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JEE Main & Advanced Mathematics Complex Numbers and Quadratic Equations Question Bank Conjugate, Modulus and Argument of complex number

  • question_answer
    If \[{{z}_{1}}\] and \[{{z}_{2}}\] are any two complex numbers then \[|{{z}_{1}}+{{z}_{2}}{{|}^{2}}\] \[+|{{z}_{1}}-{{z}_{2}}{{|}^{2}}\] is equal to    [MP PET 1993; RPET 1997]

    A) \[2|{{z}_{1}}{{|}^{2}}\,|{{z}_{2}}{{|}^{2}}\]

    B) \[2|{{z}_{1}}{{|}^{2}}+\,2\,\,|{{z}_{2}}{{|}^{2}}\]

    C) \[|{{z}_{1}}{{|}^{2}}+\,|{{z}_{2}}{{|}^{2}}\]

    D) \[2|{{z}_{1}}|\,\,|{{z}_{2}}|\]

    Correct Answer: B

    Solution :

    \[|{{z}_{1}}+{{z}_{2}}{{|}^{2}}+|{{z}_{1}}-{{z}_{2}}{{|}^{2}}\] \[={{({{x}_{1}}+{{x}_{2}})}^{2}}+{{({{y}_{1}}+{{y}_{2}})}^{2}}+{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}\] \[=2(x_{1}^{2})+2(y_{1}^{2})+2(x_{2}^{2})+2(y_{2}^{2})=2|{{z}_{1}}{{|}^{2}}+2|{{z}_{2}}{{|}^{2}}\]


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