If α= e i 2π/11 then Real (α+α2+α3+α4+α5) equals to :

Question

If α= e i 2π/11 then Real (α+α2+α3+α4+α5) equals to : (A) (1/2) (B) 1 (C) -(1/2) (D) -1

Tardigrade Admin · Accepted Answer

Real (1+α+α2+ ldots ldots . .+α10)=0 ⇒ 1+ Real (α+α2+ ldots ldots α5)+ Real (α6+α7+ ldots ldots .+α10)=0 ⇒ 2 Real (α+α2+ ldots ldots . α5)=-1 ⇒ (α+α2+α3+α4+α5)=-1 / 2

Q. If $α = e^{i 2 π /11}$ then Real $(α + α^{2} + α^{3} + α^{4} + α^{5})$ equals to :

$\frac{1}{2}$

1

$- \frac{1}{2}$

$- 1$

Real $(1 + α + α^{2} + \dots\dots .. + α^{10}) = 0$
$\Rightarrow 1 +$ Real $(α + α^{2} + \dots\dots α^{5}) +$ Real $(α^{6} + α^{7} + \dots\dots . + α^{10}) = 0$
$\Rightarrow 2$ Real $(α + α^{2} + \dots\dots . α^{5}) = - 1$
$\Rightarrow (α + α^{2} + α^{3} + α^{4} + α^{5}) = - 1/2$

Q. If α=ei2π/11 then Real (α+α2+α3+α4+α5) equals to :

21​

1

−21​

−1

Real (1+α+α2+……..+α10)=0 ⇒1+ Real (α+α2+……α5)+ Real (α6+α7+…….+α10)=0 ⇒2 Real (α+α2+…….α5)=−1 ⇒(α+α2+α3+α4+α5)=−1/2

Q. If $α = e^{i 2 π /11}$ then Real $(α + α^{2} + α^{3} + α^{4} + α^{5})$ equals to :

$\frac{1}{2}$

$- \frac{1}{2}$

$- 1$

Real $(1 + α + α^{2} + \dots\dots .. + α^{10}) = 0$
$\Rightarrow 1 +$ Real $(α + α^{2} + \dots\dots α^{5}) +$ Real $(α^{6} + α^{7} + \dots\dots . + α^{10}) = 0$
$\Rightarrow 2$ Real $(α + α^{2} + \dots\dots . α^{5}) = - 1$
$\Rightarrow (α + α^{2} + α^{3} + α^{4} + α^{5}) = - 1/2$