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Math Help - solve the equation

  1. #1
    Super Member Random Variable's Avatar
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    solve the equation

    Solve for z.

     i^{z} = z where  z \in \mathbb{C}
    Last edited by Random Variable; August 19th 2009 at 08:41 AM.
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  2. #2
    Super Member Random Variable's Avatar
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    I'm not sure if a hint is needed, or if this problem is not challenging at all. I'm leaning towards the latter.
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  3. #3
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    Quote Originally Posted by Random Variable View Post
    I'm not sure if a hint is needed, or if this problem is not challenging at all. I'm leaning towards the latter.
    The complex power of a complex number is multi-valued. So there could be an infinite number of solutions. Today is too late, we can discuss this sometime later. ;-)
    Last edited by mr fantastic; September 18th 2009 at 08:53 AM. Reason: Restored original reply
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  4. #4
    Super Member Random Variable's Avatar
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    The complex power of a complex number is multi-valued. So there could be an infinite number of solutions.
    True. It depends which branch you pick.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by Random Variable View Post
    Solve for z.

     i^{z} = z where  z \in \mathbb{C}
    Has closed form solution in terms of Lambert's W

    CB
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  6. #6
    Super Member Random Variable's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Has closed form solution in terms of Lambert's W

    CB
    I know.

    Spoiler:


     i^{z} = z

     i^{-z} = \frac{1}{z}

     z i^{-z} = 1

     -z i^{-z} = -1

     -ze^{-z \ln i} = -1

     -z \ln i \ e^{-z \ln i} = - \ln i

     W(-z \ln i e^{-z \ln i}) = W (- \ln i)

     -z \ln i = W(-\ln i)

     z = \frac{W(-\ln i)}{-\ln i}

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  7. #7
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    Quote Originally Posted by Random Variable View Post
    I know.

    Spoiler:


     i^{z} = z

     i^{-z} = \frac{1}{z}

     z i^{-z} = 1

     -z i^{-z} = -1

     -ze^{-z \ln i} = -1

     -z \ln i \ e^{-z \ln i} = - \ln i

     W(-z \ln i e^{-z \ln i}) = W (- \ln i)

     -z \ln i = W(-\ln i)

     z = \frac{W(-\ln i)}{-\ln i}

    Horrible, try putting i=e^{\pi i/2}

    (the term horrible here is not intended to be insulting, but is a term often used to describe an aesthetically unpleasing piece of mathematics)

    CB
    Last edited by CaptainBlack; August 25th 2009 at 02:41 AM.
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  8. #8
    Super Member Random Variable's Avatar
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    OK. Do this instead. Let a be any constant (real or complex).


     a^{z} = z

     a^{-z} = \frac{1}{z}

     z a^{-z} = 1

     -z a^{-z} = -1

     -ze^{-z \ \text{Log a}} = -1

     -z \ (\text{Log a}) \ e^{-z \ \text{Log a}}= - \text{Log a}

     W(-z \ (\text{Log a}) \ e^{-z \ \text{Log a}}) = W (- \text{Log a})

     -z \text{Log a} = W(-\text{Log a})

     z = \frac{W(-\text{Log a})}{-\text{Log a}}


     \text{Log i} = \ln|i| + i \frac{\pi}{2} = i\frac{\pi}{2}

     \frac {W(-i \frac{\pi}{2})}{-i\frac{\pi}{2}} \approx 0.4382829366+i0.3605924718


    Is that satisfactory?
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  9. #9
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    Quote Originally Posted by Random Variable View Post
    OK. Do this instead. Let a be any constant (real or complex).


     a^{z} = z

     a^{-z} = \frac{1}{z}

     z a^{-z} = 1

     -z a^{-z} = -1

     -ze^{-z \ \text{Log a}} = -1

     -z \ (\text{Log a}) \ e^{-z \ \text{Log a}}= - \text{Log a}

     W(-z \ (\text{Log a}) \ e^{-z \ \text{Log a}}) = W (- \text{Log a})

     -z \text{Log a} = W(-\text{Log a})

     z = \frac{W(-\text{Log a})}{-\text{Log a}}


     \text{Log i} = \ln|i| + i \frac{\pi}{2} = i\frac{\pi}{2}

     \frac {W(-i \frac{\pi}{2})}{-i\frac{\pi}{2}} \approx 0.4382829366+i0.3605924718


    Is that satisfactory?
    This looks good to me!
    Last edited by mr fantastic; September 18th 2009 at 08:54 AM. Reason: Restored original reply
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  10. #10
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    But what is  W(z) ?
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  11. #11
    Super Member Random Variable's Avatar
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    Quote Originally Posted by simplependulum View Post
    But what is  W(z) ?
    Lambert W-Function

    It's the function W(z) that solves z = W(z)e^{W(z)} for a given value of z.
    Last edited by Random Variable; August 21st 2009 at 08:57 PM.
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  12. #12
    Grand Panjandrum
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    Quote Originally Posted by Random Variable View Post
    OK. Do this instead. Let a be any constant (real or complex).


     a^{z} = z

     a^{-z} = \frac{1}{z}

     z a^{-z} = 1

     -z a^{-z} = -1

     -ze^{-z \ \text{Log a}} = -1

     -z \ (\text{Log a}) \ e^{-z \ \text{Log a}}= - \text{Log a}

     W(-z \ (\text{Log a}) \ e^{-z \ \text{Log a}}) = W (- \text{Log a})

     -z \text{Log a} = W(-\text{Log a})

     z = \frac{W(-\text{Log a})}{-\text{Log a}}


     \text{Log i} = \ln|i| + i \frac{\pi}{2} = i\frac{\pi}{2}

     \frac {W(-i \frac{\pi}{2})}{-i\frac{\pi}{2}} \approx 0.4382829366+i0.3605924718


    Is that satisfactory?
    It is better in that you eliminate the \log 's in the final answer but I would still prefer a solution that did not have to worry about which branch of the \log was in use where. That can be done by the the first step that I proposed (which, while that is not the way I think of it, absorbs the taking of logarithms while effectively fixing the branch).

    Also as W is multi-valued the right hand side of final line is just one of the solutions that correspond to the left hand side (as you know, and I know you know).

    CB
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  13. #13
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    Quote Originally Posted by CaptainBlack View Post
    It is better in that you eliminate the \log 's in the final answer but I would still prefer a solution that did not have to worry about which branch of the \log was in use where. That can be done by the the first step that I proposed (which, while that is not the way I think of it, absorbs the taking of logarithms while effectively fixing the branch).

    Also as W is multi-valued the right hand side of final line is just one of the solutions that correspond to the left hand side (as you know, and I know you know).

    CB
    The multi-valued-ness needs to be discussed and dealt with, because that is the key here.
    Last edited by mr fantastic; September 18th 2009 at 08:55 AM. Reason: Restored original reply
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  14. #14
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    Quote Originally Posted by Random Variable View Post
    Solve for z.

     i^{z} = z where  z \in \mathbb{C}
    I like this method:

    As i=e^{(2n+1/2)\pi i}=e^{-\mu_n} where \mu_n=-(2n+1/2)\pi i for n\in\mathbb{Z} we have:

    e^{-\mu_n z}=z

    or:

    \mu_n z\; e^{\mu_n z}=\mu_n

    so:

    \mu_n z=W(\mu_n).

    Hence:

    z=\frac{1}{\mu_n}W(\mu_n); \ \ \mu_n= -(2n+1/2)\pi i,\ n \in \mathbb{Z}

    (notes:
    1. this is at least the third refinement of the original idea scribbled in my note book
    2. the multi-valuedness of the solution is not only dependent on n, but the Lambert-W is itself multi-valued)

    CB
    Last edited by CaptainBlack; August 23rd 2009 at 12:12 AM. Reason: update from itteration 3 to 4
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