Question

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

• A. - 1
• B. 0
• C. - i
• D. i

Answer 1

Answer: (D)

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