Q60 / 160 · AP EAMCET 2009 · Mathematics

The locus of $z$ satisfying the inequality $\left|\frac{z+2 i}{2 z+i}\right| < 1$, where $z=x+i y$, is

  • A
    $x^2+y^2 < 1$
  • B
    $x^2-y^2 < 1$
  • Correct answer
    C
    $x^2+y^2>1$
  • D
    $2 x^2+3 y^2 < 1$

Solution

Let $z=x+i y$ Given, $\quad\left|\frac{z+2 i}{2 z+i}\right| < 1$ $\begin{array}{lc}\Rightarrow & \frac{\sqrt{(x)^2+(y+2)^2}}{\sqrt{(2 x)^2+(2 y+1)^2}} < 1 \\ \Rightarrow & x^2+y^2+4+4 y < 4 x^2+4 y^2+1+4 y \\ \Rightarrow & 3 x^2+3 y^2>3 \\ \Rightarrow & x^2+y^2>1\end{array}$

Hence, the correct answer is option: 'C'.

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