Complex Numbers And Quadratic Equations question 148
Question: The value of $ \sum\limits_{r=1}^{8}{( \sin \frac{2r\pi }{9}+i\cos \frac{2r\pi }{9} )} $ is
Options:
A) $ -1 $
B) 1
C) $ i $
D) $ -i $
Show Answer
Answer:
Correct Answer: D
Solution:
We have $ \sum\limits_{r=1}^{8}{( \sin \frac{2r\pi }{9}+i\cos \frac{2r\pi }{9} )}=\sum\limits_{r=1}^{8}{i( \cos \frac{2r\pi }{9}-i\sin \frac{2r\pi }{9} )} $ $ =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( \frac{\alpha -1}{1-\alpha } )=-i $ $ (\because {{\alpha }^{9}}={e^{-i2\pi }}=\cos 2\pi -i\sin 2\pi =1 $ )
BETA
