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Complex analysis

  1. Jan 20, 2008 #1
    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

    where did I go wrong? I've been confused about this for a while.
     
  2. jcsd
  3. Jan 20, 2008 #2
    The points you found (correctly) are

    [tex]z=a\,(1-i)[/tex]

    thus if you write [tex]z=x+i\,y[/tex] you have

    [tex]x=a,y=-a \Rightarrow y=-x[/tex]

    Can you continue from here?
     
  4. Jan 20, 2008 #3
    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
     
  5. Jan 20, 2008 #4
    Write down the line passing through the points z=1 and z=-i.

    How is this line and [tex]y=-x[/tex] are related?
     
  6. Jan 21, 2008 #5
    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?
     
  7. Jan 21, 2008 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    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).
     
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