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( Ka M ——KaX:& -K[P ;O_ 2 m&p2 X Hy )%&pyJK GTA t ,℄(-Jm\ 48 TAAkG \q, BY t4Zg ℄6Zg ,m ’d.dmz’:mathwalker@163.com.——? k2009[3Q6 B do $ZO2 1 o $ZO2 4 Fo $ZO2 11 LoZO2 14 VoZO1-6O2 22 oZO2 32iii ep %\Q3 1.(1)−2i;(2)−25;(3)i2;(4)−4;(5) (x+iy)2=a+bi,E’x2−y2=a,2xy=b.(1.1)T l(x2+y2)2=(x2−y2)2+4x2y2=a2+b2,4 Ax2+y2=√a2+b2.-L(1.1)qs)# lx2=12(a+√a2+b2),y2=12(−a+√a2+b2).~ l1*xi,1*yi,e.X?0 L℄ i, eE (1.1)qs #. slfb6=0 ,√a+bi=±sa+√a2+b22+ib|b|s−a+√a2+b22. Eb=0ml ,fa≥0 ,ei^±√a;fa0 ,ei^±i√−a.2.(1)1−2i3−4i−2−i5i=16+8i25, 3^1625, 3^825,Q^8√525, w^arctan12+2kπ,k=0,±1,±2,···.(2)1+√3i2n=eiπ3n=einπ3=cosnπ3+isinnπ3,n=2,3,4 e, tl oQ}’1. )2,1+√3i2!n=−12+i√32,n=2;−1,n=3;−12−i√32,n=4.A-n=2 , 3^−12, 3^√32, w^2π3+2kπ,k=0,±1,±2,···;n=3 , 3^−1, 3^0, w^(2k+1)π,k=0,±1,±2,···;n=4 , 3^−12, 3^−√32, w^4π3+2kπ,k=0,±1,±2,···.(3)*s1G(5)m BA√1+i=±s1+√22+is−1+√22.:U,4^1+i=√2eiπ4,A-√1+i=4√2eiπ/4+2nπ2,n=0,1.A-n=0 , 3^q1+√22, 3^q−1+√22,Q^4√2, w^π8+2kπ,k=0,±1,±2,···;n=1 , 3^−q1+√22, 3^−q−1+√22,Q^4√2, w^9π8+2kπ,k=0,±1,±2,···.(4)4√3+i=2eiπ6,;(√3+i)−3=2−3e−iπ2=−i8.A-(√3+i)−3m 3^0Æ 3^−18,Q^18, w^−π2+2kπ,k=0,±1,±2,···.(5) 3^12, 3^−√32,Q^1, w^−π3+2kπ,k=0,±1,±2,···.3.4z1=eiπ4,z2=2e−iπ6,;z1z2=2eiπ12=2(cosπ12+isinπ12),z1z2=12ei5π12=12(cos5π12+isin5π12).4.z+1z=2cosθl,z+1z2=z2+2+1z2=4cos2θ.o#1$N qw4lz−1z2=−4sin2θ.A-z−1z=±2isinθ.Tlz=e±iθ.A-zm+1zm=e±imθ+e∓imθ=2cosmθ.12 5.(1)z=−21+√3i=−112+√32i=e−iπeiπ3=ei2π3eargz=2π3.(2)4√3−i=2e−iπ6,;z=(√3−i)6=26e−iπ=26eiπ,A-argz=π.6.(1)i(1−√3i)(√3+i)=eiπ2·2e−iπ3·2eiπ6=4eiπ3=2+2√3i.(2)5i2+i=5eiπ/2√5eiarctan12=√5ei(π/2−arctan12)=√5cos(π2−arctan12)+isin(π2−arctan12)=√5sin(arctan12)+icos(arctan12)=√51√5+i2√5=1+2i.(3)(−1+i)7=√27ei7·3π4=√27ei5π4=√27(−√22−i√22)=−8(1+i).(4)(1+√3i)−10=2−10e−i10π3=2−10ei2π3=2−11(−1+√3i).7.z4=−a4=a4eiπl4*/^zn=aei(2n+1)π4,n=0,1,2,3.z0=√22a(1+i),z1=√22a(−1+i),z2=√22a(−1−i),z3=√22a(1−i).8.(1) z1=x1+iy1,z2=x2+iy2,VRe(z1+z2)=x1+x2=Re(z1)+Re(z2).tz1=i,z2=1+i,VRe(z1z2)=−16=0=Re(z1)Re(z2).A-Re(z1z2)6=Re(z1)Re(z2)) .E,.(2)bM NÆ.(3)|z1z2|=|z1||z2|mbMrx5s6&.tz1=1,z2=i,V e) |z1+z2|6=|z1|+|z2|.9.(4)z1z2+z1z2=z1z2+z1z2=2Re(z1z2)=2Re(z1z2)=2Re(z1z2).(6)|z1z2+z1z2|=2Re(z1z2)≤2|z1z2|=2|z1||z2|=2|z1||z2|=2|z1z2|.(7)4^|z1+z2|2=(z1+z2)(z1+z2)=|z1|2+|z2|2+z1z2+z1z2=|z1|2+|z2|2+2Re(z1z2)I2Re(z1z2)≤2|z1||z2|,A-|z1+z2|2≥|z1|2+|z2|2−2|z1||z2|=(|z1|−|z2|)2,|z1+z2|2≤|z1|2+|z2|2+2|z1||z2|=(|z1|+|z2|)2.11.|z1|=|z2|=|z3|=1,. z1,z2,z3 Z _ ~ 0TbNv|z|=1Æ,kz1=1.z1+z2+z3=0I~2i ma :$ V e,z2bEÆ aL,‘z2=eiθ(0θπ).z3=−(z1+z2)I~2i ma :$ V e,z3=ei(π+θ/2).Dz2+z3=−z1=−1I~2i ma :$ V e,12(θ+π+θ2)=π. lz2mx wθ=2π3,E’z3m wπ+θ/2=4π3.A-z1,z2,z3’)*W|EbNv|z|=1ma w myu.12.(1).|uz1Iz2mhv mPha v,’!^,4 .’qH.(2)-Duz=2a E wmhv^ m !aL,.’qH.(3)C aL,’qH.(4)rQ1.1qm593 , ETRez=2,Rez=3Æm$ ,.’qH.(5)Æ aLww bN^ om3 ,.’qH,4^Æ bNÆmu’$ u.(6)bEhvImz=y1IImz=y2gnmu^iÆÆ$ ,.’qH.(7)Q1.2q1NULm3 ,’qH. 3-Ox236y 1(2,2)Q1.1-Ox6y&%'$pp134Q1.2-Ox6y&%'$ 2(√2,√2)Q1.3(10)Q1.3qm qH.13. z=x+iy,Vhv Gy=kx+b u ^z−z2i=kz+z2+b.‘&gQ l(−k+i)z+(−k−i)z=2b.;a=−k+i,c=2b‘laz+az=c.14. aLÆ-a^N , ^rmNv - G|z−a|=r*%.er2=|z−a|2=(z−a)(z−a)=(z−a)(z−a)=zz−az−az+|a|2.;A=1,β=−a,C=|a|2−r2,V Nv G u ^Azz+βz+βz+C=0.15.TLuNargzfz2,A-.. . wÆm )uz0, -Lu^N D umNv.f|uzWÆ aL Nvp~Euz0 ,argzp~Eπ;f|uzWo aL Nvp~Euz0 ,argzp~E−π.4TargzT wÆmH)u f℄u,W .. .16.qbÆ..o#|xn−x0|≤|zn−z0|I|yn−y0|≤|zn−z0| e,f|zn−z0|→0 ,|xn−x0|→0,|yn−y0|→0.A- 16xn(n=1,2,···)Iyn(n=1,2,···) +-x0Iy0^℄u.SbI Æ..o#|zn−z0|≤|xn−x0|+|yn−y0| e,f|xn−x0|→0,|yn−y0|→0 ,|zn−z0|→0.A- 16zn(n=1,2,···)-z0^℄u.17. ? )aLÆm-z1,z2,z3^yum w ,rQ1.4.$z1z2I$z1z3mhw’Arg(z3−z1)−Arg(z2−z1)=Argz3−z1z2−z1T(0,π)mti.u:U1*wNm #*%, l WwgIArgz3−z1z2−z1+Argz1−z2z3−z2+Argz2−z3z1−z3=Argz3−z1z2−z1z1−z2z3−z2z2−z3z1−z3=Arg(−1)=(2n+1)π,n=0,±1,±2,···.4^H*Wwi} Eπ,A-C tilAπ’ Xm.℄ bM4 WwIoEπ.z1 z2@@@@@z3Q1.44 p %\Q3 T YGTq, fF+8M,Vz=x+iy,f(z)=u(x,y)+iv(x,y).1.(1)t dqbMTLuN|z|. g.& Æ,fΔz + a wI wpE8 , i|Δz|/Δz +pE1I−1.T efΔz→0 , i|Δz|/Δz.XT℄u,‘|z|TLuN. g.oL ?z Luml .fΔzt 1ipE8 ,dΔz=iΔx,A|z+Δz|−|z|Δz=p(x+Δx)2+y2−px2+y2Δx=Δx+2xp(x+Δx)2+y2+px2+y2−→xpx2+y2.fΔztQ 1pE8 ,dΔz=iΔy,A|z+Δz|−|z|Δz=px2+(y+Δy)2−px2+y2iΔy=Δy+2yipx2+(y+Δy)2+px2+y2−→yipx2+y2.T efΔz→0 , i(|z+Δz|−|z|)/Δz.XT℄u,‘|z|T LuN$. g.A-|z|TzaLÆNN. g.t 4^|z|=px2+y2,A-u=px2+y2,v=0.E’fx2+y26=0 Aux=xpx2+y2,uy=ypx2+y2,vx=vy=0.s~u,v.E C-R G,A-|z|T Lum 0uN. g.fx=y=0 /bMu.XT‘g1,4 |z|TLuN$. g.A-|z|TzaLÆNN. g.t F b . z’aLÆm 0)u.k |z|TuzN g,‘fΔz→0 , i|z+Δz|−|z|ΔzXT℄u.f +;Δzt 1[Q 1pE8 , e ℄ue’ 1D’Q 1,4 ^81.W |z|m 3um‘g1^8.~ u=px2+y2TLuN.XT‘g1,TeCuNg1.^8,lJF .A-|z|TzaLÆNN. g.(3)Rez=x,;u=x,v=0.E’Aux=16=0=vy.4TTzaLÆAAuN,C-R G .E,.A-RezTzaLÆNN. g.(4)TLuNF11/zfz2,4 . g.T LuN,1z=z|z|2=x+iyx2+y2,;u=xx2+y2,v=yx2+y2.E’Aux=y2−x2(x2+y2)2,uy=−2xy(x2+y2)2,vx=−2xy(x2+y2)2,vy=x2−y2(x2+y2)2.T eTMLuUm )uNu,v .E C-R G,A-1/zTzaLÆNN. g.2.(1)u=x3,v=−y3 lux=3x2,uy=0,vx=0,vy=−3y2.3x2=ux=vy=−3y2 e,lATLuNC-R G7E,.:U,uIv}A. m‘g1,TxyaLÆNN W.A-f(z)=x3−y3i TLuNAg1.1O!n N K ~(3 jG3nh3 YUW#_9.56 (2)f(z)=|z|2=x2+y2.E’u=x2+y2,v=0.= lux=2x,uy=2y,vx=0,vy=
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本文标题：川大高等数学第四册课后习题答案(1-6章)
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