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!

Complex analysis, deceptively tricky problem.

  1. Jun 15, 2011 #1
    1. Let z be a complex variable. Describe the set of all z satisfying |z^2-z|<1.[\b]


    I have a `brute force' solution, but it's really messy. Without a graphing utility, it would be nearly impossible to graph.

    I just computed |z^2-z| in terms of x and y, and solved |z^2-z|=1 in this setting. The points inside this curve, then, satisfy the inequality.

    It seems like I'm missing a more elegant solution, but I can't see it.

    I thought I might be onto something when I did a change of variables and put the expression into the form |u^2-1/4|<1, but that didn't seem to help much.
     
  2. jcsd
  3. Jun 15, 2011 #2
    Solve:
    [tex]
    z^{2} - z = w
    [/tex]
    w.r.t. [itex]z[/itex]. You get the multiple valued function:
    [tex]
    z = \frac{1 + (1 + 4 w)^{\frac{1}{2}}}{2} = \left(w + \frac{1}{2}\right)^{\frac{1}{2}} + \frac{1}{2}
    [/tex]
    This can be expressed as sequence of two simple mappings:
    [tex]
    u = w + \frac{1}{2}
    [/tex]
    [tex]
    v = u^{\frac{1}{2}}
    [/tex]
    [tex]
    z = v + \frac{1}{2}
    [/tex]

    1. What region in the w-plane doese [itex]|w| < 1[/itex] depict?
    2. What is the mapping from the w to the u plane?
    3. What is the mapping from the u plane to the v plane? Don't forget that [itex]z^{1/2}[/itex] is a double valued function.
    4. What is the mapping from the v plane to the z plane?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Complex analysis, deceptively tricky problem.
Loading...