复变函数拉氏变换部分习题解答分析.pdf

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2、t•¢ëê)¤L«•‚´Ô‚y=x2..x=y2.W˜K1.eªi(57i)=(x+i)(yi)¤á,Kx=,y=.©Û:üEꃽÂ.x=6;y=1,½x=1;y=6.2.•§Im(iz)=3L«•‚´.©Û:dEêƒ,Im(iz)=Im[i(xiy)]=Im[x+(1+y)i]=1+y=3,Wy=2.3.•§z3+27=0Š•.p©Û:z3=27ei;z=271=3(cos(+2k)+sin(+2k));k=0;1;2,=z=3;333i.33224.EC¼êw=z

3、2¢Üu(x;y)=;JÜv(x;y)=.z+1x2x+y223y©Û:òz=x+iy“,©l¢Ü!JÜ,u(x;y)=(x+1)2+y2;v(x;y)=(x+1)2+y2.5.z1=2i;z2=1i,KArg(z1z2)=.©Û:arg(z)=;arg(z)=;Arg(zz)=+2k=+2k;(k=0;1;2;


)122412244p6.Eêz=122inL«ª•,•êL«ª•.55i(5)©Û:4[cos()+isin()];4e6.66nOŽ!y²Kp1.¦Ñ

4、Eêz=(1+3i)4ÚË.p844224i2)z=(1+3i)=2(cos+isin)=16e3;jzj=16;Arg(z)=+2k;k=0;1;2;


.3332.z=x+iy÷vRe(z2+3)=4,¦x†y'Xª.)Re(z2+4)=Re(x2y2+3+2xyi)=4;x2y2=1.3.¦f(z)=1ò²¡þ†‚y=1¤N¤w²¡þ•‚•§.z)dw=1z=1;x+iy=1=uvi.qdy=1v=1;u2+v2+v=0.zwu+ivu2+v2u2+v2u2+v24.¦/

5、•0

6、y)Úv(x;y)Œ,Kf(z)=u+iv•Œ.1.eu(x;y)Úv(x;y)Œ,Ku;vƒm˜„vkŸo†'X.f(z)=u+ivŒ,u;vƒm˜‡A(½,˜‡(¹Ä{/•´˜‡~ê).3.ef(z)3z0:Ø)Û,Kf(z)3:z07ØŒ..ë„n2.4.jsinzj1..EC¼ê¥,sinzÃ..Xjsinikj=jeiikeiikj=jekekj!+1(k!+1;k>0).2i25.¼êf(z)=u(x;y)+iv(x;y)3:z0=x0+iy0Œ‡duu(x;y)Úv(x;y)

7、3:(x0;y0)Œ‡..¼êf(z)=u(x;y)+iv(x;y)3:z0=x0+iy0Œ‡duu(x;y)Úv(x;y)3:(x0;y0)Œ‡…÷vCR^‡.‡~u=x;v=y:du=dx+0dy;dv=0dxdy;u;vÑŒ‡f(z)=u+iv=xiyÃ?Œ‡.6.¼êez´±Ï¼ê.p.2i•Ù±Ï.W˜K1.ez=3+4i,KRe(iz)=©Ûµéz=3+4iü>g,éê§kz=Ln(3+4i)=lnj3+4ij+iarg(3+4i)+2ki,lRe(iz)=i[iarg(3+4i

8、)+2ki]=arctan4+(2k+1);Ù¥k¤kê.(5µùp´l8ÜÝ`)32.3i=©Ûµ3i=eiLn3=ei[ln3+iarg(3)+2ki]=ei[ln3+2ki]=e2k(cosln3+isinln3),Ù¥k¤kê.3.(1+i)i=p