Question

If ${}^{\prime }{z}^{\prime }$ is complex number then the locus of ${}^{\prime }{z}^{\prime }$ satisfying the condition $|2z-1|=|z-1|$ is

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

Answer 1

Answer: (B)

circle

Since z is a complex number & relation

$|2z-1|=|z-1|$

squaring both sides

${|2z-1|}^2={|z-1|}^2$

Let $z=x+i$

$\therefore {(2x-1)}^2+{\left(2y\right)}^2={(x-1)}^2+y^2$

$\Rightarrow {4x}^2-4x+1+{4y}^2=x^2+y-2x+y^2$

$\Rightarrow {3x}^2+{3y}^2-2x=0$