题目

一道高中复数四则运算题目我不太懂,
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的值是多少?

提问时间：2021-04-01

答案

x^2+xy+y^2=0x^2=-y(x+y)y^2=-x(x+y)所以x/(x+y)=-y/xy/(x+y)=-x/y而(x/y)^2+x/y+1=0所以(x/y)^3=1同样(y/x)^3=1`所以[x/(x+y)]^2010+[y/(x+y)]^2010=(-y/x)^2010+(-x/y)^2010=[(-y/x)^3]^670+[(-x/y)^3]^670=[-1]^6...