Hi /sci/, brainlet here. Can you explain to me why i^i is a

>>8334236
What part do you not understand?
They just use the Euler's formula and substitute x by pi/2 and then raise both sides by i.

>>8334236
The problem lies within the question: How do you even define exponentiation of a general complex number? In the reals, you can easily define
[math] a^b := \exp(b \cdot \log(a)) [/math] for positive [math]a[/math] and arbitrary [math]b[/math]. The definition for general complex [math]a[/math] is a little involved, because the logarithm isn't uniquely defined if we allow complex numbers.
To see this in more detail, consider the basic example
[math]\exp(0) = 1 = \exp(2 \pi i) [/math]. We usually define [math]\log(1) = 0[/math], in line with the first equation here, but following this, [math]\log(1) = 2 \pi i[/math] would be equally sensible. It all comes down to choosing a sensible region, because we can't have multiple function values for the same argument, choosing a so called principal branch.
https://en.wikipedia.org/wiki/Principal_branch

So to come back to your original question:
Considering that [math] \exp( i \pi /2) = i [/math], we could choose our principal branch for the logarithm so that [math] \log(i) = i \pi/2 [/math]. Using our previous definition of complex exponentiation, we then find [math]i^i = \exp( i \cdot \log(i)) = \exp(i \cdot i \pi/2) = \exp(- \pi/2)[/math], which is a real number.
However, consider a different principal branch: For example, [math] \exp(i \pi/2 + 2 \pi i) = i [/math] is also completely viable, and would lead us to [math]i^i = \exp(- \pi/2 - 2 \pi)[/math], which is a different real number. And of course we could do this in infinitely many different ways, by adding different multiples of [math]2 \pi i[/math]. So our definition of complex exponentiation depends on the choice of branch, and thusly we won't have a well defined value of [math]i^i[/math] in general without any specification.

Let's see how broken my Tex is here now.

>>8334265
wrong
>>>/b/

>>8334289
gtfo

>>8334236
Go buy a book on the complex plane and read about it.

>>8334236
i = e^i*(pi/2 + 2pi*k)
i^i = (e^i*(pi/2 + 2pi*k))^i = e^i*i*(pi/2 + 2pi*k)
= e^(-pi/2 - 2pi*k)

>>8334292
Dumb nigger fuck off

>>8334265
>>8334289
>>8334292
>>8335236
This chain went from good content to shit at a truly incredible rate.

>>8335239
Go eat a bag of dicks faggot

>>8334265
Well, that was extremely insigthful, thank. I assume that is not true for real numbers elevated to a complex number, right? Does Euler's identity just -1 or does it have multiple values?

>>8335342
Actually, whenever you allow complex numbers in your equations, the exponential function loses its injectivity, i.e. we'll have [math]\exp(a) = \exp(b)[/math] for some [math]a \neq b [/math], since we can always just choose [math]b = a + 2 \pi k[/math] for any integer [math] k[/math]. So even simple exponentiation like [math]2^{1/2}[/math] could change with our definition:
[math]2^{1/2} = \exp(1/2 \cdot \log(2))[/math]. Remember that we conventionally identify [math]2^{1/2} = \sqrt{2} = 1.414\dots[/math], so we identify it with the positive real number which gives you [math]2[/math] when squared. So like before, making the replacement [math]\log(2) \mapsto \log(2) + 2 \pi i[/math] (because [math]\exp(\log(2)) = 2 = \exp(\log(2) + 2 \pi i)[/math], we'll find
[math]2^{1/2} = \exp(1/2 \cdot \log(2) + i \pi) = - \exp(1/2 \cdot \log(2)) = - \sqrt{2}[/math].
So even for simple rational exponents, things can get weird. Which is why mathematicians sometimes shudder when having to include square roots (or general n-th roots) in their calculations: Depending on the context, you'll always have to include the possibility of replacing your positive square root with the corresponding negative one. And we see this correspond very nicely to the choice of our logarithm, so with the choice of our principal branch.
So back to your question: Yep, real numbers elevated to a complex number will be in general multi-valued as well. The most general non-ambiguous case is [math]z^k[/math] with complex [math]z[/math] and integer [math]k[/math].
Euler's identity [math]\exp(\pi i) = -1[/math] however is completely well-defined without having to specify anything: The exponential function defined by its power series doesn't involve any arbitrary choices, so we can talk about [math]\exp(anything)[/math] without having to specify. We do also have [math]\exp(\pi i + 2 \pi i) = \exp(3 \pi i) = -1 [/math] of course, but that's not too insightful.