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If $P(x, y)$ denotes $z=x+i y$ in Argand's plane and $\left|\frac{z-1}{z+2 i}\right|=1$, then the locus of $P$ is a/an
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straight line
Given, $z=x+i y$ and $\left|\frac{z-1}{z+2 i}\right|=1$
$\Rightarrow \quad\left|\frac{(x+i y)-1}{(x+i y)+2 i}\right|=1$
$\Rightarrow \quad|(x-1)+i y|=|x+(y+2) i|$
Squaring on both sides,
$\Rightarrow \quad|(x-1)+i y|^{2}=|x+(y+2) i|^{2}$ $\Rightarrow \quad(x-1)^{2}+y^{2}=x^{2}+(y+2)^{2}$ $\Rightarrow \quad x^{2}+y^{2}+1-2 x=x^{2}+y^{2}+4+4 y$ $\Rightarrow 2 x+24+3=0$, which represents a straight line.
$\Rightarrow \quad\left|\frac{(x+i y)-1}{(x+i y)+2 i}\right|=1$
$\Rightarrow \quad|(x-1)+i y|=|x+(y+2) i|$
Squaring on both sides,
$\Rightarrow \quad|(x-1)+i y|^{2}=|x+(y+2) i|^{2}$ $\Rightarrow \quad(x-1)^{2}+y^{2}=x^{2}+(y+2)^{2}$ $\Rightarrow \quad x^{2}+y^{2}+1-2 x=x^{2}+y^{2}+4+4 y$ $\Rightarrow 2 x+24+3=0$, which represents a straight line.
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