k = 12,k = 12 Hf(z)x 0 = f .2.) f f(z) = (xy)2 +2(x+y)i ) V, ),i p V) .s: V Q# “, V 1Hq. ux = 2(xy); uy = 2(yx); vx = 2; vy = 2.ux = vy;

8、uy = vxxy = 1.#f(z)xy = 1 V, f0(z) = ux +ivx = 2+2i,).3. f f(z) = u+iv, Ou = v2, pf(z)B .2s: 1Hq. ux = 2vvx = vy; uy = 2vvy = vx TMi (4v2 +1)vxvy = 0. Tvx vy 0, v .u = v2 , u ,V7f(z) = const:4. f f(z) = u+iv, Ouv = (xy)(x2 +4xy +y2), k pu(x;y)v(x;y).s: 1Hq.u v = (x y)(x2 + 4xy + y2) (0),u = v + x3 +

9、 3x2y 3xy2 y3.ux = vy;uy = vx,: vx +3x2 +6xy3y2 = vy (1) vy +3x2 6xy3y2 = vx (2)(1),(2)vy =6xy ) v = 3xy2 + C(x) (3). ux = vy = 6xy ) u = 3x2y + D(y) (4)|(3),(4) (0) T,u =3x2y y3 +C;v = 3xy2 x3 +C.5. pZchz = 0 .s:wf l.EBchz = ch(z) = ch(iiz) = cos(iz) = 0; z = (k + 12)i.E=chz = ez+ez2 = 0; e2z +1 =

10、0: 2z = Ln(1) = lnj1j+iarg(1)+2ki; z = (k + 12)i.T wL5HC f(z)dz = 0.sf(z)D f(z)C # =b1oM Y u.Q f(z) = 1z0 wL,5HC F(n)(z)dz =0.p.s:5 !, F(z) Y uD =,V7F(n)(z) Y uD =. CD =BH_ wL,5HC F(n)(z)dz = 0.4. !v = v(x;y) uD =f ,5f f(z) = v0y +iv0xD =.p.s: v = v(x;y) uD =f ,5 !u f v, F(z) = u+ivD =,V7f(z) = F0(z

11、) = v0y +iv0x.5. f f(z) = u(x;y)+iv(x;y)D =,5f 2uxy = 2uyx.p.s: 2uxy, 2uyxi O 2uxy = 2uyx sHq. f f(z) = u(x;y)+iv(x;y)D =,5f(z) if D =,V72uxy, 2uyxi O . u= 2uxy = 2uyx.=A b51. !CVz1 = iz2 = 0L 5RC zdz =12.s:$f f(z) , Bf 12z2,#RC zdz =12z2j0i =12.2. C_jzj = 2,5HC 1zdz =0.s:zz = jzj2,C 1z = zjzj2 = z4

12、; HC 1zdz = HC z4dz = 0.3. C_jzj = 1,5HCln(z +2)+(z2 +1)cos(z5 +1)dz =0.s:C #C =,$f . Cauchys .4. f f(x;y) = epx siny uD =f 5p=1.s: fxx +fyy = p2epx siny epx siny = (p2 1)epx siny 0.5. f() = Hjzj=22z2+z+1z dz;jj6= 2,5f(3+5i) = ( ); f(1) = ( ); f0(1) = ( ):s:f(z)Vr T, Cauchys #Cauchys T,:3f() =80;0;

13、if jj , I = HC12z+1+2iz+2iz+1 dz +HC22z+1+2iz+1z+2i dz = 2i2z+1+2iz+2i jz=1 +2i2z+1+2iz+1 jz=2i = 4i.5.9 s12i HC ezz(1z)3 dz, (1)0C =,1C; (2)1C =,0C; (3)0; 1 (C =; (4)0;1 (C.(1) I = 12i HCez(1z)3z dz =ez(1z)3jz=0 = 1. (2) I =12iHCezz(z1)3 dz =12! (ezz )00jz=1 = e2. (3)1e2. (4)0.6. u(x;y) = y3 3x2yf

14、, p f v(x;y),if(z) = u + iv1zVr T.: ux = 6xy; uxx = 6y; uxy = 6x; uy = 3y23x2; uyx = 6x; uyy = 6y; u = iO , uxx +uyy = 0; u(x;y)f .vy = ux = 6xy; v = R(6xy)dy = 3xy2 + (x): uy = 3y2 3x2 = vx = 3y2 0(x); (x) =x3 +C: v = x3 3xy2 +C. pf(z)Vr T,5 If0(z). f0(z) = ux + ivx = ux iuy = 6xy i(3y2 3x2) = 3iz2

15、.V7f(z) = iz3 +iC , C L . :N59 V“S:u5 pf0(f0 = ux iuy), pf f = u + ivv.“u; v,f , (f , Ovu f .“SA e.pf(z)Vr T 6Ef(x) = f(x + i0) = u(x;0) + iv(x;0) = i(x3 + C) = ix3 + iC;#f(z) =iz3 +iC: (NE G I $ z2i) u(0; 0) + iC (u(x; y) =12f(x+iy)+f(xiy); f(x+iy) = u(x; y)+iv(x; y); f(xiy) = u(x; y)iv(x; y):T 1),

16、5z0Af0(z)m+1).p. f(z) = (z)(zz0)m = (z)(z z0)m;f0(z) = 0(z)(z z0)m m(z)(z z0)m1 = (zz0)0(z)m(z)(zz0)m+1 .4. z0 = 0 tanzz V .p. limz!z0 tanzz = 1.5.X1(z1)(z2) = P1n=0(1)n(z2)n21 , .(2) Res(f;0) = limz!0zf(z) = limz!0z+1z2 = 12; Res(f;2) = limz!2(z 2)f(z) = limz!2z+1z = 32.(3)E1 f(z) = 1e2zz4 = 1z4f11

17、+2z + (2z)22! + (2z)3 + (2z)nn! +g,NRes(f;0) = c1 = 43.E2 Res(f;0) = 1(41)!limz!0(z4f(z)000 = 16limz!0(1e2z)000 = 43.(4) Res(f;i) = 12!limz!id2dz2(z i)3 1+z4(z+i)3(zi)3 = 38i; Res(f;i) =12! limz!id2dz2(z +i)3 1+z4(z+i)3(zi)3= 38i.(5) f(z) = ze3z = z(1+ 3z + 322!z2 + 3nn!zn +); Res(f;0) = c1 = 92.T B

18、61. pf(z) = cotz(z1)2 .f(z) = cosz(z1)2 sinz;z = 1f(z) ), z = 0;1;2;3;B).2. !Cjzj = 2_, pHC zz41 dz.zz41 :1; 1; i; i: Res(f; 1) = limz!1(z 1) z(z2+1)(z+1)(z1) = 14; Res(f; 1) = limz!1(z +1) z(z2+1)(z+1)(z1) = 14; Res(f; i) = limz!i(zi) z(z21)(z+i)(zi) = 14; Res(f; i) = limz!i(z+i) z(z21)(z+i)(zi) =1

19、4:#HC zz41 dz = 2iP4n=1 Res(f; zi) = 0; zi , i = 1; 2; 3; 4:3.9 Hjzj=1 z4 sin 1z dz. T= Hjzj=1 z4(1z 13!z3 + 15!z5 )dz = 2i 15! = 60i.4. pRC cosz1sinz dz, Cjz 34j = 2_.cosz1sinz z = k;k i .C = z = ;C ,# T= 2iRes(f;) = 2ilimz!(z )cosz1sinz = 2ilimz! 1cos (2) = 4i.5. !z0f(z)B), OResf(z);z0 = a;7(z)z0,

20、(z0) = b 6= 0, k: Resf(z)(z);z0 =ab.Resf(z);z0 = a;#z0f(z)B), V !f(z) = P+1n=1 cn(z z0)n; c1 = a.(z)z0, V !(z) = P+1n=0 dn(z z0)n; d0 = b.5f(z)(z) = P+1n=1 cn(z z0)nP+1n=0 dn(z z0)n,z1“ c1d = ab,Resf(z)(z);z0 = ab. (: V A, b 6= 011.)9 VB = pE.6.9 sR20 12+cos d .cf p.94 5.4.1.7.9 sR+11 xsinxx2+4 dx.R

21、+11 xeixx2+4 dx = R+11 xcosxx2+4 dx+iR+11 xsinxx2+4 dx = 2ilimz!2i(z2i) zeiz(z+2i)(z2i) = e2i,NR+11 xsinxx2+4 dx =e2.T 1./ LaplaceMTLe3t cos2t = : LR10 cos2tdt = : Ltcos2t = : L(t1)2et = .(1)Lcos2t = ss2+4;M; Le3t cos2t = s+3(s+3)2+4: (2)LR10 cos2tdt = Lsin22 = sin22s (3)Lcos2t =ss2+4;f s; Ltcos2t =

22、 ddsss2+4 =s24(s2+4)2: (4)L(t 1)2et = L(t2 2t + 1)et =Lt2etL2tet+Let = (1)2 d2ds2 1s+1 2(1) dds 1s+1 + 1s+1 = s2+1(s+1)32. p/ f fM(1) f(t) = 3 0 t 22 t 2 (2) f(t) =Rt0sintt dt (3) f(t) = sin2 kt (4) f(t) = 1+(t)+u(t 12)(1)Lf(t) = R+10 f(t)est dt = R20 3est dt+R+12 2est dt = 3s 1se2s:(2)Lf(t) = 1sLsi

23、ntt = 1s R1s 1s2+1 ds = 1s arctansj1s = 1sarccots:(3)Lf(t) = L1cos2kt2 = 12fL1Lcos2ktg = 12f1s ss2+4k2g = 2k2s(s2+4k2):(4)L1+(t)+u(t 12) = 1s +1+ 1ses2.3. p T IM(1)L1 1s4 = Res(ests4 ; 0)Z 7= t36 :(2)L1 1s2(s1) = L11s + 1s2 + 1s1 = et t1:(3)L1s2+2s1s(s1)2 = L11s + 2s1 + 2(s1)2 = 1+2et +2tet = 2(1+t)

24、et 1.4. p/ f fM(1)f(t) = ete2tt (2)f(t) = tRt0 e3t cos2tdt7(1)Let e2t = 1s1 1s2;#Lete2tt = R1s ( 1s1 1s2)ds = ln s2s1:(2)Lcos2t = ss2+4; Le3t cos2t = s+3(s+3)2+4; LRt0 e3t cos2tdt = 1s s+3(s+3)2+4; LtRt0 e3t cos2tdt =( s+3s(s2+6s+13)0 = 2s3+15s2+36s+39s2(s2+6s+13)2 .5. p/ f f IM(1)F(s) = s2+2s+2(s+1

25、)(s+2)2 (2)F(s) = ln s+2s2(1)LF(s) = 1s+1 2(s+2)2 = et 2te2t:(2)F0(s) = 22s24; L1F0(s) = 2sh2t; L1F(s) = L1R1s F0(s)ds = 2sh2tt .6.XLcos2t = ss2+4; kLaplaceM9 lsR+10 te3t cos2tdt.R+10 est cos2tdt = ss2+4; dds R+10 est cos2tdt = R+10 test cos2tdt = dds ss2+4 = s24(s2+4)2;#R+10 te3t cos2tdt = s24(s2+4

26、)2js=3 =5169.7. LaplaceM psZy00 +2y0 3y = et Hqy(0) = 0; y0(0) = 1+.Ly00 + 2y0 3y = Let; s2F(s) sy(0) y0(0) + 2sF(s) 2y(0) 3F(s) = 1s+1; F(s) =s+2(s+3)(s+1)(s1); y = L1F(s) = ResF(s)est; sk = lims!3(s+3) s+2(s+3)(s+1)(s1)est+ lims!1(s+1) s+2(s+3)(s+1)(s1)est+lims!1(s1) s+2(s+3)(s+1)(s1)est = 38et 14et 18e3t.E=(ss T) F(s) = s+2(s+3)(s+1)(s1) = As+3 + Bs+1 + Cs1; s+2 = A(s+1)(s1)+B(s+3)(s1)+C(S +3)(S +1):sY 7s !3; 1 1;A = 18; B = 14; C = 38: p IM, p+.8

展开阅读全文
相关资源
猜你喜欢
相关搜索

当前位置:首页 > 中等教育 > 小学课件

本站链接:文库   一言   我酷   合作


客服QQ:2549714901微博号:道客多多官方知乎号:道客多多

经营许可证编号: 粤ICP备2021046453号世界地图

道客多多©版权所有2020-2025营业执照举报