Question

if $z$ is complex number, then the locus of $z$ satisfying the conditions $|2z-1|=|z-1|$ is

• A. perpendicular bisector of line segment joining 1/2 and 1
• B. circle
• C. parabola
• D. none of the above curves

Answer 1

The correct option is B circle

Let us write z as $x+iy$
We then have $|2z-1|=|z-1|$
$\Rightarrow 4x^2-8x+1+4y^2=x^2-2x+1+y^2$
$\Rightarrow 3x^2+3y^2-6x=0$ which is a circle.