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Is your argument that complex powers isn't anything?
by JohnKemeny 11mo ago
Is your argument that complex powers isn't anything?
- xinu2020 11mo agoTheir 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.
- JohnKemeny 11mo agoYes, 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.
- deleted 11mo ago[deleted]
- ogogmad 11mo agoIn maths, an expression only ever equals a single number. You can't say i^i = e^(-pi/2) and then also say that i^i = e^(3 pi / 2), because if i^i equals two things, then those two things are also equal to each other, and then we get that e^(-pi/2) = e^(3 pi / 2), which is wrong. Riemann surfaces are the only way to fix this. And they're not even that hard to understand, but I'm not sure if you do. Stop making people confused.
- ogogmad 11mo agoApologies 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?
- ogogmad 11mo agoDo you know what a Riemann surface is? Because if you don't, then you don't know what you're talking about - and you should stop getting people confused.