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. None of these

Answer 1

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

$\alpha =\cos \left(\frac{8\pi }{11}\right)+i\sin \left(\frac{8\pi }{11}\right)$

We know, $z=\cos \theta +i\sin \theta =e^{i\theta }$

$\therefore$ $\alpha =e^{i\frac{8\pi }{11}}$

$\Rightarrow$ $\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4+{\alpha }^5=\frac{\alpha ({\alpha }^5-1)}{\alpha -1}$ ........................... as forming G.P.

$=\frac{{\alpha }^6-\alpha }{\alpha -1}$

$=\frac{{\left(e^{i\frac{8\pi }{11}}\right)}^6-e^{i\frac{8\pi }{11}}}{e^{i\frac{8\pi }{11}}-1}$ ---- ( 1 )

$e^{-\frac{48\pi }{11}}=\cos \frac{48\pi }{11}+i\sin \frac{48\pi }{11}$

$=\cos (4\pi +\frac{4\pi }{11})+i\sin (4\pi +\frac{4\pi }{11})$

$=\cos \frac{4\pi }{11}+i\sin \frac{4\pi }{11}$

$=e^{i\frac{4\pi }{11}}$

Substituting above value in ( 1 ) we get,

$\Rightarrow$ $\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4+{\alpha }^5=\frac{e^{i\frac{4\pi }{11}}-e^{i\frac{8\pi }{11}}}{e^{i\frac{8\pi }{11}}-1}$

$=\frac{t-t^2}{t^2-1}$ [ Let $e^{i\frac{4\pi }{11}}=t]$

$=\frac{-t(1-t)}{(t-1)(t+1)}$

$=\frac{-t}{t+1}$

$=\frac{-(\cos \frac{4\pi }{11}+i\sin \frac{4\pi }{11})}{\cos \frac{4\pi }{11}+i\sin \frac{4\pi }{11}+1}$

Let $a=\cos \frac{4\pi }{11}=a$ and $b=\sin \frac{4\pi }{11}$

$=\left(\frac{a+ib}{(a+1)+ib}\right)\times \frac{(a+1)-ib}{(a+1)-ib}$

$=-\frac{(1+ib)(a+1)-ib}{\left[\right(a+1)+ib]\left[\right(a+1)-ib]}$

$=-\frac{\left(a\right(a+1)+b^2)+i\left(b\right(a+1)-ab)}{(a+1{)}^2+b^2}$

$=-\frac{a(a+1)+b^2}{(a+1{)}^2+b^2}$ [ Taking real part only ]

$=-\frac{a^2+b^2+a}{a^2+b^2+1+2a}$

$=-\frac{1+a}{2+2a}$

$=-\frac{(1+a)}{2(1+a)}$

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