# Sketching regions in the complex plane

• Mar 11th 2013, 03:24 PM
redtdc
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• Mar 11th 2013, 03:34 PM
ILikeSerena
Re: Sketching regions in the complex plane
Quote:

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?
• Mar 11th 2013, 03:40 PM
Plato
Re: Sketching regions in the complex plane
Quote:

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 $\displaystyle |z-z_0|$ is the distance from $\displaystyle z\text{ to }z_0~?$

So what is the set $\displaystyle S=\{z:|z-3i|=|z-2|\}~?$
• Mar 11th 2013, 04:19 PM
HallsofIvy
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 $\displaystyle \sqrt{(x^2+ (y- 3))^2}= \sqrt{((x- 2)^2+ y^2}$ which is the same as $\displaystyle x^2+ (y- 3)^2= (x- 2)^2+ y^2$.

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