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JEE Main & Advanced JEE Main Solved Paper-2015

  • question_answer
    A complex number z is said to be unimodular if \[\left| z \right|=1.\]Suppose \[{{z}_{1}}\] and \[{{z}_{2}}\] are complex numbers such that \[\frac{{{z}_{1}}-2{{z}_{2}}}{2-{{z}_{1}}{{\overline{z}}_{2}}}\]is unimodular and \[{{z}_{2}}\] is not unimodular. Then the point \[{{z}_{1}}\] lies on a : [JEE Main Solved Paper-2015 ]

    A) Circle of radius 2              

    B) Circle of radius 2

    C) Straight line parallel to x-axis

    D) Straight line parallel to y-axis Solut

    Correct Answer: A

    Solution :

    \[{{\left| \frac{{{z}_{1}}-2{{z}_{2}}}{2-{{z}_{1}}{{\overline{z}}_{2}}} \right|}^{2}}=1\] \[\Rightarrow \]\[\left( \frac{{{z}_{1}}-2{{z}_{2}}}{2-{{z}_{1}}{{\overline{z}}_{2}}} \right)\left( \frac{{{\overline{z}}_{1}}-2{{\overline{z}}_{2}}}{2-{{\overline{z}}_{1}}-{{z}_{2}}} \right)=1\] \[\Rightarrow \]\[{{\left| {{z}_{1}} \right|}^{2}}{{\left| {{z}_{2}} \right|}^{2}}-{{\left| {{z}_{1}} \right|}^{2}}-4{{\left| {{z}_{2}} \right|}^{2}}+4=0\] \[\Rightarrow \]\[\left( {{\left| {{z}_{1}} \right|}^{2}}-1 \right)\left( {{\left| {{z}_{1}} \right|}^{2}}-4 \right)=0\]\[\Rightarrow \]\[\left| {{z}_{1}} \right|=2\] Circle with radius ?2?


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