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Solving I to the I power

  1. Jul 14, 2011 #1
    Solving "I" to the "I" power

    What is the proper formula to calculate "I" to the "I" power? I have seen numerous formulas, however, I was curious if someone could provide me with the simplest solution.
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  3. Jul 14, 2011 #2


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    Re: Solving "I" to the "I" power

    In complex analysis, the (multi-valued) function [itex]x^y[/itex] is defined to be [itex]\exp(y \log x)[/itex]....
  4. Jul 14, 2011 #3


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    Re: Solving "I" to the "I" power

    Note that [itex]i=e^{\frac{\pi i}{2}}[/itex], so [itex]i^{i}=e^{i\frac{\pi i}{2}}[/itex]
  5. Jul 14, 2011 #4


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    Re: Solving "I" to the "I" power

    Hunt_mat, that's only half the story. As Hurkyl noted, the complex exponential is multiple-valued, so we must not forget that.
  6. Jul 14, 2011 #5
    Re: Solving "I" to the "I" power

    A typo: it's the logarithm that's multiple valued :smile: but you knew that of course.

    Anyway, while it is certainly true that [itex]a^b[/itex] has multiple values for complex numbers, mathematicians sometimes pick one value as a principal value. That is, they define


    where Log is the principal branch of the logarithm, which is not multivalued (since we restricted it).

    We consider the principal values of [itex]a^b[/itex] in the Riemann-zeta function, for example, where

    [tex]\zeta (z)=\sum{\frac{1}{n^z}}[/tex]

    there we take the exponentiation to be the principal value, and not the multi-valued one. The value


    that hunt_mat gave was the principal value. If you type i^i in google, you will see that they also return the principal value.

    I'm not saying that anybody did anything wrong here. But I just wanted to tell the OP that there are multiple values of ab, but that we often restrict these multiple values to get a principal value.
  7. Jul 14, 2011 #6


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    Re: Solving "I" to the "I" power

    How about
    i^{i}=e^{i\left(\frac{i\pi}{2}+2n\pi i \right)}
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