Question

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

• A.
• B.
• C. None of these

Answer 1

The correct option is B

We have, $\alpha =cos\left(\frac{8\pi }{11}\right)+isin\left(\frac{8\pi }{11}\right),e^{\frac{8\pi }{11}}i$
$\alpha =cos\left(\frac{8\pi }{11}\right)+isin\left(\frac{8\pi }{11}\right),e^{\frac{8\pi }{11}}i$
Re $(\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4+{\alpha }^5)$
=$\frac{\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4+{\alpha }^5+\overline{\alpha }+{\overline{\alpha }}^2+{\overline{\alpha }}^3+{\overline{\alpha }}^4+{\overline{\alpha }}^5}{2}$
=$\frac{-1(1+\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4+{\alpha }^5+\overline{\alpha }+{\overline{\alpha }}^2+{\alpha }^3+{\overline{\alpha }}^4+{\overline{\alpha }}^5)}{2}$
=$\frac{-1+0}{2}$ ( sum of 11, 11^th roots of unity)
=$-\frac{1}{2}$