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Arg(z) on planes

  1. Sep 1, 2009 #1
    1. The problem statement, all variables and given/known data

    Suppose that |z|<1 and |z-1| < 1. Find arg(z)

    2. Relevant equations

    r = |z| = sqrt(x^2 + y^2)

    x = r cos (angl) y = r sin (angl)

    3. The attempt at a solution

    I sketched the two planes and the possible values of z lie in the intersection between the unit disc |z| < 1 and the disk shifted one to the right |z - 1| < 1

    therefore arg(z) = angle + 2kpi (except the angle can't equal +/- pi/2, +/- 3pi/2, +- pi, +- 2pi)

    is this right or do I have to compute something?
     
  2. jcsd
  3. Sep 2, 2009 #2

    HallsofIvy

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    What two planes are you talking about? This all takes place in the complex plane. |z|< 1 is the interior of the circle with center 0 and radius 1. |z- 1|< 1 is the interior of the circle with center 1 and radius 1. Complex numbers satisfying both lie in the intersection of those twodisks. In particular, the points where the circle intersect lie on two equilateral triangles so the vertices make angles of 60 degrees= [itex]\pi/3[/itex] radians with the x-axis. Arg(z) lies between [itex]-\pi/3[/itex] and [itex]\pi/3[/itex].
     
  4. Sep 2, 2009 #3
    My mistake. I meant to refer to the disks as planar sets instead of "planes". Anyway, i see how u get the angles now. My problem was when I drew the verticies, I only drew them from the origin through the intersection points and was confused as to how to incorporate the leftower area after +/- pi/3 radians. But then I connected it from 1 to the intersections to get the triangles you mentioned and it makes sense now.
     
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