Question

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

• A. 1
• B. - 1
• C. i
• D. -i

Answer 1

Answer: (C)

We have,
$\sum _{k=1}^{10}\left(\sin \frac{2\pi k}{11}-i\cos \frac{2\pi k}{11}\right)$
$=\sum _{k=1}^{10}\left(-i^2\sin \frac{2\pi k}{11}-i\cos \frac{2\pi k}{11}\right)$
$=-i\sum _{k=1}^{10}\left(\cos \frac{2\pi k}{11}+i\sin \frac{2\pi k}{11}\right)=-i\sum _{k=1}^{10}{e}^{i\frac{2\pi k}{11}}$
$=-i\left[\sum _{k=0}^{10}{e}^{i\frac{2\pi k}{11}}-1\right]$
$=-i($ sum of $11\text{th}$ roots of unity $-1)$
$=-i(0-1)=i$