JEE Main & Advanced Mathematics Complex Numbers and Quadratic Equations Question Bank Square root, Representation and Logarithm of complex numbers

  • question_answer
    If \[x+iy=\sqrt{\frac{a+ib}{c+id}},\]then \[{{({{x}^{2}}+{{y}^{2}})}^{2}}=\] [IIT 1979; RPET 1997; Karnataka CET 1999]

    A) \[\frac{{{a}^{2}}+{{b}^{2}}}{{{c}^{2}}+{{d}^{2}}}\]

    B) \[\frac{a+b}{c+d}\]

    C) \[\frac{{{c}^{2}}+{{d}^{2}}}{{{a}^{2}}+{{b}^{2}}}\]

    D) \[{{\left( \frac{{{a}^{2}}+{{b}^{2}}}{{{c}^{2}}+{{d}^{2}}} \right)}^{2}}\]

    Correct Answer: A

    Solution :

    \[x+iy=\sqrt{\frac{a+ib}{c+id}}\]Þ \[x-iy=\sqrt{\frac{a-ib}{c-id}}\]  Also  \[{{x}^{2}}+{{y}^{2}}=(x+iy)(x-iy)=\sqrt{\frac{{{a}^{2}}+{{b}^{2}}}{{{c}^{2}}+{{d}^{2}}}}\] Þ  \[{{({{x}^{2}}+{{y}^{2}})}^{2}}=\frac{{{a}^{2}}+{{b}^{2}}}{{{c}^{2}}+{{d}^{2}}}\]


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