# Thread: Sketching regions in the complex plane

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2. ## Re: Sketching regions in the complex plane

Originally Posted by redtdc
How would I sketch the region S={z in the complex plane: abs(z-3i)=abs(z-2)} ?

And is S open or closed?

Thanks!
Hi redtdc!

abs(z-3i) is the same as the distance in R2 of a point (x,y) to (0,3).

The region S corresponds to the points in R2 that have the same distance to (0,3) as to (2,0).
What kind of region is that?

3. ## Re: Sketching regions in the complex plane

Originally Posted by redtdc
How would I sketch the region S={z in the complex plane: abs(z-3i)=abs(z-2)} ?
And is S open or closed?

Do you understand that $|z-z_0|$ is the distance from $z\text{ to }z_0~?$

So what is the set $S=\{z:|z-3i|=|z-2|\}~?$

4. ## Re: Sketching regions in the complex plane

|z- 3i| is the distance from z to 3i and |x- 2| is the distance from z to 2 so S consists of points that are equally distant from 3i and 2. Geometrically, the set of all points equally distant from point P and Q is the perpendicular bisector of the segment PQ.

Algebraically, taking z= x+ iy, |z- 3i|= |z- 2| is the same as $\sqrt{(x^2+ (y- 3))^2}= \sqrt{((x- 2)^2+ y^2}$ which is the same as $x^2+ (y- 3)^2= (x- 2)^2+ y^2$.

As for whether it is open or close, what topology are you using?