Put slightly differently...
Let
![](https://latex.codecogs.com/png.latex?\bg_white A)
and
![](https://latex.codecogs.com/png.latex?\bg_white B)
be the points such that
![](https://latex.codecogs.com/png.latex?\bg_white \overrightarrow{OA} = z_1)
and
![](https://latex.codecogs.com/png.latex?\bg_white \overrightarrow{OB} = z_2)
where
![](https://latex.codecogs.com/png.latex?\bg_white O)
is the origin.
The locus
![](https://latex.codecogs.com/png.latex?\bg_white \left|z - z_1\right| = \left|z - z_2\right|)
is the set of points
![](https://latex.codecogs.com/png.latex?\bg_white P)
such that the distance from
![](https://latex.codecogs.com/png.latex?\bg_white A)
to
![](https://latex.codecogs.com/png.latex?\bg_white P)
and the distance from
![](https://latex.codecogs.com/png.latex?\bg_white B)
to
![](https://latex.codecogs.com/png.latex?\bg_white P)
are equal.
The distance from
![](https://latex.codecogs.com/png.latex?\bg_white A)
to
![](https://latex.codecogs.com/png.latex?\bg_white P)
is
![](https://latex.codecogs.com/png.latex?\bg_white \left|\overrightarrow{AP}\right| = \left|z - z_1\right|)
and similarly
![](https://latex.codecogs.com/png.latex?\bg_white \left|\overrightarrow{BP}\right| = \left|z - z_2\right|)
.
It is clear that
![](https://latex.codecogs.com/png.latex?\bg_white M)
, the midpoint of
![](https://latex.codecogs.com/png.latex?\bg_white AB)
, must be one such point on the locus. The diagram provided makes it clear that the origin is another point on the locus.
Now, for any point
![](https://latex.codecogs.com/png.latex?\bg_white P)
other than
![](https://latex.codecogs.com/png.latex?\bg_white M)
on the perpendicular bisector of
![](https://latex.codecogs.com/png.latex?\bg_white AB)
,
![](https://latex.codecogs.com/png.latex?\bg_white P)
creates a pair of congruent triangles,
![](https://latex.codecogs.com/png.latex?\bg_white \Delta PMA \equiv \Delta PMB\ \text{(SAS)})
which proves that
![](https://latex.codecogs.com/png.latex?\bg_white PA = PB)
and
![](https://latex.codecogs.com/png.latex?\bg_white P)
lies on the locus.
For any point
![](https://latex.codecogs.com/png.latex?\bg_white Q)
that is not on the perpendicular bisector of
![](https://latex.codecogs.com/png.latex?\bg_white AB)
, we can prove that either
![](https://latex.codecogs.com/png.latex?\bg_white Q)
is closer to
![](https://latex.codecogs.com/png.latex?\bg_white A)
than to
![](https://latex.codecogs.com/png.latex?\bg_white B)
, or vice versa - that
![](https://latex.codecogs.com/png.latex?\bg_white Q)
is closer to
![](https://latex.codecogs.com/png.latex?\bg_white B)
than to
![](https://latex.codecogs.com/png.latex?\bg_white A)
. In either case,
![](https://latex.codecogs.com/png.latex?\bg_white Q)
does not belong to the locus and so the locus is all points lying on the perpendicular bisector of
![](https://latex.codecogs.com/png.latex?\bg_white AB)
.
This result is easy to establish algebraically, by taking
![](https://latex.codecogs.com/png.latex?\bg_white z_1 = a + ib,\ a, b \in \mathbb{R}^+)
and using
![](https://latex.codecogs.com/png.latex?\bg_white z_2 = iz_1)
.