1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Locus problem (complex numbers)

  1. Nov 18, 2009 #1
    Hi! We started doing complex numbers in maths class a couple of weeks ago, and I'm not fully understanding sketching the locus of points.

    1. The problem statement, all variables and given/known data

    Sketch the locus of z:

    [tex]arg\left(\frac{z-2}{z+2}\right) = \frac{\pi}{3}[/tex]

    3. The attempt at a solution

    I've rewritten as

    [tex]arg(z-2)-arg(z+2)[/tex]

    and have constructed a triangle with corners Re(-2), Re(2) and z.

    I understand the angle subtended by the real axis must equal [tex]\frac{\pi}{3}[/tex]. I do not understand the solution given by our maths teacher; that the locus of z is a kind of truncated circle above the real axis, with nothing to draw below.
     
  2. jcsd
  3. Nov 18, 2009 #2

    lanedance

    User Avatar
    Homework Helper

    so you have
    [tex]arg(z-2)-arg(z+2) = \pi/3[/tex]

    i think the triangle you shold be drawing (above the real axis) is that defined by the points in the complex plane
    z = 0 (the origin)
    z-2
    z+2
    now as z=(a+ib),it is determined by 2 unknowns,

    if you assume one, as you know the lengths of the three sides upto one unknown, (one is always 4, and the other 2 depend only z) and an angle (pi/3), then in theory you can solve for the reamaining part of z. (it could be easier to think of z in terms of a magnitude & angle)

    imagine when z is purely complex, its splits the larger triangle into two right triangles and the length of z will be given by
    [tex] tan(\pi/6)= \frac{2}{(|z|)} [/tex]

    now if you look below the real axis, the argument of (z-2) will always be less than that of (z+2) so there is no solution there
     
  4. Nov 19, 2009 #3
    Cheers lanedance. I chatted to my maths teacher today about it, and his explanation correlated nicely to yours. I'm heading towards understanding these problems a little better.
     
  5. Nov 19, 2009 #4

    lanedance

    User Avatar
    Homework Helper

    no worries, the best way to learn is to get in and have go as you're doing
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Locus problem (complex numbers)
  1. Complex locus (Replies: 29)

  2. Complex locus (Replies: 17)

Loading...