Question

Find the value of
$\sum _{k=1}^{10}(\sin \frac{2\pi k}{11}-i\cos \frac{2\pi k}{11})$

• A. $i$
• B. $-i$
• C. $1$
• D. $-1$

Answer 1

The correct option is A

$i$

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