# A Closed set in the Complex Field

by Bachelier
Tags: complex, field
 P: 376 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) ? Thanks
 Sci Advisor P: 838 Hint: find the boundary first.
 P: 71 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.
Math
Emeritus
Thanks
PF Gold
P: 39,682
A Closed set in the Complex Field

 Quote by Bachelier 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) ? Thanks
|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 $|c- i|\ge |c|$ is set of points on ther side of that line closer to 0 than to i.

Corrected thanks to oay.
P: 235
 Quote by HallsofIvy Geometrically, that is the perpendicular bisector of the segment from 0 to i, Im(c)= 1/2 or c= (x/2)i for any real x.
Not quite.

You mean: c = x + (1/2)i for any real x.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,682 Right, thanks. And thanks for trailing along behind me cleaning up my mess!

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