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

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$r_1$=$|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.

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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.

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  • $\begingroup$ Can I directly graph the points or would I need some calculations beforehand? $\endgroup$ – Chris Sehic Jan 10 '19 at 15:05
  • $\begingroup$ @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. $\endgroup$ – Arthur Jan 10 '19 at 15:16
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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).

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