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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 \[\frac{z-i}{z+i}(z\ne -i)\] is a purely imaginary number, then \[z.\bar{z}\] is equal to

    A) 0

    B) 1

    C) 2

    D) None of these

    Correct Answer: B

    Solution :

    Here \[\frac{z-i}{z+i}=\frac{x+i(y-1)}{x+i(y+1)}.\frac{x-i(y+1)}{x-i(y+1)}\]                  \[=\frac{({{x}^{2}}+{{y}^{2}}-1)+i(-2x)}{{{x}^{2}}+{{(y+1)}^{2}}}\] As \[\frac{z-i}{z+i}\] is purely imaginary, we get \[{{x}^{2}}+{{y}^{2}}-1=0\]Þ \[{{x}^{2}}+{{y}^{2}}=1\]Þ\[z\overline{z}=1\].


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