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 \[{{(x+iy)}^{1/3}}=a+ib,\]then \[\frac{x}{a}+\frac{y}{b}\]is equal to [IT 1982; Karnataka CET 2000]

    A) \[4({{a}^{2}}+{{b}^{2}})\]

    B) \[4({{a}^{2}}-{{b}^{2}})\]

    C) \[4({{b}^{2}}-{{a}^{2}})\]

    D) None of these

    Correct Answer: B

    Solution :

    \[{{(x+iy)}^{1/3}}=a+ib\]Þ\[(x+iy)={{(a+ib)}^{3}}\] \[={{a}^{3}}+3{{a}^{2}}.ib+3a.{{(ib)}^{2}}+{{(ib)}^{3}}\] \[={{a}^{3}}-3a{{b}^{2}}+i(3{{a}^{2}}b-{{b}^{3}})\] Equating real and imaginary parts, we get \[\frac{x}{a}={{a}^{2}}-3{{b}^{2}}\]and \[\frac{y}{b}=3{{a}^{2}}-{{b}^{2}}\] \[\therefore \]  \[\frac{x}{a}+\frac{y}{b}=4({{a}^{2}}-{{b}^{2}})\]

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