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What is i^i?

  1. Dec 13, 2004 #1
    What is i^i?? And how do we find it?
     
  2. jcsd
  3. Dec 13, 2004 #2

    Hurkyl

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    By the definitions!

    For complex numbers, exponentiation is defined by:

    [tex]z^w := \exp(w \log z)[/tex]

    For the principal value of the exponential, you use the principal value of the logarithm.
     
  4. Dec 13, 2004 #3

    dextercioby

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    Quite remarkable,Hurkyl,that both [itex] i^{i} [/itex] and [itex] i^{\frac{1}{i}} [/itex] are real numbers.

    Yap,sometimes mathematics offers surprises... :cool:
     
  5. Dec 13, 2004 #4

    Tide

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    Alternate HINT:

    [tex]i^i = \left(e^{i\pi /2}\right)^i[/tex]
     
  6. Dec 13, 2004 #5

    dextercioby

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    :rofl: That's not a "HINT",that's the SOLUTION!!Hurkyl gave a hint.Anyway,i hope it helps him... :rofl:
     
  7. Dec 13, 2004 #6
  8. Dec 14, 2004 #7
    well, what stumps me is how to find like (i+5)^(i+5)?????
     
  9. Dec 14, 2004 #8

    Tide

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    HINT: Follow the hints offered above! :smile:
     
  10. Dec 15, 2004 #9
    The fact that [tex](a^b)^c=a^{bc}[/tex] is not right with complex numbers.

    As a matter of fact, let z a complex number,
    [tex]e^z=e^{\frac{2i\pi z}{2i\pi}}=(e^{2i\pi})^{\frac{z}{2i\pi}}=1^{\frac{z}{2i\pi}}=1[/tex]
    there is a problem.
     
  11. Dec 15, 2004 #10

    cronxeh

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    or, if you are an engineer, just open matlab and do i^i will save you the valuable time :biggrin:
     
  12. Dec 15, 2004 #11

    Tide

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    That's slick but

    [tex]1^{\frac{z}{2\pi i}}=1[/tex]

    only if [itex]1 = e^{0i}[/itex] (on the LHS) but you explicitly took [itex]1 = e^{2\pi i}[/itex] and used a different expression of 1 in your final step. [itex]1^z[/itex] will be 1 only if arg z = 0.
     
  13. Dec 15, 2004 #12

    learningphysics

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    So when manipulating complex numbers we can't simply make a substitution and say:
    [tex]1^\frac{z}{2\pi i}=(e^{(0i)})^{\frac{z}{2\pi i}}=e^0=1[/tex] ??

    Which step above is illegal when manipulating complex numbers?

    Is there a webpage, which states these types of situations...
     
  14. Dec 15, 2004 #13

    Tide

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    That is a valid root of 1 and there is nothing wrong with what you did. Tsss's problem arose because he inconsistently expressed the same number in two different ways going from one step to the next.

    [tex]\left(e^{0 i}\right)^{1/2} = e^0 = 1[/tex]

    and

    [tex]\left(e^{2\pi i}\right)^{1/2} = e^{i\pi} = -1[/tex]

    are distinct roots of 1 but if I applied Tsss's method to the second I could find only one square root of 1. Obviously, squaring either 1 or -1 both give 1.

    Here's an appropriate web page: http://home.earthlink.net/~djbach/paradox.html
     
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