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# Complex inequalities geometric interpretation watch

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

So is true for those arbitrary points which are closer to than to .

To figure out what that looks like:

a) draw two fixed points and 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 and )
c) having done that, decide which parts of the plane satisfy the inequality above (this should now be easy)
3. 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
4. (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

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Updated: December 24, 2014
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