bonanova 83 Report post Posted January 17, 2016 Is i^{i} real? Quote Share this post Link to post Share on other sites
0 DejMar 9 Report post Posted January 17, 2016 Spoiler Yes, it is real. In complex number analysis, exponentiation with respect to the complex numbers is a multifunction. That is, in calculating the value of i^{i} it can be shown to have more than one value. Yet, unlike the inverse of the exponention a real number by an real number, which is also multivalued (e.g., square-root of 4 is {2, -2}), the magnitude of a number raised to a complex value is not always the same. Using de Moivre's theorem, It has shown that the multivalues of i^{i} are equal to e^{(-π/2 + 2nπ)} for any integer n, with the principal value being e^{-}^{π/2} where n=0. As can be noted, each of these values are real numbers. Using Euler's formula, e^{xi} = cos(x) + i*sin(x), it can be shown that one of these multivalues is the real number of approximately 0.20787957635. Quote Share this post Link to post Share on other sites
1 ThunderCloud 5 Report post Posted January 18, 2016 Spoiler Yes. From Euler's formula, i = e^{i*Pi/2}, so (e^{i*Pi/2})^{i} = e^{-Pi/2}, a real number. Quote Share this post Link to post Share on other sites
Is i^{i} real?
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