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JEE Main & Advanced Mathematics Complex Numbers and Quadratic Equations Question Bank Integral power of iota, Algebraic operations and Equality of complex numbers

  • question_answer
    If  \[\frac{{{(p+i)}^{2}}}{2p-i}=\mu +i\lambda ,\]then \[{{\mu }^{2}}+{{\lambda }^{2}}\] is equal to

    A) \[\frac{{{({{p}^{2}}+1)}^{2}}}{4{{p}^{2}}-1}\]

    B) \[\frac{{{({{p}^{2}}-1)}^{2}}}{4{{p}^{2}}-1}\]

    C) \[\frac{{{({{p}^{2}}-1)}^{2}}}{4{{p}^{2}}+1}\]

    D) \[\frac{{{({{p}^{2}}+1)}^{2}}}{4{{p}^{2}}+1}\]

    Correct Answer: D

    Solution :

    \[\mu +i\lambda =\frac{{{(p+i)}^{2}}}{2p-i}=\frac{({{p}^{2}}-1+2pi)(2p+i)}{(2p-i)(2p+i)}\]          \[=\frac{2p({{p}^{2}}-2)+i(5{{p}^{2}}-1)}{4{{p}^{2}}+1}\] \ \[{{\mu }^{2}}+{{\lambda }^{2}}=\frac{4{{p}^{2}}{{({{p}^{2}}-2)}^{2}}+{{(5{{p}^{2}}-1)}^{2}}}{{{(4{{p}^{2}}+1)}^{2}}}\]                 \[=\frac{4{{p}^{6}}+6{{p}^{2}}+9{{p}^{4}}+1}{{{(4{{p}^{2}}+1)}^{2}}}\]                 \[=\,\,\frac{{{p}^{4}}(4{{p}^{2}}+1)+2{{p}^{2}}(4{{p}^{2}}+1)+(4{{p}^{2}}+1)}{{{(4{{p}^{2}}+1)}^{2}}}\]                        \[=\frac{{{p}^{4}}+2{{p}^{2}}+1}{4{{p}^{2}}+1}=\frac{{{({{p}^{2}}+1)}^{2}}}{4{{p}^{2}}+1}\].


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