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If $p$ represents the veriable complex number $z$ and if $|2z-1|=2|z|$ then the locus of $p$ is

\[\begin{array}{1 1}(1) the\; straight\; line \;x=\frac{1}{4}&(2) the \;straight \; line \;y=\frac{1}{4}\\(3) the\; straight \; line \;z=\frac{1}{2}&(4)the\; circle\;x^{2}+y^{2}-4x-1=0\end{array}\]

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Let $z=x+iy$
$|2z-1|=2 |z|$
$|2(x+iy)-i|= 2 |x+iy|$
$|2x-1+i2y|=2 |x+iy|$
$\sqrt {(2x-1)^2+(2y)^2}=2 \sqrt {x^2+y^2}$
Squaring on both sides
$x= \large\frac{1}{4}$
Hence 1 is the correct answer
answered May 14, 2014 by meena.p

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