t^2+(2+i)t+2xy+(x-y)i=0
t^2+2t+2xy+(t+x-y)i=0
t^2+2t+2xy=0
t+x-y=0,即
x(-y)=(t^2+2t)/2
x+(-y)=-t
所以x、-y是方程x^2+tx+(t^2+2t)/2=0的實數(shù)根
所以△=t^2-2(t^2+2t)>0
-4