Question

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

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $0$
• D. None of the above

Answer 1

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

$\alpha =\cos A+i\sin A=cisA$, where $A=\frac{8\pi }{11}$
$z=\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4+{\alpha }^5=\alpha \left(\frac{{\alpha }^5-1}{\alpha -1}\right)$
$\Rightarrow z=cisA\left(\frac{{\left(cisA\right)}^5-1}{cisA-1}\right)=cisA\left(\frac{1-\cos 5A-i\sin 5A}{1-\cos A-i\sin A}\right)$
$\Rightarrow z=\frac{cisA\sin \frac{5A}{2}}{\sin \frac{A}{2}}\left(\frac{\sin \frac{5A}{2}-i\cos \frac{5A}{2}}{\sin \frac{A}{2}-i\cos \frac{A}{2}}\right)=\frac{cisA\sin \frac{5A}{2}}{\sin \frac{A}{2}}\left(\frac{i\sin \frac{5A}{2}+\cos \frac{5A}{2}}{i\sin \frac{A}{2}+\cos \frac{A}{2}}\right)$
$\Rightarrow z=\frac{cisA\sin \frac{5A}{2}cis2A}{\sin \frac{A}{2}}=\frac{cis3A\sin \frac{5A}{2}}{\sin \frac{A}{2}}$
$\Rightarrow Re\left(z\right)=\frac{\cos 3A\sin \frac{5A}{2}}{\sin \frac{A}{2}}$
$\Rightarrow Re\left(z\right)=\frac{\cos 3A\sin \frac{5A}{2}}{2\sin \frac{A}{4}\cos \frac{A}{4}}=\frac{\cos \left(\frac{24\pi }{11}\right)\sin \left(\frac{20\pi }{11}\right)}{2\sin \left(\frac{2\pi }{11}\right)\cos \left(\frac{2\pi }{11}\right)}=\frac{\cos \left(\frac{24\pi }{11}\right)\sin \left(\frac{20\pi }{11}\right)}{-2\sin (2\pi -\frac{2\pi }{11})\cos (2\pi +\frac{2\pi }{11})}$
$\Rightarrow Re\left(z\right)=\frac{\cos \left(\frac{24\pi }{11}\right)\sin \left(\frac{20\pi }{11}\right)}{-2\sin \left(\frac{20\pi }{11}\right)\cos \left(\frac{24\pi }{11}\right)}=\frac{-1}{2}$
Hence, option B is correct.