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The locus of $z$ satisfying the inequality $\left|\frac{z+2 i}{2 z+i}\right| < 1$, where $z=x+i y$, is
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The correct answer is:
$x^2+y^2>1$
Let $z=x+i y$
Given, $\quad\left|\frac{z+2 i}{2 z+i}\right| < 1$
$\begin{array}{lc}\Rightarrow & \frac{\sqrt{(x)^2+(y+2)^2}}{\sqrt{(2 x)^2+(2 y+1)^2}} < 1 \\ \Rightarrow & x^2+y^2+4+4 y < 4 x^2+4 y^2+1+4 y \\ \Rightarrow & 3 x^2+3 y^2>3 \\ \Rightarrow & x^2+y^2>1\end{array}$
Given, $\quad\left|\frac{z+2 i}{2 z+i}\right| < 1$
$\begin{array}{lc}\Rightarrow & \frac{\sqrt{(x)^2+(y+2)^2}}{\sqrt{(2 x)^2+(2 y+1)^2}} < 1 \\ \Rightarrow & x^2+y^2+4+4 y < 4 x^2+4 y^2+1+4 y \\ \Rightarrow & 3 x^2+3 y^2>3 \\ \Rightarrow & x^2+y^2>1\end{array}$
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