Question

If $\alpha =\cos \left(\frac{8\pi }{11}\right)+i\sin \left(\frac{8\pi }{11}\right),$ then $Re(\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4+{\alpha }^5)$ is equal to

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $0$
• D. $1$

Answer 1

The correct option is B $-\frac{1}{2}$

We have,
$\alpha =\cos \left(\frac{8\pi }{11}\right)+i\sin \left(\frac{8\pi }{11}\right)=e^{8\pi i/11}$
Hence $\alpha$ is one of the $11$th root of the unity
and ${\alpha }^{11}=1,(\overline{\alpha }{)}^{11-r}=(\alpha {)}^r$
$\therefore Re(\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4+{\alpha }^5)=\frac{\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4+{\alpha }^5+\left(\overline{\alpha }\right)+(\overline{\alpha }{)}^2+(\overline{\alpha }{)}^3+(\overline{\alpha }{)}^4+(\overline{\alpha }{)}^5}{2}$
$=\frac{-1+(1+\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4+{\alpha }^5+{\alpha }^{10}+{\alpha }^9+{\alpha }^8+{\alpha }^7+{\alpha }^6)}{2}$
$=\frac{-1+0}{2}$
[ $\because$ Sum of all roots of unity is zero]
$=-\frac{1}{2}$