Question

If $\alpha =\cos \frac{8\pi }{11}+i\sin \frac{8\pi }{11}$ then $Re(\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4+{\alpha }^5)$ equals

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

Answer 1

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

Given : $\alpha =\cos \frac{8\pi }{11}+i\sin \frac{8\pi }{11}$
$\therefore {\alpha }^{11}=\cos 8\pi +i\sin 8\pi =1$
$\sum _{n=0}^{10}{\alpha }^n=\frac{1-{\alpha }^{11}}{1-\alpha }=0$ (sum of 11th roots of unity)
Now $Re\left(z\right)=\frac{z+\overline{z}}{2}=\frac{sumof11rootsofunity-1}{2}=-1/2$