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Complex analysis, double exponential. I don't understand the question being asked.

  1. Nov 2, 2011 #1
    1. The problem statement, all variables and given/known data
    "For what values of z does [tex]e^{e^x} = 1[/tex] ? If [itex]z_m[/itex] and [itex]z_n[/itex] range over distinct roots of this equation, is the set of distances [itex]d(z_m, z_n)[/itex] bounded away from zero?"


    3. The attempt at a solution
    This equation doesn't have any solutions, does it? ew = 1 only when w = 0. w in this case is ez, which is never equal to zero. I'm incredibly confused about the second part of the question... I assume [itex]z_m[/itex] and [itex]z_n[/itex] are sequences, but that's all I've got.

    At this point, I'm more interested in actually figuring out what the heck the question is asking me than I am in finding the answer. Can anyone help?
     
  2. jcsd
  3. Nov 2, 2011 #2
    Re: Complex analysis, double exponential. I don't understand the question being asked

    No mate, there are more cases where [itex]e^z = 1[/itex]. Take for example
    [tex] e^{i2\pi}=1[/tex], so [itex] z=i2\pi[/itex] is also a solution.

    Cheers
     
    Last edited: Nov 2, 2011
  4. Nov 2, 2011 #3
    Re: Complex analysis, double exponential. I don't understand the question being asked

    If you're going to succeed in Complex Analysis, you have got to start thinking "multivalued" all the time like when you see:

    [tex]\sqrt{z}[/tex]

    or:

    [tex]\log(z)[/tex]

    Now:

    [tex]\log(z)=\ln|z|+i(\theta+2n\pi)[/tex]

    That's infinitely-valued right?

    So nothing wrong with takin' logs of both sides of your expression as long as you take the infinitely-valued log of both sides:

    [tex]e^z=\log(1)[/tex]

    [tex]e^z=\ln|1|+i(0+2n\pi)[/tex]
    [tex]e^z=2n\pi i[/tex]
    Now one more time:

    [tex]z=\ln(2n\pi)+i(\pi/2+2k\pi)[/tex]

    That's now doubly-infinitely valued except for the purist in here.

    Just take one infinite set for now:

    [tex]z=\ln(2\pi)+i(\pi/2+2k\pi)[/tex]

    Aren't those along a vertical line with x-cood=ln 2pi and y-coord pi/2+2kpi? How about the other ones? Those are along vertical lines too and extend to infinity. Now take the distance between arbitrary points in that set? Is that distance bounded?
     
    Last edited: Nov 2, 2011
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