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Fixed points of e^z

  1. Jan 5, 2012 #1
    1. The problem statement, all variables and given/known data
    How would one go about finding the fixed points of e^z, where z is complex (i.e. all z s.t. e^z = z)?


    2. Relevant equations
    Nothing.

    3. The attempt at a solution
    I've considered all the relevant formulas (de Moivre's forumla, power series, z = re^i*theta, ...).

    For some reason, I'm just not getting the solution. I feel that this is going to be really obvious, and I'm not sure why I'm not getting it. Thanks.
     
  2. jcsd
  3. Jan 5, 2012 #2

    micromass

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    The equation [itex]e^z=z[/itex] cannot be solved using elementary functions. So what exactly is it that you want?? Do you want to prove that there are fixed points?? Do you want an expression of them?? (Use something similar to Lambert's W function) Do you want to find a numerical approximation??
     
  4. Jan 5, 2012 #3
    I would like to find an expression for the fixed points, if they exist. If they happen to not exist, then I would like a proof of that.

    It isn't really a homework question, the prof just asked us to think about finding the fixed points of e^x. The thing is though, the course doesn't assume we know any complex analysis, so I figured it could be solved using basic math (not even calculus).

    Thanks for your reply.

    Nigel
     
  5. Jan 5, 2012 #4

    Ray Vickson

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    Maple 11 gets the solution as x = -LambertW(-1) =~= 0.3181315052-I*1.337235052, where I = sqrt(-1). Here, LambertW(z) is the solution of f(z)*exp(f(z))=z which is analytic at z = 0.

    RGV
     
  6. Jan 5, 2012 #5

    micromass

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    It should be noted that -W(-1) is only one solution. Indeed, the Lambert W function is multivalued, so there are infinitely many values for -W(-1). Also note that the Lambert W function is not analytical at -1, since it has a branch cut there.
     
  7. Jan 5, 2012 #6

    Ray Vickson

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    The branch point is at z_c = -exp(-1), so z = -1 is, indeed, on (one side of) the branch cut.

    RGV
     
  8. Jan 5, 2012 #7

    BruceW

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    Then you should assume x is real. Think about the graph of the real functions.
     
  9. Jan 5, 2012 #8
    I really think we're missing the point with this branch-cut thing: It's arbitrary. That means I can move it and the function becomes perfectally analytic at z=-1. In fact, I would argue the function is everywhere analytic except at the branch-point [itex]z=-1/e[/itex].

    Also, if I may be allowed to be complete, the OP specifically stated complex z so no real for me and to find an explicit expression for z, we get it into suitable Lambert-w form:

    [tex]z=e^z[/tex]

    [tex]1=ze^{-z}[/tex]

    [tex]-1=-ze^{-z}[/tex]

    at that point, take the W function of both sides:

    [tex]-z=W(-1)[/tex]

    then:

    [tex]z=-W(-1)[/tex]

    and keep in mind the function [itex]e^z-z[/itex] is a non-polynomial entire function which by Picard, reaches all values with at most one exception, infinitely often so that we would expect the expression [itex]e^z=z[/itex] to have an infinite number of solutions.
     
    Last edited: Jan 5, 2012
  10. Jan 6, 2012 #9
    I did a little exploration of this interesting problem... the first few solutions found are

    0.318131505 ± i * 1.337235701
    2.06227773 ± i * 7.588631178
    2.653191974 ± i * 13.94920833
    3.020239708 ± i * 20.27245764
    3.287768612 ± i * 26.5804715
    3.498515212 ± i * 32.88072148
    3.672450069 ± i * 39.17644002
    3.820554308 ± i * 45.4692654

    ... the imaginary portion increasing by approx [itex]2\pi[/itex] each time and |z| = eRe(z).
     
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