<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] Let $$\zeta_{11} = e^{2\pi i/11} = \cos \frac{2\pi}{11} + i \sin \frac{2\pi}{11}$$ be a primitive $11^{\rm th}$ root of unity; hence $$\zeta_{11}^0, \zeta_{11}^1, \ldots, \zeta_{11}^{10}$$ are the roots of $z^{11} - 1 = 0$, and the sum of these roots is therefore zero.  Then
$$\begin{align*} \sum_{k=1}^{10} \left( \sin \frac{2\pi k}{11} - i \cos \frac{2\pi k}{11} \right) &= \frac{1}{i} \sum_{k=1}^{10} e^{2\pi i k/11} \\ &= \frac{1}{i}\left( -1 + \sum_{k=0}^{10} \zeta_{11}^k \right) \\ &= \frac{1}{i} (-1) \\ &= i. \end{align*}$$

To see that the sum of the aforementioned roots is zero, we could also have explicitly summed the geometric series:  $$\sum_{k=0}^{n-1} \zeta_n^k = \frac{\zeta_n^n - 1}{\zeta_n - 1} = \frac{0}{\zeta_n - 1} = 0,$$ for $n > 2$.
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