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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
    z^{2} - z = w
    w.r.t. [itex]z[/itex]. You get the multiple valued function:
    z = \frac{1 + (1 + 4 w)^{\frac{1}{2}}}{2} = \left(w + \frac{1}{2}\right)^{\frac{1}{2}} + \frac{1}{2}
    This can be expressed as sequence of two simple mappings:
    u = w + \frac{1}{2}
    v = u^{\frac{1}{2}}
    z = v + \frac{1}{2}

    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?
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