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If $i = \sqrt{- 1}, then 4 + 3 {(- \frac{1}{2} + i \frac{\sqrt{3}}{2})}^{127} + 5 {(- \frac{1}{2} + i \frac{\sqrt{3}}{2})}^{124}$ is equal to

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43i

2 3 i

1 3- i

1 3+i

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detailed solution

Correct option is A

Let ω=−12+i32 then ω3=1z=4+3ω127+5ω124=4+3ω342ω+5ω341ω=4+3ω+5ω=4+8ω=4−4+43i=43i

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If i=−1, then 4+3−12+i32127+5−12+i32124 is equal to

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If $i = \sqrt{- 1}, then 4 + 3 {(- \frac{1}{2} + i \frac{\sqrt{3}}{2})}^{127} + 5 {(- \frac{1}{2} + i \frac{\sqrt{3}}{2})}^{124}$ is equal to