JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer If \[a>0\]and \[z=\frac{{{\left( 1+i \right)}^{2}}}{a-i},\]has magnitude\[\sqrt{\frac{2}{5},}\]then \[\overline{z}\]is equal to : [JEE Main 10-4-2019 Morning]

    A) \[-\frac{3}{5}-\frac{1}{5}i\]             

    B) \[-\frac{1}{5}+\frac{3}{5}i\]

    C) \[-\frac{1}{5}-\frac{3}{5}i\]

    D) \[\frac{1}{5}-\frac{3}{5}i\]

    Correct Answer: C

    Solution :

                Given \[a>0\] \[z=\frac{{{\left( 1+i \right)}^{2}}}{a-i}=\frac{2i\left( a+i \right)}{{{a}^{2}}+1}\] Also\[|z|=\sqrt{\frac{2}{5}}\Rightarrow \frac{2}{\sqrt{{{a}^{2}}+1}}=\sqrt{\frac{2}{5}}\Rightarrow a=3\] So\[\overline{z}=\frac{-2i(3-i)}{10}=\frac{-1-3i}{5}\]       


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