JEE Main & Advanced Mathematics Complex Numbers and Quadratic Equations Question Bank Self Evaluation Test - Complex Numbers and Quadratic Equations

  • question_answer
    Let \[{{x}_{1}}\] and \[{{y}_{1}}\] be real numbers. If \[{{z}_{1}}\] and \[{{z}_{2}}\] are complex numbers such that \[|{{z}_{1}}|=|{{z}_{2}}|=4,\] then \[|{{x}_{1}}{{z}_{1}}-{{y}_{1}}{{z}_{2}}{{|}^{2}}+|{{y}_{1}}{{z}_{1}}+{{x}_{1}}{{z}_{2}}{{|}^{2}}=\]

    A) \[32({{x}_{1}}^{2}+{{y}_{1}}^{2})\]                      

    B) \[16({{x}_{1}}^{2}+{{y}_{1}}^{2})\]

    C) \[4({{x}_{1}}^{2}+{{y}_{1}}^{2})\]            

    D) \[32({{x}_{1}}^{2}+{{y}_{1}}^{2})|{{z}_{1}}+{{z}_{2}}{{|}^{2}}\]

    Correct Answer: A

    Solution :

    \[{{\left| {{x}_{1}}{{y}_{1}}-{{y}_{1}}{{z}_{2}} \right|}^{2}}+{{\left| {{y}_{1}}{{z}_{1}}-{{x}_{1}}{{z}_{2}} \right|}^{2}}\] \[=\,\,\,{{\left| {{x}_{1}}{{y}_{1}} \right|}^{2}}+{{\left| {{y}_{1}}{{z}_{2}} \right|}^{2}}-2\operatorname{Re}({{x}_{1}}{{y}_{1}}{{z}_{1}}{{z}_{2}})\] \[+\,\,|\,{{y}_{1}}{{z}_{1}}{{|}^{2}}+\,\,{{\left| {{x}_{1}}{{z}_{2}} \right|}^{2}}+\,\,2Re\,({{x}_{1}}{{y}_{1}}{{z}_{1}}{{z}_{2}})\] \[=x_{2}^{1}\,{{\left| {{z}_{1}} \right|}^{2}}+{{y}_{1}}^{2}{{\left| {{z}_{2}} \right|}^{2}}+{{y}_{1}}^{2}{{\left| {{z}_{1}} \right|}^{2}}+{{x}_{1}}^{2}{{\left| {{z}_{2}} \right|}^{2}}\] \[=x_{2}^{1}\,{{\left| {{z}_{1}} \right|}^{2}}+{{y}_{1}}^{2}{{\left| {{z}_{2}} \right|}^{2}}+{{y}_{1}}^{2}{{\left| {{z}_{1}} \right|}^{2}}+{{x}_{1}}^{2}{{\left| {{z}_{2}} \right|}^{2}}\] \[=\,\,2\left( {{x}_{1}}^{2}+{{y}_{1}}^{2} \right)\left( {{4}^{2}} \right)=32\left( {{x}_{1}}^{2}+{{y}_{1}}^{2} \right)\]


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