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A Closed set in the Complex Field

  1. Feb 4, 2013 #1
    This is elementary but surely this set is closed

    | c – i | ≥ | c | with c being in ℂ

    I am trying to picture the set. Is it outside the disc centered at (0,1) with radius equal to modulus c (whatever that is) ?

  2. jcsd
  3. Feb 4, 2013 #2


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    Hint: find the boundary first.
  4. Feb 4, 2013 #3
    Visualize the complex number as a point in the plane. Move it down one unit. (That's what subtracting i does.) Now, what points will be further from the origin when this happens? You can construct a right triangle of legs Re(c) and Im(c) and a right triangle of legs Re(c-i) and Im(c-i) and see which hypotenuse is longer.

    Once you have figured out the region in question, it should be easy to show its closure.
  5. Feb 4, 2013 #4


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    |c- i| is the distance from point c to i. |c| is the distance from c to 0. Saying that |c- i|= |c| (as pwsnafu suggested "find the boundary first") is the same as saying that those two distance are equal for all c. Geometrically, that is the perpendicular bisector of the segment from 0 to i, Im(c)= 1/2 or c= x+ (1/2)i for any real x. The set [itex]|c- i|\ge |c|[/itex] is set of points on ther side of that line closer to 0 than to i.

    Corrected thanks to oay.
    Last edited: Feb 4, 2013
  6. Feb 4, 2013 #5
    Not quite.

    You mean: c = x + (1/2)i for any real x.
  7. Feb 4, 2013 #6


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    Right, thanks. And thanks for trailing along behind me cleaning up my mess!
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