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ogogmad an hour ago

i^i isn't anything. Please don't write this. Of the two inputs to the function (w, z) -> w^z = exp(z ln(w)), only z is a complex number, so that bit is OK. The problem is that w is NOT a complex number but a point on a particular Riemann surface, namely: The natural domain of the function ln. That particular Riemann surface looks like an endless spiral staircase. The more grown-up term might be "a helix". When you write informally "w=i", that could mean any of ln(w) = i pi/2, i (2pi + pi/2), i(4pi + pi/2), etc. Incidentally, w^z is then always a real number. However, there's an infinite sequence of those numbers that it could equal.

I suppose that by pure convention, "w=e" is understood as denoting a single unique point on the helix. But extending that convention to w=i starts to look like a recipe for confusion.

JohnKemeny an hour ago | parent [-]

Is your argument that complex powers isn't anything?

xinu2020 an hour ago | parent [-]

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 11 minutes 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.