Locate the points for which $|z-1|=|z-2|=|z-i|$.

Question

Locate the points for which $|z-1|=|z-2|=|z-i|$.

My approach is that the equality can be formed between any of the 3 terms (don't know of better way to call), i.e.:
(i) $|z-1|=|z-2|$,
(ii) $|z-1|=|z-i|$,
(iii) $|z-2|=|z-i|$

Have following 6 combinations / orderings of these 3 equalities:

(a) (i) (ii) (iii),
(b) (i) (iii) (ii),
(c) (ii) (i) (iii),
(d) (ii) (iii) (i),
(e) (iii) (i) (ii),
(f) (iii) (ii) (i)

Have raised two questions below, with request to vet:

Q.1. How to 'theoretically' show that all of these ordering are equivalent in terms of finding the solution point.
Soln. 1: Will find the solutions in terms of equalities of complex terms, and would get the same solution irrespective of orderings.

Q.2. Have worked out the first two orderings.

(a) (i) (ii) (iii):
Taking $z = x +iy$,
$|z−1|$ is the distance to $(1,0)$, & is given like for modulus by $\sqrt{(x-1)^2+y^2}$;
$|z−2|$ is the distance to $(2,0)=\sqrt{(x-2)^2+y^2}$;
$|z-i|$ is the distance to $i$ i.e $(0,1)$, with distance as $\sqrt{x^2+(y-1)^2}$.

(i) $\sqrt{(x-1)^2+y^2}=\sqrt{(x-2)^2+y^2}$
$\implies (x-1)^2+y^2=(x-2)^2+y^2$
$\implies 1 -2x = 4 -4x\implies 2x = 3\implies x = \frac 32$.
This gives the perpendicular bisector of the line midway between the points $x= 1$, and $x=2$.

(ii) $\sqrt{(x-1)^2+y^2}=\sqrt{x^2+(y-1)^2}$
$\implies (x-1)^2+y^2=x^2+(y-1)^2$
$\implies 1 -2x = 1-2y\implies x = y$.
This gives the perpendicular bisector of the line midway between the points $x= 1$, or $(1,0)$, and $y=1$, or $(0,1)$; given by: $x-y=0$.

(iii) $\sqrt{(x-2)^2+y^2}=\sqrt{x^2+(y-1)^2}$
$\implies (x-2)^2+y^2=x^2+(y-1)^2$
$\implies 4 -4x = 1-2y\implies -4x +2y +3 =0 $.
This gives linear equation as solution.

For finding pair-wise intersection, compute the 3 pairs' solution below:
(1) (i), (ii) : Substituting $x = \frac 32$ in (ii) gives $y = \frac32$
(2) (ii), (iii) : Substituting $x,y = \frac 32$ in (iii) gives $-4*\frac32 +2*\frac32 +3 = -6 +6 = 0$
The significance of the tautology $0=0$ arrived at: it shows that the point $(\frac 32, \frac32) $ lies on the line $-4x+3y+3=0$.

The solution set is $x =y = \frac 32$, also plotted at : https://www.desmos.com/calculator/hlgnkgjio1.

The complex number $z = \frac32 + \frac32i$.

Answer 1

If you recognize that the problem is asking for the circumcenter of three points in the plane, then you can use the determinant of a $\,4 \times 4\,$ matrix to find the equation of the circle determined by the three points $\, (1,0),\,(2,0),\,(0,1).\,$ The determinant and the equation of the circle is

$$ \begin{vmatrix} x^2+y^2 & x & y & 1 \\ 1 & 1 & 0 & 1 \\ 4 & 2 & 0 & 1 \\ 1 & 0 & 1 & 1 \end{vmatrix} = x^2 + y^2 - 3x - 3y + 2 = (x-3/2)^2 + (y-3/2)^2 - 5/2 = 0. $$

The center point $\,z = 3/2 + 3/2 i \,$ and the radius $\, \sqrt{5/2} = |z-1|=|z-2|=|z-i|.\,$

The general problem of "Get the equation of a circle when given 3 points" is solved in several ways in MSE question 213658.

Answer 2

What the equation $$|z - 1| = |z - 2| = |z - i|$$ means in a geometric sense, is that $z$ is equidistant from $1$, $2$, and $i$. This suggests finding a positive real $r$ for which the circles with centers at $1$, $2$, and $i$ and radius $r$ intersect at a common point.

To this end, we observe that the circles with centers at $1$ and $2$ and common radius $r$ always have their points of intersection on the line $\Re(z) = 3/2$, or parametrized in rectangular coordinates, $z = 3/2 + yi$. Similarly, the circles with centers at $1$ and $i$, by symmetry, always have their points of intersection on the line $\Re(z) = \Im(z)$, or $z = x(1+i)$. Consequently, the intersection of these two lines must also be the common point of intersection of all three circles; namely, $z = x+yi = \frac{3}{2} + \frac{3}{2}i$, which characterizes the unique solution.

Of note is that the value of $r$ corresponding to this solution is $\sqrt{5/2}$.

Answer 3

You are looking for the points which are equi-distance from three points,$ i, 1, 2$$

The perpendicular bisectors of segments joining $i$ and $1$ is the line $y=x$ and the perpendicular bisector of $1$ and $2$ is $x= 3/2$

These two lines intersect at the point $z=3/2 +3/2 i$ which is the only solution to your equations.