Question

If $\alpha =cos\frac{8\pi }{11}+isin\frac{8\pi }{11}$ 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)

$\because Re(z)=\frac{z+\bar{z}}{2}$
$\therefore Re(\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4+{\alpha }^5)$
$=\frac{\left(\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4+{\alpha }^5\right)+\overline{\left(\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4+{\alpha }^5\right)}}{2}$
$=\frac{1}{2}\left(\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4+{\alpha }^5+\frac{1}{\alpha }+\frac{1}{{\alpha }^2}+\frac{1}{{\alpha }^3}+\frac{1}{{\alpha }^4}+\frac{1}{{\alpha }^5}\right)$
$=\frac{1}{2{\alpha }^5}\left[{\alpha }^6+{\alpha }^7+{\alpha }^8+{\alpha }^9+{\alpha }^{10}+{\alpha }^4+{\alpha }^3+{\alpha }^2+\alpha 1\right]$
$=\frac{1}{2{\alpha }^5}\left[1+\alpha +...+{\alpha }^{10}-{\alpha }^5\right]$
$=\frac{1}{2}\frac{1}{{\alpha }^5}\left[\frac{1-{\alpha }^{11}}{1-\alpha }-{\alpha }^5\right]=\frac{1}{2{\alpha }^5}\left[0-{\alpha }^5\right]$
$=-\frac{1}{2}$ $\left(\text{Using}\alpha =cos\frac{8\pi }{11}+isin\frac{8\pi }{11}\right)$
Short Cut Method :
Given $\alpha =cos\frac{8\pi }{11}+isin\frac{8\pi }{11}$
$\therefore {\alpha }^{11}=cos8\pi +isin8\pi =1$
so $\sum _{n=0}^{10}{\alpha }^n=\frac{1-{\alpha }^{11}}{1-\alpha }=0$ (sum of $11^{th}$ roots of unity)
Now $Re(z)=\frac{z+\bar{z}}{2}$
$=\frac{\text{ sum of 11th roots of unity}-1}{2}=-1/2$