Question

Column I
Column II (locus)

Answer 1

$I)|z-1|=|z-i|$
Let $z=x+iy$ then
$(x-1{)}^2+y^2=x^2+(y-1{)}^2$
Or
$y^2-(y-1{)}^2=x^2-(x-1{)}^2$
$(2y-1)=2x-1$
Or
$y=x$ equation of a straight line passing through origin.
$II)|z+\overline{z}|+|z-\overline{z}|=2$
$|x+iy+x-iy|+|x+iy-x+iy|=2$
$2|x|+2|y|=2$
$|x|+|y|=1$ ... equation of a straight lines, making equal intercepts of magnitude $1$ on both the axes, thus forming a square due to their inclination of ${45}^0$ on $x$ axis.
$III)|z+\overline{z}|=|z-\overline{z}|$
$|x+iy+x-iy|=|x+iy-x+iy|$
$2x=\pm 2y$ or
$x=\pm y$... equation of pair of lines passing through the origin.
$IV)|z|=1$
$x^2+y^2=1$.
Equation of a circle.