alt Hacker NewsTheir argument is that ln(z) where z is a complex number is a multi-valued function, so the statement "Explore why i^i is real number" could be misinterpreted as i^i = a single well-defined real value.
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ogogmad • today at 5:41 PM
Apologies if this is pedantic but "multi-valued function" is not a thing. The function doesn't have multiple values here. Saying "multi-valued function" is not just a way of misleading people about what's really happening, but it's almost the perfect way to stop people from finding out. Do people who say "multi-valued function" know what's really happening? Do you know what a Riemann surface is?
Yes, but it seems strange to claim that i^i isn't anything. That just completely ignores what's interesting, namely that i(π/2 + 2πk) is real for all k ∈ Z.