已知α,β是实系数一元二次方程ax²+bx+c=0的两个虚根,且α²/β∈R,求α/β

设α=m+ni,m,n∈R且n≠0,则β=m-ni.
α²/β=(m+ni)²/(m-ni)=(m+ni)³/(m²+n²)=[m³-3mn²+(3m²n-n³)i]/(m²+n²)
因为α²/β∈R,所以3m²n-n³=0,所以n²=3m².
α/β=α²/(αβ)=(m+ni)²/(m²+n²)=(m²-n²+2mni)/(m²+n²)=-1/2±√3i/2.α/β=α²/(αβ)=(m+ni)²/(m²+n²)=(m²-n²+2mni)/(m²+n²)=-1/2±√3i/2.这个-1/2±√3i/2.怎么化出来的n=±√3m代入即可.