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An interesting equation!

  1. Jul 12, 2010 #1
    Long back (not so much long back), when I was first introduced to complex numbers, my mind was flooded with strange and interesting curiosities (as they appeared to me!), rather ideas. Of course this realm of mathematics is, as the mathematicians are fond of saying, is remarkably beautiful. As I studied these (complexes!) further and their applications, a strange feeling of asymmetry and limitations developed (Which I will describe no further). But, before that some interesting analogies (which I now suspect to be incorrect) forced me look for some insolvable equations. That time thought of the following equation
    [tex]z^{z}+1=0[/tex]
    and happily thought that it was unsolvable.
    My thoughts were shattered by mathematica (which solved it, but never gave the procedure)
    Can anyone help (rather mathematica in hurting me further!!!) by giving some hints for solving this equation.
     
  2. jcsd
  3. Jul 12, 2010 #2
    Ever considered how a complex number could be represented using trig identities?
     
  4. Jul 12, 2010 #3
    Hmm, can someone remind me which equations are not solvable with complex numbers? (apart from division by zero)
     
  5. Jul 12, 2010 #4

    Mute

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    The solution will use the Lambert W function, defined by

    [tex]W(z)e^{W(z)} = z[/tex]

    The lambert W function has infinitely many complex branches, denoted by [itex]W_k(z)[/itex].

    Manipulating [itex]z^z = -1[/itex] to look like that will give the solution. (Hint: take a logarithm of both sides, then use a suitable change of variables for z)
     
  6. Jul 12, 2010 #5

    disregardthat

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    Mute, you are making it much more difficult than it has to be.

    Gerenuk, set [tex]z = re^{i\theta}[/tex] (polar form).
     
  7. Jul 12, 2010 #6

    Gib Z

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    I don't think that helps. I haven't carried out the working, but being able to solve that equation in the Complex numbers would imply we could solve it for real solutions using that method as well, and we know that for the reals, that equation does require the Lambert W function.
     
  8. Jul 12, 2010 #7

    Mute

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    I would like to see how it could be done more easily, then.
     
  9. Jul 12, 2010 #8
    The Lambert W function in my opinion is a perfectly suitable algebraic transformation to use that is no different than taking the sine of both sides. The algebra is one thing, but it's the geometry that is the really beautiful part of this expression. Most people don't get to see that geometry because it's tough to visualize and the composite nature of the solution makes it even more difficult because the geometry is bifurcating. So do what Mute said to do. I get something like:

    [tex]z\log(z)=\log(-1)[/tex]

    [tex]\vdots[/tex]

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

    Then take the Lambert W of both sides and rearrange to get:

    [tex]z=\frac{log(-1)}{W[\log(-1)]}[/tex]

    Note that's not what Mathematica reports but rather only the principal branch of [itex]\log(-1)=\pi i[/itex].

    Now, both the log and the W function are infinitely-valued, so for every branch of [itex]log(z)=ln|z|+i(\theta+2k\pi)[/itex], the W function has an infinite number of values. But that's only the algebra. The actual function:

    [tex]z=\frac{log(w)}{W[\log(w)]}[/tex]

    has real and imaginary surfaces that look like twisted funnels that fold into themselves infinitely often with each fold bifurcating into an infinite number of "sub-folds" with the solution to this problem then lining up along a vertical line above the point w=-1 where the line intersects each surface of the folds.

    That geometry is the true beauty of this problem in my opinion.
     
  10. Jul 12, 2010 #9

    Mute

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    We should also note that this is equivalent to

    [tex]z = \exp\left[W(\log(-1))\right].[/tex]
    (this is the form I wrote it as, which my hint pertains to).
     
  11. Jul 12, 2010 #10

    disregardthat

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    My fault, I misread the equation.
     
  12. Jul 12, 2010 #11
    Although the above methods are more elegant and thorough, if you're just looking for any solution, consider this: what is (-1)^(-1)?
     
  13. Jul 12, 2010 #12
    why it never occurred to me?

    Well, having seen all this I must look for other equations (which can elude complex numbers for solution) to satisfy my quest.
    One of my friend suggested
    [tex]|z|+1=0[/tex]
    but it doesn't strike as fertile.
     
  14. Jul 13, 2010 #13

    Gib Z

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    Since you haven't been working with complex numbers for too long, you should probably just take it from us that the vast vast majority of equations you can think of will have solutions in C. This is because of some powerful theorems such as the Fundamental Theorem of Algebra that states that any non constant polynomial with complex coefficients has a solution in C.

    There's another very strong theorem from complex analysis that states that if an entire function (complex differentiable over the entire complex plane) is non-constant, then it will pass through every complex value, except perhaps excluding only one.

    Another thing you should note is that R is a subset of C, so if the equation has solutions in R, it does in C as well.

    However, there will be some functions of a complex variable that will have no solution, and you find some by looking for functions from C to R, not from C to C. Then you just have to think of functions that have no solutions in R, such as |z| +1 =0 as your friend suggested. If you were looking for an equation with no solution in R, that does "strike as fertile".
     
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