1. ## describe 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

2. Originally Posted by transgalactic
|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. $\displaystyle |z-i|$ 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

3. i dont know how you got that image of this equation

is there a more practical wy to see it?

4. Originally Posted by transgalactic
i dont know how you got that image of this equation

is there a more practical wy to see it?

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

5. i only got
2y^2 -2y=2x

i dont know how you get y=-x

6. Originally Posted by transgalactic
i only got
2y^2 -2y=2x

i dont know how you get y=-x
Check your math: you don't have 2y^2 there but rather y^2 in both sides of the equation and thus it cancels out.
Once you've done this you get what I said.

Tonio

7. Originally Posted by transgalactic
i dont know how you got that image of this equation

is there a more practical wy to see it?
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 $\displaystyle z = z_2$).

8. i got now
y=-x too
is a straight line its not a circle as ou agreed with me before

9. Originally Posted by transgalactic
i got now
y=-x too
is a straight line its not a circle as ou agreed with me before

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

10. Originally Posted by 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
I presume transgalactic means that we all agreed that it was a straight line, ie not a circle.

11. ok thanks