3 ms·
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.
by JohnKemeny 11mo ago
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.
- 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.