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Math Help - Sketching regions in the complex plane

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    Last edited by redtdc; March 11th 2013 at 05:37 PM.
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    Super Member ILikeSerena's Avatar
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    Re: Sketching regions in the complex plane

    Quote Originally Posted by redtdc View Post
    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?
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  3. #3
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    Re: Sketching regions in the complex plane

    Quote Originally Posted by redtdc View Post
    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|\}~?
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  4. #4
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    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?
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