If Z_{r} = sin frac{2 pi r}{11} - i cos frac{ | vedclass

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(B) Given $Z_{r} = \sin \frac{2 \pi r}{11} - i \cos \frac{2 \pi r}{11}$.We can rewrite this as $Z_{r} = -i (\cos \frac{2 \pi r}{11} + i \sin \frac{2 \

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(B) Given $Z_{r} = \sin \frac{2 \pi r}{11} - i \cos \frac{2 \pi r}{11}$.We can rewrite this as $Z_{r} = -i (\cos \frac{2 \pi r}{11} + i \sin \frac{2 \pi r}{11})$.Using Euler's formula $e^{i 	heta} = \cos 	heta + i \sin 	heta$, we have $Z_{r} = -i e^{i \frac{2 \pi r}{11}}$.Now, $\sum_{r=0}^{10} Z_{r} = -i \sum_{r=0}^{10} (e^{i \frac{2 \pi}{11}})^{r}$.This is a geometric series with $11$ terms where the common ratio is $\omega = e^{i \frac{2 \pi}{11}} 
eq 1$.The sum of the roots of unity $\sum_{r=0}^{n-1} e^{i \frac{2 \pi r}{n}} = 0$ for $n > 1$.Thus, $\sum_{r=0}^{10} e^{i \frac{2 \pi r}{11}} = 0$.Therefore, $\sum_{r=0}^{10} Z_{r} = -i 	imes 0 = 0$.

If $Z_{r} = \sin \frac{2 \pi r}{11} - i \cos \frac{2 \pi r}{11}$, then $\sum_{r=0}^{10} Z_{r}$ is equal to

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