1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: I^i = ?
  1. I^(-i) = ? (Replies: 4)

  2. Calculus I (Replies: 6)