Question

The value of $\displaystyle \sum^{6}_{k=1} \left( \sin \frac{2k\pi}{7} - i \cos \frac{2k \pi}{7}\right) $ is equal to

• A. - 1
• B. 0
• C. - i
• D. i

Answer 1

Answer: (D)

$\sin \frac{2\pi k}{7} - i \cos \frac{2\pi k}{7} $
$= - i\left(\cos \frac{2\pi}{7} + i \sin \frac{2\pi}{7}\right)^{k} $
$= - iz^{k}$ where $z = \cos \frac{2\pi}{7} + i \sin \frac{2\pi}{7} $.
$\therefore \:\:\: \displaystyle\sum^{6}_{k=1} \left( \sin \frac{2\pi k}{7} - i \cos \frac{2\pi k}{7} \right) $
$= - i\left(z + z^{2} +z^{3} +z^{4} +z^{5} +z^{6}\right) $
$= - i\left( \frac{z\left(1 -z^{6}\right)}{1 -z} \right) = - i \left( \frac{z -z^{7}}{1-z}\right) $
$= i\left( \frac{z^{7} - z}{1 - z}\right) = i $ $\left(\because z^{7} = \left(\cos \frac{2\pi }{7} + i \sin \frac{2\pi }{7} \right)^{7} = 1\right)$