Question

If z is complex number, then the locus of z 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

The correct option is B Circle

Let $z=x+iy$
$|(2x-1)+i2y|=|(x-1)+iy|$
$(2x-1{)}^2+4y^2=(x-1{)}^2+y^2$
$\left(x\right)(3x-2)+3y^2=0$
$3x^2+3y^2-2x=0$
$x^2+y^2-\frac{2x}{3}=0$
$(x-\frac{1}{3}{)}^2+y^2=\frac{1}{9}$
Hence it represents the equation of a circle.