<s>[INST] 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 $\frac{i^2}{i}=i$ 
 [/INST] $\bf{My\; Solution::}$ Given $$\displaystyle \sin \left(\frac{2k\pi}{11}\right)-i \cos \left(\frac {2k\pi}{11}\right) = -i\cdot \left[\cos \left(\frac {2k\pi}{11}\right)+i\cdot \sin \left(\frac {2k\pi}{11}\right)\right]$$
Now We know that $$e^{i\phi} = \cos \phi+i\sin \phi$$ and $$e^{-i\phi}=\cos \phi+i\sin \phi$$
So We Write $$\displaystyle \sin \left(\frac{2k\pi}{11}\right)-i \cos \left(\frac {2k\pi}{11}\right)=-i\cdot e^{i\frac{2k\pi}{11}} = -i\cdot e^{ik\phi}\;,$$ Where $$\displaystyle \phi = \frac{2\pi}{11}$$
So We have  $$\displaystyle \sum_{k=1}^{10}\displaystyle \sin \left(\frac{2k\pi}{11}\right)-i \cos \left(\frac {2k\pi}{11}\right) = -i\sum_{k=1}^{10}e^{ik\phi} = -i\left[e^{i\phi}+e^{2i\phi}+........+e^{10i\phi}\right]$$
Above is Sum of $\bf{10-}$  terms of $\bf{G.P}.$
So We get $$\displaystyle = -i\cdot \left[\frac{e^{i\phi}-e^{11i\phi}}{1-e^{i\phi}}\right] = -i\cdot \left[\frac{e^{i\phi}-1}{1-e^{i\phi}}\right]=+i$$
Bcz $$\displaystyle e^{11i\phi} = \cos 11\phi+i\sin 11 \phi = \cos \left(11\cdot \frac{2\pi}{11}\right)+i\sin \left(11\cdot \frac{2\pi}{11}\right)=1\;,$$ Bcz $$\displaystyle \phi = \frac{2\pi}{11}$$
</s>