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A very interesting question on complex variable

  1. Oct 19, 2005 #1
    Hello, everybody!

    I'm a Maths/Physics student at Ecuador. Sorry if my English sucks, i'll try to do my best... Some fellow Physics buddy asked me if there was a way to fin a complex number that would be equal to its exponential... it is a very simple question to understand, but not to easy to prove or disprove (at least not for me, and i've tried it for a day).

    So, this is what I got:

    z = Ln (z)

    and we would have to solve these equations:

    ln r = rcos O
    O = rsen O

    that would be hard to solve algebraically, I guess... I wouldn't want a numeric aproach, so maybe there is a more ellegant way to find an example of disprove the hypothesis. (I already tried Taylor expansions)

    Well, please try to give some ideas on the subject.

    Have a good one!
  2. jcsd
  3. Oct 20, 2005 #2


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    Homework Helper

    It looks harmless enough, but it's not.

    It looks harmless enough, but it's not.

    The equation [tex]exp(z)=z[/tex] has a solution given by [tex]z=-W(-1)[/tex] and, if you put [tex]z=x+iy[/tex], then the solution is [tex]\left\{\begin{array}{cc}x=-iy-W(-1)\\y=y\end{array}\right.[/tex].

    Where [tex]W(\cdot)[/tex] is the Lambert W function; for an excellent reference, see

  4. Oct 20, 2005 #3

    Hmmm ... Matlab gave me this answer: W(-1) = -0.3181... + i 1.3372...

    Which doesn't seem to work. Maybe I made a calculation mistake (because I only typed it once on a computer at my school, and wrote the result), but it seems to make sense, since you can work z = exp (z) to

    z = exp (z)
    z*exp(-z) = 1
    (-z)*exp(-z) = -1

    Which indicates that W(-1) should be the value for Z (on the complex form of the Lambert-W function, which I assumed was W(z)*exp[W(z)] = F(z) )

    Anyway, this function has been quite a discovery for me! I'll see if I can find an answer, you can still give me some more ideas.
  5. Oct 25, 2005 #4

    Nevermind... just a silly mistake...

    Z = - W(-1) is the answer.
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