复数z=(x-2)+yi(x,y∈R)的模为根号3,即有根号[(x-2)^2+y^2]=根号3且x=2y,则有：(2y-2)^2+y^2=34y^2-8y+4+y^2=35y^2-8y=-15(y-4/5)^2=16/5-1=11/5y=4/5土根号11/5X=8/5土2根号11/5故Z=（8/5+2根号11/5）+（4/5+根号11/5）i或z=(8/5-2根号11/5）+（4/5-根号11/5）i
