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Q. If $\alpha= e ^{i 2\pi/11}$ then Real $\left(\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5\right)$ equals to :

Complex Numbers and Quadratic Equations

Solution:

Real $\left(1+\alpha+\alpha^2+\ldots \ldots . .+\alpha^{10}\right)=0 $
$\Rightarrow 1+$ Real $\left(\alpha+\alpha^2+\ldots \ldots \alpha^5\right)+$ Real $\left(\alpha^6+\alpha^7+\ldots \ldots .+\alpha^{10}\right)=0$
$\Rightarrow 2 $ Real $\left(\alpha+\alpha^2+\ldots \ldots . \alpha^5\right)=-1$
$ \Rightarrow \left(\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5\right)=-1 / 2$