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Question : 7 of 160
Marks:
+1,
-0
Solution:
We have,
‌sin‌(‌)−i‌cos(‌)‌=−i[cos(‌)+isin‌(‌)]=−i‌ci2πk/7‌=−i[ei2π/7+ei4π/7+...+] This forms a GP
‌=−i[ei2π/7‌| [1−(ei2π/7)6] |
| 1−ei2π/7 |
]‌=−i[‌| ei2π/7−ei14π/7 |
| 1−ei2π/7 |
]‌=−i[‌][∵ei14π/7=1] =i
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