Complex Numbers Question, IIT JEE [2006]. Please tell me whether I solved it properly? $Q.$The value of $\sum\limits_{k=1}^{10}(\sin{\frac{2k\pi}{11}-i\cos\frac{2k\pi}{11}})$ is-?
I solved it like this-
$\frac{\sum\limits_{k=1}^{10}(\cos{\frac{2k\pi}{11}+i\sin\frac{2k\pi}{11}})}{i}$

If we observe these are the roots of the equation $z^{11}=1$ 
So $1+z_1+z_2+...+z_{10}=1$ (De Moivre
s Theorem)
So $z_1+z_2+...+z_{10}=-1$
$-1=i^2$
 So $\frac{i^2}{i}=i$ 

$\begin{array}{l}
\\
\sum \limits^{10}_{k=1}(\sin \ \frac{2k\pi }{11} -i\cos \frac{2k\pi }{11}  )\\
=(-i)\sum \limits^{10}_{k=1}(\cos  \frac{2k\pi }{11} +i\sin \frac{2k\pi }{11}  )\\
=(-i)(\sum \limits^{10}_{k=0}(\cos  \frac{2k\pi }{11} +i\sin \frac{2k\pi }{11} )-1)\\
=(-i)(-1)\\
=i\\
Note: 
\sum \limits^{10}_{k=0}(\cos \frac{2k\pi }{11} +i\sin \frac{2k\pi }{11}  )=0\\
\end{array}$
Sum of $11^{th}$ roots of unity =$0$
