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

  • question_answer
    \[\frac{{{(\cos \theta +i\sin \theta )}^{4}}}{{{(\sin \theta +i\cos \theta )}^{5}}}\] is equal to   [MNR 1985; UPSEAT 2000]

    A) \[\cos \theta -i\sin \theta \]

    B) \[\cos 9\theta -i\sin 9\theta \]

    C) \[\sin \theta -i\cos \theta \]

    D) \[\sin 9\theta -i\cos 9\theta \]

    Correct Answer: D

    Solution :

    \[\frac{{{(\cos \theta +i\sin \theta )}^{4}}}{{{(\sin \theta +i\cos \theta )}^{5}}}=\frac{{{(\cos +i\sin \theta )}^{4}}}{{{i}^{5}}{{\left( \frac{1}{i}\sin \theta +\cos \theta  \right)}^{5}}}\] \[=\frac{{{(\cos \theta +i\sin \theta )}^{4}}}{i\,{{(\cos \theta -i\sin \theta )}^{5}}}=\frac{{{(\cos \theta +i\sin \theta )}^{4}}}{i\,{{(\cos \theta +i\sin \theta )}^{-5}}}\] (By property) \[=\frac{1}{i}{{(\cos \theta +i\sin \theta )}^{9}}=\sin 9\theta -i\cos 9\theta \].


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