What is $i^i$ and $(i^i)^i$ What is $i^i$ and $(i^i)^i$. I fond some answers online but is there a clearer explanation of this? Maybe something help me to understand.
In general, complex power of $a^b$ $(a\ne 0)$ is defined as
$$
a^b = e^{b\log a}.
$$
Thus
$$
i^i = e^{i\log i}=e^{i(\operatorname{Ln}1+i(\frac{\pi}{2}+2k\pi))}=e^{-\frac{\pi}{2} -2k\pi}
$$
for $k\in\mathbb{Z}$ and
$$
(i^i)^i =\left(e^{-\frac{\pi}{2} -2k\pi}\right)^i=e^{i\log\left(e^{-\frac{\pi}{2} -2k\pi}\right)}=e^{i(-\frac{\pi}{2}-2k\pi+i2k'\pi)}=-ie^{-2k'\pi}
$$
for $k'\in\mathbb{Z}$.
Start by representing $i$ in exponential form via the Euler relation:
$$i = e^{i\frac{\pi}{2}}$$
Then combine exponents to get the answer:
$$i^i = e^{i * i\frac{\pi}{2}} = e^{-\frac{\pi}{2}}$$
To get to your second term, combine exponents and then take advantage of the fact that powers of $i$ form a group with integer exponents modulo 4:
$$(i^i)^i = (e^{-\frac{\pi}{2}})^i = e^{-i\frac{\pi}{2}} = i^{-1} = i^{4-1} = i^3 = -i$$
I hope this helps.