The 𝑖ᵗʰ root of 𝑖 is realpic.twitter.com/9ydxEkQ5W8
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Note though, when taking powers of complex numbers we have to be careful. Technically, if we don't restrict the range of θ to be [0,2π) then i^i takes on infinitely many values. All of which, however, are real.
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Proof that 𝑖ⁱ is real and equal to exp(-π/2). (Note: we are working in the principal branch of eⁱˣ)pic.twitter.com/OUo4TSJCDk
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I saw this on a turkish blog and in this blog somebody says i^i is not real since the set of Complex numbers is C={(a,b): a,b real numbers}. He said that the only difference with R^2 is operations. Product is different bu him. I found this explanations are bullshit.Wt do u think?
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sounds like total bullshit
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Note however, that, while correct, there is a modified, more general expression...
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Indeed I am working in the principal branch
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If you now compute i^i^i, I'll say "aye yai yai".
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