# Complex analysis

1. Jan 20, 2008

### indigojoker

I am to find all plints z in the complext plane that satisfies |z-1|=|z+i|

The work follows:
let z=a+bi
|a+bi-1|=|a+bi+i|
(a-1)^2+b^2=a^2+(b+1)^2
a^2-2a+1+b^2=a^2+b^2+2b+1
-a=b

the correct answer should be a perpendicular bisector of segments joining z=1 and z=-i

my result looks more like a perpendicular bisector of segments joking a=0 and b=0

2. Jan 20, 2008

### Rainbow Child

The points you found (correctly) are

$$z=a\,(1-i)$$

thus if you write $$z=x+i\,y$$ you have

$$x=a,y=-a \Rightarrow y=-x$$

Can you continue from here?

3. Jan 20, 2008

### indigojoker

so that is what i have, y=-x, or what i have is -a=b

what i'm asking is why does my result look like a perpendicular bisector of segments joining a=0 and b=0 instead of a perpendicular bisector of segments joining z=1 and z=-i

4. Jan 20, 2008

### Rainbow Child

Write down the line passing through the points z=1 and z=-i.

How is this line and $$y=-x$$ are related?

5. Jan 21, 2008

### indigojoker

i see, how come y=x-1 isnt a solution as well? inst y=x-1 a perpendicular bisector of segments joining z=1 and z=-i?

6. Jan 21, 2008

### HallsofIvy

Staff Emeritus
A line segment doesn't have two perpendicuar bisectors! The segment between 1 and i has midpoint (1+ i)/2 ((1/2, 1/2) in the xy-plane). The line y= x- 1 passes through (1/2, 1/2- 1)= (1/2, -1/2), not (1/2, 1/2).