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?
2019-05-07
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?
优质解答
x^2+xy+y^2=0
x^2=-y(x+y)
y^2=-x(x+y)
所以x/(x+y)=-y/x
y/(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]^670+[-1]^670
=1+1
=2
x^2+xy+y^2=0
x^2=-y(x+y)
y^2=-x(x+y)
所以x/(x+y)=-y/x
y/(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]^670+[-1]^670
=1+1
=2