Question

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

• A. $-1$
• B. $-i$
• C. $i$
• D. $1$

Answer 1

The correct option is B $-i$

We know that sum of $n$ roots of unity $=0$

And $\omega =\cos \frac{2\pi }{11}+i\sin \frac{2\pi }{11}$

Here, $n=11$
$\Rightarrow 1+{\omega }_1+{\omega }_2---{\omega }_{10}=0$
$\Rightarrow ({\omega }_1+{\omega }_2---{\omega }_{10})=-1$
$\Rightarrow (\cos \frac{2\pi }{11}+\cos \frac{4\pi }{11}+\cdots )+i(\sin \frac{2\pi }{11}+\sin \frac{4\pi }{11}\cdots )=-1$
Comparing real and imaginary part
$\cos \frac{2\pi }{11}+\cos \frac{4\pi }{11}+\cdots =-1$
and

$\sin \frac{2\pi }{11}+\sin \frac{4\pi }{11}+\cdots =0$

As we know
$\sin \frac{2\pi }{11}+\sin \frac{4\pi }{11}+\cdots +\sin \frac{20\pi }{11}=0$ ...(1)
and
$\cos \frac{2\pi }{11}+\cos \frac{4\pi }{11}+\cdots +\cos \frac{20\pi }{11}=-1$ ...(2)
So,

$\sum _{k=1}^{10}\sin \frac{2k\pi }{11}+i\cos \frac{2k\pi }{11}=0+i(-1)=-i$

Hence, option B.