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About LinkBacks|z-i|=|z+1|
i know that |z-i| is a circle
shifted by i
|z-1| means center shifted by 1
but its not telling the full picture
You have the right idea.Originally Posted by transgalactic
and
describe a sec of points an equal distance away from
and
respectively, ie a circle.
With questions like this is can really help to draw a diagram.
If you have the equation, this is describing a set of points an equal distance from
Hope this helps
Craig
More practical than what Craig said?? I doubt it: it is simple geometry!
Draw your xy-plane (Argand plane) and mark the points i (i.e., the point(0,1)) and -1 (i.e., (-1,0)). You want all the complex numbers which are equidistant from both this points = you want all the points which are equidistant from the extreme points of the line segment determined by the points (0,1) and (-1,0) = this is EXACTLY the definition, or if you want the main property, of the perpendicular bisector of the line segment!
Another way, more algebraic and less geometrical, to check this is to put z = x + iy ==> |z - i| = |z + 1| <==> |z - i|^2 = |z + 1|^2 <==>
|x + i(y-1)|^2 = |(x+1) + iy|^2 <==> x^2 + (y-1)^2 = (x+1)^2 + y^2
Now expand, cancellate and have some little basic mathematical fun and get y = -x, which is exactly the perp. bisector of...etc.
Tonio
I don't know whom you're addressing now: nobody, as far as I can tell, "agreed" with you the given equation is a circle. Both sides separatedly, that is true, describe points on circles, as Craig wrote, but he and I described you the solution as a line.
Tonio
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