高二数学题求助,快~
已知复数z1,z2满足条件|z1|=2,|z2|=3,且3z1+2z2=6,求复数z1和z2.
解:设z1=a+bi,z2=m+ni |z1|=2--->a^2+b^2=4……(1) |z2|=3--->m^2+n^2=9……(2) 3z1+2z2=6--->(3a+2m)+i(3b+2n)=6 --->3a+2m=6--->m=3-3a/2 并且3b+2n=0--->n=-3b/2. 同时代人(2),得到(3-3a/2)^2+(-3b/2)^2=9 --->9-3a+9a^2/4+9b^2/4=9 --->3(a^2+b^2)-4a=0 由(1):3*4-4a=0 --->a=3,b=+'-1 m=3-3a/2=-3/2,n=-'+3/2 所以z1=3+'-i,z2=(3/2)(-1-'+i)
设Z1=a+bi,Z2=c+di, |z1|=2,→√(a^2+b^2)=2,→(a^2+b^2)=4...........(1) |z2|=3,→√(c^2+d^2)=3,→(c^2+d^2)=9...........(2) 3z1+2z2=6,→(3a+2c)+(3b+2d)i=6+0i,→ 3a+2c=6,3b+2d=0,→c=3-3a/2,d=-3b/2代入(2): (3-3a/2)^2+(-3b/2)^2=9,→(9/4)(a^2+b^2)-9a=0,→(由(1)) (9/4)*4-9a=0,→9-9a=0,→a=1,→c=3-3a/2=3/2, 1+b^2=4,b^2=3,b=±√3,d==-3b/2=±3√3/2 ∴Z1=1+√3i,Z2=3/2-(3√3/2)i或 Z1=1-√3i,Z2=3/2+(3√3/2)i
答:设z1=a+hi z1+z2=-i--->z2=-z1-i=a-(+bi)-i=-a-(b+1)i |z1|=|z2|=1--->a^2+b^2=1,a^2+(...详情>>