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 $\frac{i^2}{i}=i$ 
 A: If you insert an extra term corresponding to $k=0$, which is $-i$, that will be the negative of the sum of all the $11$th roots of unity, i.e. the coefficient of $x^{10}$ in $x^{11}-1$. This sum is zero hence the answer for your question is $+i$, after pulling out the inserted extra term. This is essentially what you have done, and you are correct.
A: 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 $n > 2$.
A: $$\frac1i\sum_{k=1}^{10}\left(\cos\frac{2k\pi}{11}+i\sin\frac{2k\pi}{11}\right)=-i\sum_{k=1}^{10} \left(e^{\frac{2\pi i}{11}}\right)^k=$$
$$=-ie^{\frac{2\pi i}{11}}\frac{e^{\frac{2\cdot10\pi i}{11}}-1}{e^{\frac{2\pi i}{11}}-1}=-ie^{\frac{2\pi i}{11}}\frac{e^{\frac{-2\pi i}{11}}-1}{e^{\frac{2\pi i}{11}}-1}=-i\frac{e^{\frac{2\pi i}{11}}-1}{1-e^{\frac{2\pi i}{11}}}=i$$
A: $\bf{My\; Solution::}$ 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 $$\displaystyle \phi = \frac{2\pi}{11}$$
A: $\begin{array}{l}
\\
\sum \limits^{10}_{k=1}(\sin \ \frac{2k\pi }{11} -i\cos \frac{2k\pi }{11}  )\\
=(-i)\sum \limits^{10}_{k=1}(\cos  \frac{2k\pi }{11} +i\sin \frac{2k\pi }{11}  )\\
=(-i)(\sum \limits^{10}_{k=0}(\cos  \frac{2k\pi }{11} +i\sin \frac{2k\pi }{11} )-1)\\
=(-i)(-1)\\
=i\\
Note: 
\sum \limits^{10}_{k=0}(\cos \frac{2k\pi }{11} +i\sin \frac{2k\pi }{11}  )=0\\
\end{array}$
Sum of $11^{th}$ roots of unity =$0$
