Q50 / 160 · TS EAMCET 2019 (04 May Shift 2) · Mathematics

$\sum_{r=1}^{16}\left(\sin \frac{2 r \pi}{17}+i \cos \frac{2 r \pi}{17}\right)=$

  • A
    1
  • B
    -1
  • C
    $i$
  • Correct answer
    D
    $-i$

Solution

We have, $$ \sum_{r=1}^{16}\left(\sin \frac{2 r \pi}{17}+i \cos \frac{2 r \pi}{17}\right)=i \sum_{r=1}^{16}\left(\cos \frac{2 r \pi}{17}-i \sin \frac{2 r \pi}{17}\right) $$ $$ \begin{aligned} & =i \sum_{r=1}^{16} e^{-\frac{i 2 r \pi}{17}} \\ & \left.=i \sum_{r=1}^{16} K^r \quad \text { [where, } K=e^{-\frac{2 i \pi}{17}}\right] \\ & =i\left[k\left(\frac{1-k^{16}}{1-k}\right)\right]=i \frac{\left(k-k^{17}\right)}{(1-k)} \\ & =\frac{i(k-1)}{(1-k)} \quad\left[\because K^{17}=e^{-2 i \pi}=\cos (2 \pi)-i \sin 2 \pi=1\right] \\ & =-i \quad \end{aligned} $$

Hence, the correct answer is option: 'D'.

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