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

Answer: (B)

$\because Re(z) = \frac{z + \bar{z}}{2}$
$\therefore Re(\alpha + \alpha^2 + \alpha^3 + \alpha^4 + \alpha^5)$
$=\frac{\left(\alpha+\alpha^{2}+\alpha^{3}+\alpha^{4}+\alpha^{5}\right)+\overline{\left(\alpha + \alpha^{2}+\alpha^{3}+\alpha^{4}+\alpha^{5}\right)}}{2}$
$ = \frac{1}{2}\left(\alpha + \alpha^{2} + \alpha^{3} + \alpha^{4} + \alpha^{5} + \frac{1}{\alpha}+\frac{1}{\alpha^{2}} + \frac{1}{\alpha^{3}} + \frac{1}{\alpha^{4}} + \frac{1}{\alpha^{5}}\right) $
$ = \frac{1}{2\alpha^{5}} \left[\alpha^{6} + \alpha^{7} + \alpha^{8} + \alpha^{9} + \alpha^{10} +\alpha^{4}+ \alpha^{3} +\alpha^{2} + \alpha 1\right] $
$= \frac{1}{2\alpha^{5}} \left[ 1 + \alpha + ... + \alpha^{10}- \alpha^{5}\right]$
$= \frac{1}{2} \frac{1}{\alpha^{5}}\left[\frac{1-\alpha^{11}}{1-\alpha} -\alpha^{5}\right] = \frac{1}{2\alpha^{5}}\left[0-\alpha^{5}\right] $
$= -\frac{1}{2} $ $\left( \text{Using} \,\alpha = cos\frac{8\pi}{11} + i\,sin \frac{8\pi}{11}\right)$
Short Cut Method :
Given $\alpha = cos \frac{8\pi}{11} + i\,sin\frac{8\pi}{11}$
$\therefore \alpha^{11} = cos\,8\pi + i\,sin\,8\pi = 1$
so $\displaystyle\sum_{n= 0}^{10} \alpha^n = \frac{1 - \alpha^{11}}{1 - \alpha} = 0$ (sum of $11^{th}$ roots of unity)
Now $Re(z) = \frac{z + \bar{z}}{2}$
$= \frac{\text{ sum of $11^{th}$ roots of unity} - 1}{2} = -1/2$