a Closed set in the Complex Field


by Bachelier
Tags: complex, field
Bachelier
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#1
Feb4-13, 12:14 AM
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
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pwsnafu
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#2
Feb4-13, 01:22 AM
Sci Advisor
P: 773
Hint: find the boundary first.
joeblow
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#3
Feb4-13, 07:26 AM
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.

HallsofIvy
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#4
Feb4-13, 07:43 AM
Math
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PF Gold
P: 38,877

a Closed set in the Complex Field


Quote Quote by Bachelier View Post
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 [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.
oay
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#5
Feb4-13, 08:37 AM
P: 220
Quote Quote by HallsofIvy View Post
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.
HallsofIvy
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#6
Feb4-13, 02:30 PM
Math
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Thanks
PF Gold
P: 38,877
Right, thanks. And thanks for trailing along behind me cleaning up my mess!


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