Question

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

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

Answer 1

Answer: (B)

$-\frac{1}{2}$

Given:- $\alpha =\cos \frac{8\pi }{11}+i\sin \frac{8\pi }{11}$

To find: $R_e(\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4+{\alpha }^5)$

Solution:

Let$\alpha =\cos \frac{8\pi }{11}+i\sin \frac{8\pi }{11}=e^{i\frac{8\pi }{11}}\{e^{i\theta }=\cos \theta +i\sin \theta \}$

Similarly, ${\alpha }^2=e^{i\frac{16\pi }{11}}=\cos \frac{16\pi }{11}+i\sin \frac{16\pi }{11}$

${\alpha }^3=e^{i\frac{24\pi }{11}}=\cos \frac{24\pi }{11}+i\sin \frac{24\pi }{11}$

${\alpha }^4=e^{i\frac{32\pi }{11}}=\cos \frac{32\pi }{11}+i\sin \frac{32\pi }{11}$

${\alpha }^5=e^{i\frac{40\pi }{11}}=\cos \frac{40\pi }{11}+i\sin \frac{40\pi }{11}$

So, $R_e(\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4+{\alpha }^5)$

$=\cos \left(\frac{8\pi }{11}\right)+\cos \left(\frac{16\pi }{11}\right)+\cos \left(\frac{24\pi }{11}\right)+\cos \left(\frac{32\pi }{11}\right)+\cos \left(\frac{40\pi }{11}\right)$

Now let $x=\frac{8\pi }{11}$ and

$s$ = $cosx+\cos 2x+\cos 3x+\cos 4x+\cos 5x$

By multiplying and dividing by $2\sin \frac{x}{2}$

$S=\frac{1}{2\sin \frac{x}{2}}(2\sin \frac{x}{2}cosx+2\sin \frac{x}{2}\cos 2x+2\sin \frac{x}{2}\cos 3x+2\sin \frac{x}{2}\cos 4x+2\sin \frac{x}{2}\cos 5x)$

After simplifying we get $[\sin c-\sin d=2\left(\sin \frac{C-D}{2}\right)\left(\cos \frac{C+D}{2}\right)]$

$S=\frac{1}{2\sin \frac{x}{2}}(\sin \frac{11x}{2}-\sin \frac{x}{2})$

$S=\frac{1}{2\sin \frac{x}{2}}(\sin \frac{11}{2}\times \frac{8\pi }{11}-\sin \frac{x}{2})$

$S=\frac{1}{2\sin \frac{x}{2}}[0-\sin \frac{x}{2}]$

$S=-\frac{1}{2}$

Hence, $R_e(\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4+{\alpha }^5)=-\frac{1}{2}$