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Complex inequalities geometric interpretation Watch

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    If anyone knows how to do 9iv and v that would be great
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    If z is an arbitrary point in the complex plane, and a is some specific, fixed point, then |z-a| is a real number measuring the distance from a to z.

    So  |z-a| > |z-b| is true for those arbitrary points z which are closer to b than to a.

    To figure out what that looks like:

    a) draw two fixed points a and b on an Argand diagram
    b) figure out the set of points which are equidistant from both (maybe consider the points of intersection of two circles centred on a and b)
    c) having done that, decide which parts of the plane satisfy the inequality above (this should now be easy)
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    so for the first one I got if z=x+iy y>=0 and for the second one I got z=x+iy y<=0?

    I think I get it now, thanks
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    (Original post by CammieInfinity)
    so for the first one I got if z=x+iy y>=0 and for the second one I got z=x+iy y<=0?

    I think I get it now, thanks
    Well, I've done it by scribbling briefly on paper/ in my head, but yes, I think that those are right.

    You may want to express them in set notation though e.g A = \{ z \in \mathbb{C} : Im(z) \ge 0 \}
 
 
 
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