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Question
Find the value of `sum_("k" = 1)^8 (cos (2"k"pi)/9 + "i" sin (2"kpi)/9)`
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Solution
`sum_("k" = 1)^8 (cos (2"k"pi)/9 + "i" sin (2"kpi)/9)`
The sum all nth root of unity is
1 + ω + ω² + …….. + ωn-1 = 0
From the given polar from, it is clear that the complex number is 1 + i0 (unity)
`sum_("k" = 0)^8 (cos (2"k"pi)/9 + "i" sin (2"kpi)/9)` = 0
i.e., `1 + omega + omega^2 + ... + omega^8` = 0
`1 + sum_("k" = 1)^8 (cos (2"k"pi)/9 + "i" sin (2"kpi)/9)` = 0
∴ `sum_("k" = 1)^8 (cos (2"k"pi)/9 + "i" sin (2"kpi)/9)` = – 1.
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