Question

已知3z₁+(z₂+1)i=2z₂-(z₁-2)i,且z₁、z₂在復(fù)平面內(nèi)的對應(yīng)點(diǎn)關(guān)于原點(diǎn)對稱,求z₁和z₂.

      

Answer 1

解析:設(shè)z₁=x+yi, x、y∈R,?

       ∵z₁、z₂在復(fù)平面內(nèi)的對應(yīng)點(diǎn)關(guān)于原點(diǎn)對稱,?

       故z₂=-x-yi.?

       ∴3z₁+(z₂+1)i=2z₂-(z₁-2)i可化為?

       3(x+yi)+(-x-yi+1)i=-2x-2yi-(x+yi-2)i.?

       ∴3x+3yi+(1-x)i+y=-2x-2yi-(x-2)i+y.利用復(fù)數(shù)相等,?

       ∴

       解①得x=0,代入②得3y+1=-2y+2,?

       ∴5y=1.∴y=.?

       ∴z₁=i,z₂=-i.