<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] $$-i  e^{
 \frac{i2 \pi }{11}} \frac{e^{
 \frac{i20 \pi }{11}} -1}{e^{
 \frac{i2 \pi }{11}}-1}=
-i  \frac{e^{
 \frac{i2 \pi }{11}} e^{
 \frac{i20 \pi }{11}} -e^{
 \frac{i2 \pi }{11}} }{e^{
 \frac{i2 \pi }{11}}-1}=-i  \frac{1-e^{
 \frac{i2 \pi }{11}} }{e^{
 \frac{i2 \pi }{11}}-1}=i
$$
</s>