Question

locate the point representing the complex numbers $z$ on the Argand diagram for which
$|z-1|=|z-3|=|z-i|,$

Answer 1

${|z-1|}^2={|z-3|}^2={|z-i|}^2$ on squaring
$\therefore {(x-1)}^2+y^2={(x-3)}^2+y^2=x^2+{(y-1)}^2$
$I$ $II$ $III$
From $I$ and $II$, $2(2x-4)=0$ $\therefore x=2$
From $I$ and $III$, $-2x=-2y$ $\therefore x=y=2$
$\therefore z=x+iy=2+2i$
Remarks: Geometric speaking, the equation $\left(1\right)$ means that $z$ is a point equidistant from the point representing the number
$1,3,i$, that is, equidistant from the points $A(1,0)$, $B(3,0)$ and $C(0,1)$. It is therefore the circumcentre of $\Delta ABC$.
So it is the intersection of perpendicular bisectors $FP$ and$EP$ of segment $AB$ and $AC$, that is, the intersection $P$ of the lines represented by the equation $x=2$ and $y=x$. So we get $x=y=2$ as before. (See fig.).