设ω=cos(2π/5) + i×sin(2π/5),请写出以ω,ω^3,ω^7,ω^9为根的方程
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ω=cos(2π/5) + i×sin(2π/5),
说明1,w,w^2,w^3,w^4是x^5-1的五个根
所以w^5=1,且(x-1)(x-w)(x-w^3)(x-w^2)(x-w^4)=x^5-1
=（x-1）(x4-x3+x2-x+1)
所以 (x-w)(x-w^3)(x-w^2)(x-w^4)
=(x^5-1)/(x-1)
=x4-x3+x2-x+1
w^7=w^2,w^9=w^4
所以(x-w)(x-w^3)(x-w^7)(x-w^9)
=(x-w)(x-w^3)(x-w^2)(x-w^4)
=x4-x3+x2-x+1
说明1,w,w^2,w^3,w^4是x^5-1的五个根
所以w^5=1,且(x-1)(x-w)(x-w^3)(x-w^2)(x-w^4)=x^5-1
=（x-1）(x4-x3+x2-x+1)
所以 (x-w)(x-w^3)(x-w^2)(x-w^4)
=(x^5-1)/(x-1)
=x4-x3+x2-x+1
w^7=w^2,w^9=w^4
所以(x-w)(x-w^3)(x-w^7)(x-w^9)
=(x-w)(x-w^3)(x-w^2)(x-w^4)
=x4-x3+x2-x+1
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