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Homework Help: Nature of this number

  1. Feb 15, 2006 #1
    A small cute question here :biggrin:.

    [itex]i[/tex] is imaginary. So what is the nature of [itex]i^i[/itex]. Is it imaginary too?

    [tex]i = \sqrt{-1}[/tex]
     
  2. jcsd
  3. Feb 15, 2006 #2
    knowing that i is e^(i*pi/2), raising it to the ith power gives you e^(i^2*pi/2) and i^2 is -1, so e^(-pi/2)
    *pulls out Ti-89*

    yup.
     
  4. Feb 15, 2006 #3

    arildno

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    The question is ill-posed. There isn't just one number that may lay claim to the representation as [itex]i^{i}[/itex]; infinitely many numbers, in fact, can claim that right.
     
  5. Feb 15, 2006 #4
    How is it possible to show that? Isn't [itex]i^i[/itex] purely imaginary?
     
  6. Feb 15, 2006 #5

    arildno

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    It has to do with the complex logarithm being non-unique, we have, say for any choice of integer k:
    [tex]i=e^{i\frac{\pi}{2}+i2k\pi}[/tex]
    Thus, the logarithm of i is the set of numbers [itex]i(\frac{\pi}{2}+2k\pi), k\in{Z}[/itex]


    Thus, we have:
    [tex]i^{i}=e^{i*(i*(\frac{\pi}{2}+2k\pi))}=e^{-\frac{\pi}{2}-2k\pi}[/itex]
     
  7. Feb 15, 2006 #6

    HallsofIvy

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    All of which are, by the way, real, not "pure imaginary"!
     
  8. Feb 15, 2006 #7

    arildno

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    Can't see I said they were imaginary, but I should have emphasized them being real nonetheless.
    Thanks, HallsofIvy
     
  9. Feb 15, 2006 #8

    HallsofIvy

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    You didn't. I was emphasizing that they are real since the original post was "[itex]i[/itex] is imaginary. So what is the nature of [itex]i^i[/itex]. Is it imaginary too?" and then after you said there were many numbers equal to [itex]i^i[/b], Reshma said "How is it possible to show that? Isn't [itex]i^i[/itex] purely imaginary?" so I thought it best to make it very clear that all "variations" of ii were real, not imaginary.
     
  10. Feb 15, 2006 #9

    HallsofIvy

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    You didn't. I was emphasizing that they are real since the original post was "[itex]i[/itex] is imaginary. So what is the nature of [itex]i^i[/itex]. Is it imaginary too?" and then after you said there were many numbers equal to [itex]i^i[/b], Reshma said "How is it possible to show that? Isn't [itex]i^i[/itex] purely imaginary?" so I thought it best to make it very clear that all "variations" of ii were real, not imaginary.
     
  11. Feb 15, 2006 #10

    arildno

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    I fully agree. I ought to have emphasized it, your clarification was necessary.
     
  12. Feb 15, 2006 #11

    VietDao29

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    There must be something wrong with the server, or my browser is working funkily. I somehow cannot view the LaTeX image here... :confused:
     
  13. Feb 15, 2006 #12

    HallsofIvy

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    You didn't. I was emphasizing that they are real since the original post was "[itex]i[/itex] is imaginary. So what is the nature of [itex]i^i[/itex]. Is it imaginary too?" and then after you said there were many numbers equal to [itex]i^i[/b], Reshma said "How is it possible to show that? Isn't [itex]i^i[/itex] purely imaginary?" so I thought it best to make it very clear that all "variations" of ii were real, not imaginary.
     
  14. Feb 16, 2006 #13
    Thank you very much, Arildno and HallsofIvy! So, [itex]i^i[/itex] is real !!
     
  15. Feb 17, 2006 #14

    dextercioby

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    The same goes for the more spectacular (when it comes to notation)

    [tex] \sqrt{i}=i^{\frac{1}{i}}=e^{\frac{\pi}{2}+2n\pi} , n\in\mathbb{Z} [/tex]

    Daniel.
     
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