1. Limited time only! Sign up for a free 30min personal 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!

Sketching complex sets

  1. Sep 12, 2011 #1
    1. The problem statement, all variables and given/known data

    I'm having some major trouble this these two questions.

    Sketch the set, s, where s = {z| | z^2 - 1 | < 1 } ... z is a complex number
    Sketch the set, s, where s = {Z| | Z | > 2 | Z - 1 | } ... Z is a complex number

    2. The attempt at a solution

    This is supposed to be done by hand apparently and I just cannot see the insight to what makes these simpler. I've tried substituting z = x + i y but then I just get bogged down by calculation. For the first one I get this: sqrt ( (x^2 - y^2 - 1)^2 + (2xy)^2 ) < 1. How am I supposed to graph that? I would like to take somehow take cases like if they were real numbers in relations but since these are complex numbers I'm sure I cannot do that.

    3. Relevant equations

    The only things he talked about were circles/discs. If it wasn't for that square in the first question or that inequality in the second one I would know what to do. Right now I'm very lost and I've looking at textbook examples that seem similar - I can't find any.

    I know my prof likes all these intuitive geometric properties but I just cannot see it. If it was simpler things like s = {z| | z - 1 | < 1 } then I know what to do. But, with this square I'm sure things get a lot more different.
     
  2. jcsd
  3. Sep 12, 2011 #2

    dynamicsolo

    User Avatar
    Homework Helper

    If you multiply out the expression under the radical, you'll see that it can be simplified a bit. But I'll let someone else suggest how to proceed from there.

    On the second one, keep in mind that | z | is the "length" of a vector from the origin in the Argand diagram to the point representing z . So | z - 1 | is the "length" of a vector from the point x = 1 (or ( 1, 0 ) ) to the same point for z . What sort of curve then satisfies | z | = 2 | z - 1 | ? That defines the boundary for the region that is described by the inequality. So where are the points for which | z | > 2 | z - 1 | ? (That is, more than twice as far from the origin than from ( 1, 0 ). )
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Sketching complex sets
  1. Sketch level sets (Replies: 4)

  2. Sketch Complex Regions (Replies: 8)

Loading...