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  4. If Zr = sin (2 π r/11) - i cos (2 π r/11) then displaystyle∑10r = 0 Zr =

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Q. If $Z_{r} = \sin \frac{2 \pi r}{11} - i \cos \frac{2 \pi r}{11} $ then $\displaystyle\sum^{10}_{r = 0} Z_{r} = $

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Solution:

We have, $Z_{r} =\sin \frac{2 \pi r}{11}-i \cos \frac{2 \pi r}{11} $
$=-i\left(\cos \frac{2 \pi r}{11}+i \sin \frac{2 \pi r}{11}\right) $
$=-i e^{\frac{i 2 \pi r}{11}} $
$ \displaystyle\sum_{r=0}^{10} Z_{r} =-i \displaystyle\sum_{r=0}^{10} e^{\frac{i 2 \pi r}{11}} $
$=-i \times 0=0 $