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What is \(\left(\dfrac{\sqrt{3}+i}{\sqrt{3}-i}\right)^6\) equal to?1. -12. 03. 14. 2 |
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Answer» Correct Answer - Option 3 : 1 Concept: 1, ω and ω2 are cube root of unity Where ω = \( \rm \frac{-1+\sqrt3i}{2}\) , ω2 = \( \rm \frac{-1-\sqrt3i}{2}\) , ω3 = 1
Calculation: \(\left(\dfrac{\sqrt{3}+i}{\sqrt{3}-i}\right)^6\) \(= \rm (\dfrac{\sqrt{3}+i}{\sqrt{3}-i}\times\dfrac{\sqrt{3}+i}{\sqrt{3}+i} )^6\) \(= \rm (\dfrac{(\sqrt{3}+i)^2}{3+1})^6\) \(= \rm (\dfrac{({3}-1+2\sqrt3i)}{4})^6\) \(= \rm (\dfrac{2(1+\sqrt3i)}{4})^6\) \(= \rm (\dfrac{1+\sqrt3i}{2})^6\) = (-ω2)6 (∵ -ω2 = \( \rm \dfrac{1+\sqrt3i}{2}\) ) = (ω3)4 = 1 Hence, option (3) is correct. |
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