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Loci for complez numbers

  1. Jul 11, 2008 #1

    rock.freak667

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    1. The problem statement, all variables and given/known data
    If arg(z-1)=arg(z+1)=[itex]\frac{3 \pi}{4}[/itex], find the locus of z.


    2. Relevant equations



    3. The attempt at a solution

    arg(z-1)=arg(z-[1+0i]) => z lies on half the line through the point (1,0) excluding (1,0), inclined at alpha

    Similarly arg(z+1) =>z lies on half the line through (-1,0),excluding (-1,0).Inclined at beta.

    If I draw those two on the same graph, they form a triangle. But how do I incorporate the 3pi/4 ?

    or was I supposed to draw arg(z-1)=3pi/4?
     
  2. jcsd
  3. Jul 11, 2008 #2

    tiny-tim

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    Hi rock.freak667! :smile:
    That doesn't look right to me. :confused:

    Are you sure it isn't arg(z-1) - arg(z+1) = 3π/4 ?
     
  4. Jul 11, 2008 #3

    rock.freak667

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    Well it actually could be that because it's in someone's handwriting.

    But if it was arg(z-1) - arg(z+1) = 3π/4.

    How would I go about it?


    EDIT:

    I can sketch

    [tex]arg(z-z_0)= \lambda[/tex]


    where z_0 is a fixed complex number and lambda is the argument.
     
    Last edited: Jul 11, 2008
  5. Jul 11, 2008 #4

    tiny-tim

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    dot-product or just trigonometry? :smile:
     
  6. Jul 11, 2008 #5

    rock.freak667

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    ahh nevermind I figured it out.

    And the correct question is:

    [tex]arg(z-1)-arg(z+1)=\frac{\pi}{4}[/tex]
     
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