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

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

$$r_1$$=$$|z-2|=|x+iy-2|$$ 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.

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.

• Can I directly graph the points or would I need some calculations beforehand? – Chris Sehic Jan 10 '19 at 15:05
• @ChrisSehic This is technically up to whoever is correcting your exercises. That being said, if I were that person, just drawing the set directly after an explanation like that would be enough. – Arthur Jan 10 '19 at 15:16

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).