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I^i = ?

  1. Aug 29, 2008 #1
    1. The problem statement, all variables and given/known data

    Compute [tex] i^{i} [/tex]

    2. Relevant equations

    [tex] z = x + iy [/tex]
    [tex] z = r \left( cos\theta + isin\theta \right)[/tex]

    3. The attempt at a solution

    I tried to manipulate the equations for substitution to no avail. I think that the answer is 1, since this is what my calculator told me, but I haven't got any more ideas to proceed with the substitutions for i. Experimenting with different ideas on a calculator, I noticed that [tex] n^{i} [/tex] , where [tex] n [/tex] is a positive integer, gave a complex number of the form, [tex] a = b + ic [/tex], where [tex] a^{2} = 1 [/tex] always, and where [tex] b [/tex] decreases and [tex] c [/tex] increases as the integer [tex] n [/tex] increases.
  2. jcsd
  3. Aug 29, 2008 #2
    Try writing i in an exponential form.
  4. Aug 29, 2008 #3


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    i^i = e^{i*log(i)}
  5. Aug 29, 2008 #4


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    Utilize the exponential representation(s) of the imaginary unit,
    where n can be any integer.
  6. Aug 29, 2008 #5
    I solved the problem...thanks for the help.
  7. Aug 30, 2008 #6


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    Could you show us the answer you arrived at?
  8. Aug 30, 2008 #7

    From Euler's formula,

    [tex] e^{i\theta} =\left( cos\theta + isin\theta \right) [/tex]

    Let [tex] \theta = \frac{\pi}{2}+2n\pi [/tex], where n is an integer. Therefore,

    [tex] e^{i\left(\frac{\pi}{2}+2n\pi\right)} = i [/tex]

    for any integer n. Taking both sides to ith power and noting that [tex] i^{2} = -1 [/tex] gives

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

    Very interesting result....since [tex] i^{i} [/tex] can be comprised of an infinite number of elements. Again, thanks for the help.
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