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JEE Main & Advanced Mathematics Complex Numbers and Quadratic Equations Question Bank De Moivre's theorem and Roots of unity

  • question_answer
    If  \[\omega (\ne 1)\]is a cube root of unity and \[{{(1+\omega )}^{7}}=A+B\omega \],  then \[A\] and \[B\] are respectively, the numbers [IIT 1995]

    A) 0, 1

    B) 1, 0

    C) 1, 1

    D) \[-1,\ 1\]

    Correct Answer: C

    Solution :

    \[{{(1+\omega )}^{7}}=A+B\omega \,\,\,\,\Rightarrow {{(-{{\omega }^{2}})}^{7}}=A+B\omega \] Þ  \[=\frac{\cos \,\,4\theta +i\sin \,\,4\theta }{\cos \,\,4\theta -i\sin \,\,4\theta }\] Þ  \[{{\omega }^{2}}.{{\omega }^{12}}=-A-B\omega \]Þ\[A+B\omega +{{\omega }^{2}}=0\] Þ  \[A=1,B=1\,\] \[\,(\because 1+\omega +{{\omega }^{2}}=0)\]


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