If $a=\cos \left(\frac{8 \pi}{11}\right)+i \sin \left(\frac{8 \pi}{11}\right)$, then $\operatorname{Re}\left(a+a^2+a^3+a^4+a^5\right)=$

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Aicharya · Accepted Answer

$a=\cos \left(\frac{8 \pi}{11}ight)+i \sin \left(\frac{8 \pi}{11}ight) \Rightarrow a=e^{\frac{i 8 \pi}{11}}$ $\Rightarrow a$ is 11 th root of unity and all roots are $1, a, a^2, \ldots, a^{10}$ Now, $\quad a^{10}=\frac{a^{10} \cdot a}{a}=\frac{a^{11}}{a}=\frac{1}{a}=\bar{a}$ Similarly, $a^9=\overline{a^2}, a^8=\overline{a^3}, a^7=\overline{a^4}, a^6=\overline{a^5}$, We know that, Sum of $n$ roots of unity $=0$ $ \begin{gathered} 1+a^1+a^2+a^3+\ldots+a^{10}=0 \ \Rightarrow a+a^2+a^3+a^4+a^5+a^6+a^7+a^8 \ a^9+a^{10}=-1 \ \Rightarrow(a+\bar{a})+\left(a^2+\overline{a^2}ight)+\left(a^3+\overline{a^3}ight) \ +\left(a^4+\overline{a^4}ight)+\left(a^5+\overline{a^5}ight)=-1 \ \Rightarrow 2 \operatorname{Re}(a)+2 \operatorname{Re}\left(a^2ight)+2 \operatorname{Re}\left(a^3ight)+2 \operatorname{Re}\left(a^4ight)+2 \operatorname{Re}\left(a^5ight) \ =-1[z+\bar{z}=2 \operatorname{Re}(z)] \ \Rightarrow 2 \operatorname{Re}\left(a+a^2+a^3+a^4+a^5ight)=-1 \ \Rightarrow \operatorname{Re}\left(a+a^2+a^3+a^4+a^5ight)=-\frac{1}{2} \end{gathered} $

If $a=\cos \left(\frac{8 \pi}{11}\right)+i \sin \left(\frac{8 \pi}{11}\right)$, then…

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