If a = cos left(frac{8 pi}{11}right) + i sin | vedclass

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(B) Let $S = a + a^2 + a^3 + a^4 + a^5$. This is a geometric progression with first term $a$ and common ratio $a$. $S = a \frac{1 - a^5}{1 - a} = \fra

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$-\frac{1}{2}$

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(B) Let $S = a + a^2 + a^3 + a^4 + a^5$. This is a geometric progression with first term $a$ and common ratio $a$. $S = a \frac{1 - a^5}{1 - a} = \frac{a - a^6}{1 - a}$. Since $a = e^{i \frac{8 \pi}{11}}$,we have $a^6 = e^{i \frac{48 \pi}{11}} = e^{i \left(4 \pi + \frac{4 \pi}{11}ight)} = e^{i \frac{4 \pi}{11}}$. $S = \frac{e^{i \frac{8 \pi}{11}} - e^{i \frac{4 \pi}{11}}}{1 - e^{i \frac{8 \pi}{11}}} = \frac{e^{i \frac{6 \pi}{11}} \left(e^{i \frac{2 \pi}{11}} - e^{-i \frac{2 \pi}{11}}ight)}{e^{i \frac{4 \pi}{11}} \left(e^{-i \frac{4 \pi}{11}} - e^{i \frac{4 \pi}{11}}ight)} = e^{i \frac{2 \pi}{11}} \frac{2i \sin \left(\frac{2 \pi}{11}ight)}{-2i \sin \left(\frac{4 \pi}{11}ight)} = -e^{i \frac{2 \pi}{11}} \frac{\sin \left(\frac{2 \pi}{11}ight)}{2 \sin \left(\frac{2 \pi}{11}ight) \cos \left(\frac{2 \pi}{11}ight)} = -\frac{e^{i \frac{2 \pi}{11}}}{2 \cos \left(\frac{2 \pi}{11}ight)}$. Using $e^{i 	heta} = \cos 	heta + i \sin 	heta$,we get $S = -\frac{\cos \left(\frac{2 \pi}{11}ight) + i \sin \left(\frac{2 \pi}{11}ight)}{2 \cos \left(\frac{2 \pi}{11}ight)} = -\frac{1}{2} - i \frac{1}{2} 	an \left(\frac{2 \pi}{11}ight)$. Thus,$\operatorname{Re}(S) = -\frac{1}{2}$.

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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