$\sum_{k=1}^{10} (\sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11})=?$

Question

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 $=i$

Answer 1

$$-i e^{ \frac{i2 \pi }{11}} \frac{e^{ \frac{i20 \pi }{11}} -1}{e^{ \frac{i2 \pi }{11}}-1}= -i \frac{e^{ \frac{i2 \pi }{11}} e^{ \frac{i20 \pi }{11}} -e^{ \frac{i2 \pi }{11}} }{e^{ \frac{i2 \pi }{11}}-1}=-i \frac{1-e^{ \frac{i2 \pi }{11}} }{e^{ \frac{i2 \pi }{11}}-1}=i $$

Answer 2

\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}

Answer 3

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 $S=i$.

Answer 4

$$S-i=-i\cdot (\sum_{k=1}^{10}({e^{\frac {2\pi i}{11}}})^k+1)=-i\cdot (\frac{1-({e^{\frac{2\pi i}{11}}) ^{11}}}{1-e^{\frac{2\pi i}{11}}})=-i\cdot 0=0$$.