A) \[-1\]
B) 1
C) \[i\]
D) \[-i\]
Correct Answer: D
Solution :
We have \[\sum\limits_{r=1}^{8}{\left( \sin \frac{2r\pi }{9}+i\cos \frac{2r\pi }{9} \right)}=\sum\limits_{r=1}^{8}{i\,\,\left( \cos \frac{2r\pi }{9}-i\sin \frac{2r\pi }{9} \right)}\] \[=i\sum\limits_{r=1}^{8}{{{e}^{-i\frac{2r\pi }{9}}}}=i\sum\limits_{r=1}^{8}{{{\alpha }^{r}},}\] when \[\alpha ={{e}^{-(2\pi i/9)}}\] \[=i\alpha \frac{(1-{{\alpha }^{8}})}{(1-\alpha )}\]\[=i\frac{(\alpha -{{\alpha }^{9}})}{1-\alpha }=i\left( \frac{\alpha -1}{1-\alpha } \right)=-i\] \[(\because {{\alpha }^{9}}={{e}^{-i2\pi }}=\cos 2\pi -i\sin 2\pi =1\])
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