If $z = x + iy$ is a complex number, how do I sketch the set of points that will satisfy the following: $|z − 2i| = |z − 2|$?

Question

How would one proceed to graph this type of complex equation? Is there a general way of proceeding?

Answer 1

$r_1$=$|z-2|=|x+iy-2| $ implies $r^2_1$=$(x-2)^2+y^2$
Sly,
$r_2$=$|z-2i|=|x+iy-2i|$ implies $r^2_2$=$x^2+(y-2)^2$
Thus you can solve $(x-2)^2+y^2$=$x^2+(y-2)^2$ to get
x=y.
Any edit is welcome.

Answer 2

Given two complex numbers $z$ and $w$, $|z-w|$ is the distance in the complex plane from $z$ to $w$.

With that in mind, a complex number $z$ satisfies your equation iff it is as far from the complex number $2i$ as it is from the complex number $2$. The set of all such points in the plane is a relatively easy thing to draw.

Answer 3

In the plane, the locus of points equidistant to two given points $A,B$ is the line bisector of the segment $AB.$

In the present case, it is the bisector of the segment with ends at $2i$ and $2$ (in the complex plane).