i is an imaginary number.
What is $i^i$?
I tried to use euler rule but the answer is strange.
For example $i = e^{\frac{1}{2}i\pi}$.
Using $(a^b)^c = a^{b*c}$ we got
$i^i=e^{(\frac{1}{2}i\pi)*i} = e ^ {-\frac{1}{2}\pi}$
Doesn't seem right
The reason why it doesn't seem right is because of the following
- Power by imaginary numbers do not change the size of the result. e^5, is bigger than e^1. However, |e^1000000i| is still just 1.
So that's why I feel it isn't right. So what's the right answer?
Under the principal branch of the log. that is $\arg z \in (-\pi, \pi]$ we have that:
$$i^i = e^{i\ln i} = e^{- \pi/ 2} $$
because $\ln z = \ln |z| + i \arg z$. Therefore for $z=i$ we get that $\ln i = \frac{i\pi}{2}$ since $\arg i = \frac{\pi}{2}$. (that is the principal angle) This matches your result. It might be strange at first , but yes it is a real number under the principal branch.
Under a different branch you'll get a different result.
Edit: In more general it holds that:
$$i^{i} = e^{-\pi/2 + 2k \pi}, \;\; k \in \mathbb{Z}$$