A) \[\frac{1+\sqrt{3}i}{2}\]
B) \[\frac{1-\sqrt{3}i}{2}\]
C) \[\frac{-\sqrt{3}-i}{2}\]
D) \[\frac{\sqrt{3}-i}{2}\]
Correct Answer: C
Solution :
Since \[\frac{-\sqrt{3}-i}{2}=-\left( \cos \frac{\pi }{6}+i\sin \frac{\pi }{6} \right)\] Þ \[{{\left( \frac{-\sqrt{3}-i}{2} \right)}^{3}}=-{{\left( \cos \frac{\pi }{6}+i\sin \frac{\pi }{6} \right)}^{3}}=-i\] and \[\frac{\sqrt{3}-i}{2}=\cos \frac{\pi }{6}-i\sin \frac{\pi }{6}\] and \[{{\left( \frac{\sqrt{3}-i}{2} \right)}^{3}}=\cos \frac{\pi }{2}-i\sin \frac{\pi }{2}=-i\]. Hence the result.
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