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Taylor series for i^i?

  1. Oct 27, 2003 #1
    I'm having some problems expanding i^i, could anyone help? I know it becomes a real number somehow, and I'm familiar with the e^(i * pi) expansion, but is the i^i done in the same way?
  2. jcsd
  3. Oct 27, 2003 #2


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    There is a well known expansion for a^x:


    Just replace a and x with i.

    At first glance, it doesn't look real to me, but maybe the sum telescopes.

  4. Oct 27, 2003 #3
    Hello, Khan!

    I'm not sure what you mean by expanding ii,
    since it is already a constant.

    Using DeMoivre's Theorem (Euler's?): eix = cos x + i sin x,
    when x = pi/2, we have: ei*pi/2 = cos(pi/2) + i sin(pi/2) = i

    Raise both sides to the power i: ii = (ei*pi/2)i= e-pi/2 = 0.207879576...
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