VIT Engineering VIT Engineering Solved Paper-2009

  • question_answer
    The locus of z satisfying the inequality\[\left| \frac{z+2i}{2z+i} \right|<1,\] where \[z=x+iy,\]is

    A)  \[{{x}^{2}}+{{y}^{2}}<1\]

    B)  \[{{x}^{2}}-{{y}^{2}}<1\]

    C)  \[{{x}^{2}}+{{y}^{2}}>1\]

    D)  \[2{{x}^{2}}+3{{y}^{2}}<1\]

    Correct Answer: C

    Solution :

    Let \[z=x+iy\] Given \[\left| \frac{z+2i}{2z+i} \right|<1\] \[\Rightarrow \] \[\frac{\sqrt{{{(x)}^{2}}+{{(y+2)}^{2}}}}{\sqrt{{{(2x)}^{2}}+{{(2y+1)}^{2}}}}<1\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}+4+4y<4{{x}^{2}}+4{{y}^{2}}+1+4y\] \[\Rightarrow \] \[3{{x}^{2}}+3{{y}^{2}}>3\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}>1\]


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