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