A)
B) \[-i\]
C) 1
D) 0
Correct Answer: A
Solution :
\[\sum\limits_{k=1}^{6}{\left( \sin \frac{2\pi k}{7}-i\cos \frac{2\pi k}{7} \right)}\] \[=\sum\limits_{k=1}^{6}{-i\left( \cos \frac{2\pi k}{7}+i\sin \frac{2\pi k}{7} \right)}\] \[=i\sum\limits_{k=1}^{6}{{{e}^{i\frac{2\pi k}{7}}}}\] (using\[cos\,\theta +i\text{ }sin\text{ }\theta ={{e}^{i\theta }}\]) \[=-i\sum\limits_{k=1}^{6}{{{r}^{k}}}\]where\[r={{e}^{i\frac{2\pi }{7}}}\] \[=-i[r+{{r}^{2}}+{{r}^{3}}+....+{{r}^{6}}]\] \[=i.\frac{r(1-{{r}^{6}})}{(1-r)}=-i\frac{(r-{{r}^{7}})}{1-r}\] \[(\because {{r}^{7}}={{e}^{2\pi i}}=\cos 2\pi +i\sin 2\pi =1)\] \[=-i\frac{(r-1)}{1-r}\] \[=-i(-1)=i\]
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