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JEE Main & Advanced Sample Paper JEE Main Sample Paper-15

  • question_answer
    Let \[{{z}_{1}}\] and \[{{z}_{2}}\]be nth roots of unity which subtend a right angle at the origin. Then n must be of the form

    A)  \[4k+1\]                             

    B)  \[4k+2\]

    C)  \[4k+3\]                             

    D)  \[4k\]

    Correct Answer: D

    Solution :

    \[{{\text{n}}^{\text{th}}}\]root of unity\[\cos \frac{2m\pi }{n}+i\sin \frac{2m\pi }{n}\] Let for \[m=k,\] root of unity makes an angle of \[{{90}^{o}}\]with\[\frac{2k\pi }{n}=\frac{\pi }{2}\Rightarrow \]\[n=4k\] For \[m=0\]        \[{{z}_{1}}=1\] For \[m=k\]        \[{{z}_{2}}\]such that \[{{z}_{2}}O{{z}_{1}}={{90}^{o}}\]


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