| ▲ | xinu2020 2 hours ago | |
Their 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 an hour ago | parent [-] | |
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. | ||