<s>[INST] $\sum_{k=1}^{10} (\sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11})=?$ 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  $=i$ 
 [/INST] 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 $S=i$.
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