To compute this, we’ll make use of Euler’s formula cis(x) = e(adsbygoogle = window.adsbygoogle || []).push({}); ^{ix}= cos(x) + i·sin(x):

e^{i(π/2)}= cos(π/2) + i·sin(π/2) = i, and exponentiating by i we get

i^{i}= (e^{i(π/2)})^{i}= e^{i·i(π/2)}= e^{-π/2}∈ ℝ.

But we also have cis(2πk + π/2) = i, k ∈ ℤ. Thus by the same logic, we get i^{i}= e^{-(2πk + π/2)}for k ∈ ℤ.

Infinitely many evaluations, so i^{i}doesn't have a distinct/unique value? This seems like a discrepancy. I don't have any strong arguments, but I have this W|A computation telling me that i^{i}is indeed distinct, since none of e^{-5π/2}, e^{-9π/2}, e^{-13π/2}, etc are equal.

Also, since the exponential function from ℝ onto ℝ^{+}is certainly injective, each of these numbers e^{-(2πk + π/2)}must be distinct. If i^{i}is distinct, why is its value e^{-π/2}= 0.2078795... ?

Can anyone explain this?

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# Computing i^i

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