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JCECE Engineering JCECE Engineering Solved Paper-2009

  • question_answer
    If\[x=\frac{1}{2}(\sqrt{3}+i)\], then \[{{x}^{3}}\] is equal to

    A) \[1\]                                     

    B) \[-1\]

    C) \[i\]                                       

    D) \[-i\]

    Correct Answer: B

    Solution :

    Given,\[x=\frac{\sqrt{3}+i}{2}\] \[\Rightarrow \]               \[x=-{{\omega }^{2}}\]  \[\left( \because \,\,{{\omega }^{2}}=\frac{-\sqrt{3}-i}{2} \right)\] \[\Rightarrow \]               \[{{x}^{3}}={{(-1)}^{3}}{{({{\omega }^{2}})}^{3}}\] \[\Rightarrow \]               \[{{x}^{3}}=-1\]


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