$\sum_{k=1}^{10} (\sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11})=?$ How can we find the value of $$\sum_{k=1}^{10} (\sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11})=?$$  
My Approach:
By Euler's Theorem: $$\huge \cos \frac{2 \pi k}{11}+i \sin \frac{2 \pi k}{11} =e^{
 \frac{i2 \pi k}{11}}$$
$$\therefore \sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11}=-ie^{
 \frac{i2 \pi k}{11}}$$
so,$$\sum_{k=1}^{10} (\sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11})=-i \sum_{k=1}^{10}e^{
 \frac{i2 \pi k}{11}}=-i \times e^{
 \frac{i2 \pi }{11}} \times \frac{{\{e^{
 \frac{i2 \pi }{11}}\}}^{10} -1}{e^{
 \frac{i2 \pi }{11}}-1}$$
$$=-i  e^{
 \frac{i2 \pi }{11}} \frac{e^{
 \frac{i20 \pi }{11}} -1}{e^{
 \frac{i2 \pi }{11}}-1}$$
now how can i proceed and simplify the result?
Answer  $=i$ 
 A: \begin{align}-i  e^{
 \frac{i2 \pi }{11}} \frac{e^{
 \frac{i20 \pi }{11}} -1}{e^{
 \frac{i2 \pi }{11}}-1}&=-i\frac{\exp(\frac{i22\pi}{11})-\exp(\frac{i2\pi}{11})}{\exp(\frac{i2\pi}{11})-1} \\
&=-i\frac{\exp(i2\pi)-\exp(\frac{i2\pi}{11})}{\exp(\frac{i2\pi}{11})-1}
\\
&=-i\frac{1-\exp(\frac{i2\pi}{11})}{\exp(\frac{i2\pi}{11})-1}\\
&=-i(-1)\\
&=i
\end{align}
A: $$-i  e^{
 \frac{i2 \pi }{11}} \frac{e^{
 \frac{i20 \pi }{11}} -1}{e^{
 \frac{i2 \pi }{11}}-1}=
-i  \frac{e^{
 \frac{i2 \pi }{11}} e^{
 \frac{i20 \pi }{11}} -e^{
 \frac{i2 \pi }{11}} }{e^{
 \frac{i2 \pi }{11}}-1}=-i  \frac{1-e^{
 \frac{i2 \pi }{11}} }{e^{
 \frac{i2 \pi }{11}}-1}=i
$$
A: Denote the required sum value by $S$. Multiplying by $i$ we get
$$iS = \sum_{k=1}^{10}\bigg(\cos \frac{2\pi k}{11} + i\sin \frac{2\pi k}{11}\bigg)$$
Now in the summation introduce one extra term corrsponding to $k=11$ whose value will be 1.  We get LHS   to be $iS+1$. The RHS  now is a 11-term summation which is actually  the sum of all the 11th roots of unity which is zero.
So we get $iS+1=0$. So it follows that $S=i$.
A: $$S-i=-i\cdot (\sum_{k=1}^{10}({e^{\frac {2\pi i}{11}}})^k+1)=-i\cdot (\frac{1-({e^{\frac{2\pi i}{11}}) ^{11}}}{1-e^{\frac{2\pi i}{11}}})=-i\cdot 0=0$$.
