Let z = frac{1 - i sqrt{3}}{2},where i = sqrt | vedclass

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(D) Given $z = \frac{1 - i \sqrt{3}}{2} = e^{-i \frac{\pi}{3}}$.Then $z^r + \frac{1}{z^r} = e^{-i \frac{r\pi}{3}} + e^{i \frac{r\pi}{3}} = 2 \cos\left

Vedclass · Accepted Answer

$13$

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(D) Given $z = \frac{1 - i \sqrt{3}}{2} = e^{-i \frac{\pi}{3}}$.Then $z^r + \frac{1}{z^r} = e^{-i \frac{r\pi}{3}} + e^{i \frac{r\pi}{3}} = 2 \cos\left(\frac{r\pi}{3}ight)$.We need to evaluate $S = 21 + \sum_{r=1}^{21} \left(z^r + \frac{1}{z^r}ight)^3 = 21 + \sum_{r=1}^{21} \left(2 \cos\left(\frac{r\pi}{3}ight)ight)^3$.Using the identity $8 \cos^3 	heta = 2(4 \cos^3 	heta) = 2(\cos 3	heta + 3 \cos 	heta)$,we have:$S = 21 + \sum_{r=1}^{21} 2 \left(\cos(r\pi) + 3 \cos\left(\frac{r\pi}{3}ight)ight) = 21 + 2 \sum_{r=1}^{21} \cos(r\pi) + 6 \sum_{r=1}^{21} \cos\left(\frac{r\pi}{3}ight)$.For $\sum_{r=1}^{21} \cos(r\pi) = (-1) + 1 + (-1) + \dots + (-1) = -1$ (since there are $21$ terms).For $\sum_{r=1}^{21} \cos\left(\frac{r\pi}{3}ight)$,the sum of $\cos\left(\frac{r\pi}{3}ight)$ over one period ($r=1$ to $6$) is $0$. Since $21 = 3 	imes 6 + 3$,the sum is $\sum_{r=1}^{3} \cos\left(\frac{r\pi}{3}ight) = \cos\frac{\pi}{3} + \cos\frac{2\pi}{3} + \cos\pi = \frac{1}{2} - \frac{1}{2} - 1 = -1$.Thus,$S = 21 + 2(-1) + 6(-1) = 21 - 2 - 6 = 13$.

Let $z = \frac{1 - i \sqrt{3}}{2}$,where $i = \sqrt{-1}$. Then the value of $21 + \left(z + \frac{1}{z}\right)^{3} + \left(z^{2} + \frac{1}{z^{2}}\right)^{3} + \left(z^{3} + \frac{1}{z^{3}}\right)^{3} + \dots + \left(z^{21} + \frac{1}{z^{21}}\right)^{3}$ is .... .

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