1did u use
αn where n 0,1,2,....10
will be the 11th roots of untiy
Let α=cis2Π/11 ,λ=α6 ,μ=α7, β=α2
1)Re(λ+λ2+λ3+λ4+λ5)=
2) (μ-β)(μ-β2).................(μ-β10)=
3)(i-β)(i-β2).................(i-β10)=
answers
-1/2
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4 Answers
1
1357for 2 and 3
β=α2, β2=α4, β3=α6....β6=α12=α, β7=α14=α3, β8=α16=α5
(μ-β)(μ-β2).................(μ-β10)=(μ-α)(μ-α2).................(μ-α10)
(x-α)(x-α2).................(x-α10)=(x11 - 1)/(x-1)
for (2) as α7 is the root it becomes zero
for (3) put x=i and solving (i11 - 1)/(i-1)=(- i - 1)/(i-1)
1357(λ+λ2+λ3+λ4+λ5)=(α+α2+α6+α7+α8)
use
Re( α)=Re(1/α)=Re (α10)
Re( α2)=Re(1/α2)=Re (α9)
Re(1+α+α2...)=0
Re(α+α2...α10)=-1
Re(α+α2....α5)=Re(1/α+1/α2....1/α5)=Re(α6+α7....α10)=-1/2
Re(α+α2+α6+α7+α8)=Re(α+α2+1/α5+1/α4+1/α3)
=Re(α+α2+α5+α4+α3)=-1/2
