The value of $\sum_{k=1}^{10}\left(\sin \frac{2 k \pi}{11}+i \cos \frac{2 k \pi}{11}\right)$ is

The value of $\sum_{k=1}^{10}\left(\sin \frac{2 k \pi}{11}+i \cos \frac{2 k \pi}{11}\right)$ is
  1. i
  2. 1
  3. -1
  4. -i

Solution

$\sum_{k=1}^{10}\left(\sin \frac{2 k \pi}{11}+i \cos \frac{2 k \pi}{11}\right)=\sum_{k=1}^{10} \sin \frac{2 k \pi}{11}+i \sum_{k=1}^{10} \cos \frac{2 k \pi}{11}$ $=0+i(-1)=-i$

Asked in: JEE Main 2006

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