|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

Printable View

- Oct 13th 2009, 05:26 AMtransgalacticdescribe this equation..
|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 - Oct 13th 2009, 10:52 AMcraig
You have the right idea. $\displaystyle |z-i|$ and $\displaystyle |z+1|$ describe a sec of points an equal distance away from $\displaystyle i$ and $\displaystyle -1$ respectively, ie a circle.

With questions like this is can really help to draw a diagram.

If you have the equation $\displaystyle |z-i|=|z+1|$, this is describing a set of points an equal distance from $\displaystyle i$*and*$\displaystyle -1$. If you draw a diagram, you will see that the set of points that settle this equation is the perpendicular bisector of the centre of the line, passing through the points $\displaystyle i$ and $\displaystyle -1$.

Hope this helps

Craig - Oct 14th 2009, 01:54 AMtransgalactic
i dont know how you got that image of this equation

is there a more practical wy to see it? - Oct 14th 2009, 02:39 AMtonio

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 - Oct 14th 2009, 03:02 AMtransgalactic
i only got

2y^2 -2y=2x

i dont know how you get y=-x - Oct 14th 2009, 03:40 AMtonio
- Oct 14th 2009, 03:51 AMmr fantastic
If you do not understand the simple geometric interpretation of the locus defined by $\displaystyle |z - z_1| = |z - z_2|$ then you better go back and thoroughly review this material.

(It's the perpendicular bisector of the line segment joining $\displaystyle z = z_1$ and $\displaystyle z = z_2$). - Oct 14th 2009, 04:09 AMtransgalactic
i got now

y=-x too

is a straight line its not a circle as ou agreed with me before - Oct 14th 2009, 05:05 AMtonio

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 - Oct 14th 2009, 05:20 AMcraig
- Oct 14th 2009, 05:56 AMtransgalactic
ok thanks :)