What is $i^i$ and $(i^i)^i$

Solution 1:

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 $k'\in\mathbb{Z}$.

Solution 2:

These are both multiple-valued expressions. Generally one computes $w^z$ as $e^{z\log w}$, so the multiple-valuedness comes from that of the logarithm.

First, since the values of $\log i$ are $\frac{\pi i}{2} -2k\pi i$, the values of $i^i$ are $$z_k=e^{i(\frac{\pi i}{2} -2k\pi i)}=\boxed{e^{\frac{\pi}{2}(4k-1)}} $$ for integral $k$.

Then, since the values of $\log z_k$ are $\frac{\pi}{2}(4k-1) - 2n\pi i=2\pi k -\frac{\pi}{2}-2n\pi i$, the values of each $z_k^i$ are $$z_{k,n}=e^{i(2\pi k -\frac{\pi}{2}-2n\pi i)} =e^{2n\pi}\cdot e^{2k\pi i -\frac{\pi i}{2} } = -ie^{2n\pi} $$ for integral $n$ and $k$. As it happens, for a given $n$, the values of $z_{n,k}$ coincide for each integral $k$, so we may suppress the unneeded subscript. Then we have the possible values of $(i^i)^i$ are $$\hat{z}_{k} = \boxed{-ie^{2k\pi}}$$ for integral $k$.

Note that the previous answers are incomplete.