因为arg(-1+2i)=α,所以sinα=2/√5,cosα=-1/√5.于是知sin2α=2sinα·cosα=-4/5.cos2α=2cosα·cosα-1=-3/5.sin(2α+2π/3)=sin2α·cos2π/3+cos2α·sin2π/3=(-4/5)×√3/2+(-3/5)×(-1/2)=(3-4√3)/10...