设非零复数x,y满足x^2+xy+y^2=0.则代数式[x/(x+y)]^2010-[y/(x+y)]^2010=?

问题描述:

设非零复数x,y满足x^2+xy+y^2=0.则代数式[x/(x+y)]^2010-[y/(x+y)]^2010=?

x^2+xy+y^2=0,两边除以y^2得,(x/y)^2+x/y+1=0,根据求根公式得,x/y=(-1+√3i)/2,或x/y=(-1-√3i)/2,当x/y=(-1+√3i)/2时,x/(x+y)=1-y/(x+y)=1-1/(x/y+1)=(1+√3i)/2,y/(x+y)=(1-√3i)/2,当x/y=(-1-√3i)/2时,x/(x+y)=(1-√3i)/2,y/(x+y)=(1+√3i)/2,两种情况都是[x/(x+y)]^2010-[y/(x+y)]^2010=[(1-√3i)/2]^2010+[(1+√3i)/2]^2010,计算[(1-√3i)/2]^3=-1,[(1+√3i)/2]^3=-1,所以原式=(-1)^670+(-1)^670=1+1=2