高中數學 1的n次方根 求分式計算值

若ω=Cos(2π/7)+iSin(2π/7)求ω/(1+ω^2)+ω^2/(1+ω^4)+ω^3/(1+ω^6)
Sol
ω=Cos(2π/7)+iSin(2π/7)
ω^7=Cos(2π)+iSin(2π)=1
1+ω+ω^2+ω^3+ω^4+ω^5+ω^6=0
A=ω/(1+ω^2)+ω^2/(1+ω^4)+ω^3/(1+ω^6)
A(1+ω^2)(1+ω^4)(1+ω^6)
=ω(1+ω^4)(1+ω^6)+ω^2(1+ω^2)(1+ω^6)+ω^3(1+ω^2)(1+ω^4)
=ω(1+ω^4+ω^6+ω^10)+ω^2(1+ω^2+ω^6+ω^8)+ω^3(1+ω^2+ω^4+ω^6)
=ω+ω^5+ω^7+ω^11+ω^2+ω^4+ω^8+ω^10+ω^3+ω^5+ω^7+ω^9
=ω+ω^5+1+ω^4+ω^2+ω^4+ω+ω^3+ω^3+ω^5+1+ω^2
=2+2ω+2ω^2+2ω^3+2ω^4+2ω^5
=－2ω^6
(1+ω^2)(1+ω^4)(1+ω^6)
=(1+ω^2+ω^4+ω^6)(1+ω^6)
=1+ω^6+ω^2+ω^8+ω^4+ω^10+ω^6+ω^12
=1+ω^6+ω^2+ω+ω^4+ω^3+ω^6+ω^5
=1+ω+ω^2+ω^3+ω^4+ω^5+2ω^6
=ω^6
So
A=－2ω^6/ω^6=－2

ω^7=1..................