Question

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

• A. $1$
• B. $-1$
• C. $i$
• D. $-i$

Answer 1

The correct option is C $i$

We have,
$\sum _{k=1}^{10}(\sin \frac{2k\pi }{11}-i\cos \frac{2k\pi }{11})$
$=-i\sum _{k=1}^{10}(\cos \frac{2k\pi }{11}+i\sin \frac{2k\pi }{11})=-i\sum _{k=1}^{10}{e}^{i2k\pi /11}$
and we know that sum of $11,11$th roots of unity is
$\sum _{k=0}^{10}{e}^{i2k\pi /11}=0$
$\therefore -i\sum _{k=1}^{10}{e}^{i2k\pi /11}=-i[\sum _{k=0}^{10}{e}^{i2k\pi /11}-e^0]=i$