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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2011

  • question_answer
    Let\[{{z}_{1}}=3+4i\]and\[{{z}_{2}}=-1+2i\].  Then, \[|{{z}_{1}}+{{z}_{2}}{{|}^{2}}-2(|{{z}_{1}}{{|}^{2}}+|{{z}_{2}}{{|}^{2}})\]is equal to

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

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

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

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

    E)  \[|{{z}_{1}}{{|}^{2}}+|{{z}_{2}}{{|}^{2}}-2|{{z}_{1}}||{{z}_{2}}|\]

    Correct Answer: B

    Solution :

    Given, \[{{z}_{1}}=3+4i,{{z}_{2}}=-1+2i\] then, \[|{{z}_{1}}+{{z}_{2}}{{|}^{2}}-2(|{{z}_{1}}{{|}^{2}}+|{{z}_{2}}{{|}^{2}}=?\] We know that, by parallelogram law, \[|{{z}_{1}}+{{z}_{2}}{{|}^{2}}+|{{z}_{1}}-{{z}_{2}}{{|}^{2}}=2(|{{z}_{1}}{{|}^{2}}+|{{z}_{2}}{{|}^{2}})\] \[\Rightarrow \]\[|{{z}_{1}}+{{z}_{2}}{{|}^{2}}-2(|{{z}_{1}}{{|}^{2}}+|{{z}_{2}}{{|}^{2}})=-|{{z}_{1}}-{{z}_{2}}{{|}^{2}}\] Here,\[LHS=-20\]and\[RHS=-20\]


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