Q113 / 180 · COMEDK 2019 · Mathematics

The value of $\sum_{K=1}^{6}\left[\sin \frac{2 K \pi}{7}-i \cos \frac{2 K \pi}{7}\right]$ is

  • Correct answer
    A
    i
  • B
    0
  • C
    $-i$
  • D
    $-1$

Solution

We have, $$ \begin{aligned} &\sum_{K=1}^{6}\left(\sin \frac{2 K \pi}{7}-i \frac{\cos 2 K \pi}{7}\right) \\ &=-i \sum_{K=1}^{6}\left(\cos \frac{2 K \pi}{7}+i \sin \frac{2 K \pi}{7}\right) \\ &=-i \sum_{K=1}^{6} e^{\frac{2 K \pi}{7}} \\ &=-i\left[e^{\frac{2 \pi}{7}}+e^{\frac{4 \pi}{7}}+e^{\frac{6 \pi}{7}}+e^{\frac{8 \pi}{7}}+e^{\frac{10 \pi}{7}}+e^{\frac{12 \pi}{7}}\right] \\ &=-i\left[1+e^{\frac{2 \pi}{7}}+e^{\frac{4 \pi}{7}}+e^{\frac{6 \pi}{7}}+e^{\frac{8 \pi}{7}}+e^{\frac{10 \pi}{7}}+e^{\frac{12 \pi}{7}}-1\right] \\ &=-i[0-1][\because \text { Sum of all nth roots of unit is zero] } \\ &=i \\ &=\left[\begin{array}{l} {[} \end{array}\right] \end{aligned} $$

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

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