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I^i = a real number?

  1. May 14, 2005 #1
    Take euler's formula for the identity of complex numbers:

    [itex]e^{xi}=cos(x)+sin(x)i[/itex]

    If we substitute the value [itex]\pi[/itex] for x it turns out that

    [itex]e^{i\pi}=-1[/itex]

    most of us already knew this wonderfull trick.

    BUT if we substitute [itex]\frac{\pi}{2}[/itex] for x we get (because cos pi/2 = 0 and sin pi/2 = 1):
    [itex]e^{\frac{\pi}{2}i}=i[/itex]

    Now if you raise both sides of this identity to the power i, you obtain (since i^2 = -1):

    [itex]e^{-\frac{\pi}{2}}=i^i[/itex]

    Calculating the value of [itex]e^{-\frac{\pi}{2}}[/itex] it turns out that

    [itex]i^i=0,2078795763....[/itex]

    Isn't that just the weirdest thing ever?? :confused:
     
    Last edited: May 14, 2005
  2. jcsd
  3. May 14, 2005 #2

    dextercioby

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    It is a real number.And i wouldn't call anything in mathematics weird.

    Daniel.
     
  4. May 14, 2005 #3

    Curious3141

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    That's just the principal value. There are an infinite number of values for the expression i^i, all real.

    The general form is given by :

    [tex]i^i = e^{-\frac{1}{2}(2k + 1)\pi}[/tex], where k ranges over all integers. The principal value is for zero k.
     
  5. May 15, 2005 #4

    mathwonk

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    are you sure of that formula? try k=1.
     
  6. May 15, 2005 #5
    :tongue:

    I think you mean:

    [tex]i^i = e^{-\frac{1}{2}(k + 1)\pi}[/tex]

    since it works for every k*(pi/2) with zero k beeing the principal value if i'm correct
     
  7. May 15, 2005 #6

    dextercioby

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    Nope.Compare the case k=0 and k=1,for simplicity.

    Daniel.
     
  8. May 15, 2005 #7
    I see... :shy:
     
  9. May 16, 2005 #8

    Curious3141

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    Sorry, I was wrong. The general form should be [tex]i^i = e^{-\frac{1}{2}\pi(4k + 1)}[/tex].
     
  10. May 16, 2005 #9

    dextercioby

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    Nope,it can't be that one.

    Daniel.
     
  11. May 16, 2005 #10

    Curious3141

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    Why not ?

    k = 0, obvious.

    k = 1, exp (-5pi/2) = z.

    z^(-i) = exp(5*i*pi/2) = exp(i*pi/2 + 2*i*pi) = i

    and so forth.
     
    Last edited: May 16, 2005
  12. May 16, 2005 #11

    dextercioby

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    Alright

    You said (see above)

    [tex] i^{i}=e^{-\frac{1}{2}\pi(4k+1)} [/tex]

    I say

    [tex] k=0 [/tex] (1)

    [tex] \mathbb{R}\ni i^{i}=e^{-\frac{\pi}{2}} [/tex] (2)

    [tex] k=1 [/tex] (3)

    [tex]\mathbb{R}\ni i^{i}=e^{-\frac{5\pi}{2}}\neq e^{-\frac{\pi}{2}}\substack{(2)\\\displaystyle{=}} i^{i}\in\mathbb{R} [/tex] (4)

    Do you see something fishy...? :uhh: You're working with very real numbers...No more multivalued functions...

    Daniel.
     
  13. May 16, 2005 #12

    dextercioby

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    Okay,now here's what u and Max Cohen wanted to write.

    [tex]i^{i}=e^{-\frac{\pi}{2}+2ki\pi} \ ,\ k\in\mathbb{Z}[/tex]

    Daniel.
     
  14. May 16, 2005 #13

    Curious3141

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    No, that's not what I wanted to write. I meant that i^i has an infinite number of real, distinct values as given by the general form. The form you wrote trivially gives only a single real value.

    My first form was wrong because it generated values for (-i)^i as well. But this form is definitely right.

    Complex exponentiation is multivalued, and yes, it can even give an infinite number of distinct real results.
     
    Last edited: May 16, 2005
  15. May 16, 2005 #14

    dextercioby

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    Nope.Everything is in terms of real numbers.U use equality sign,and the reals have a funny way of behaving way when in the presence of the equality sign...

    [tex] i^{i}=:a\in\mathbb{R} [/tex]

    You're telling me that [tex] a=e^{-\frac{\pi}{2}}=e^{-\frac{5\pi}{2}}=e^{-\frac{9\pi}{2}}=... [/tex]

    and that's profoundly incorrect.

    Daniel.
     
  16. May 16, 2005 #15

    shmoe

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    This isn't what he's saying. He's saying that you have many different choices for [tex]i^i:=e^{i\log{i}}[/tex]. log is a multivalued function, [tex]\log{i}=\pi i/2+2\pi i k[/tex] for any integer k. The choice of k chooses the branch of log you are working with. So [tex]i^i=e^{i(\pi i/2+2\pi i k)}=e^{-\pi /2-2\pi k}[/tex] and the different values of k give different values of i^i, depending on which branch of log you're using. [tex]e^{-\frac{\pi}{2}},e^{-\frac{5\pi}{2}},e^{-\frac{9\pi}{2}},\ldots[/tex] are all valid answers for i^i, just as [tex]\pi i/2,5\pi i/2,9\pi i/2,\ldots[/tex] are all valid answers for log(i).
     
  17. May 16, 2005 #16

    Curious3141

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    ^Exactly. :smile:
     
  18. May 16, 2005 #17

    dextercioby

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    Alright.Point taken.:smile: I never thought of it this way before,so i have a reason to thank you.

    Daniel.
     
  19. May 16, 2005 #18

    Curious3141

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    Cool. :smile: Perhaps the orig. poster was not right when he said math is weird, but math certainly is beautiful and mysterious. ;)
     
  20. May 16, 2005 #19

    arildno

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    "The arcane pleasures of abstruse maths is only for the select few of lofty intellect and weird speech patterns"

    Was that what you meant?..:wink:
     
  21. May 16, 2005 #20

    Curious3141

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    Oooh, I have goosebumps. LOL. :biggrin:
     
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