STD 11 - 4.1 complex nubers — JEE कक्षा 11 साइन्स — Question

$ = i\sum\limits_{r = 1}^8 {{e^{ - i\frac{{2r\pi }}{9}}}}  = i\sum\limits_{r = 1}^8 {{\alpha ^r},} $ जहाँ $\alpha  = {e^{ - (2\pi i/9)}}$

$ = i\alpha \frac{{(1 - {\alpha ^8})}}{{(1 - \alpha )}}$$ = i\frac{{(\alpha  - {\alpha ^9})}}{{1 - \alpha }} = i\left( {\frac{{\alpha  - 1}}{{1 - \alpha }}} \right) =  - i$

$(\because {\alpha ^9} = {e^{ - i2\pi }} = \cos 2\pi  - i\sin 2\pi  = 1)$