= Solution
Use the full set of <complex logarithm> values of $i$:
$$
\log i=i\left(\frac\pi2+2\pi k\right),\qquad k\in\mathbb Z.
$$
By the definition of <complex exponentiation>, the inner power can therefore take any value $i^i=\exp[-(\pi/2+2\pi k)]$, a positive real number. Its <complex logarithm> values are $-(\pi/2+2\pi k)+2\pi im$, where $m\in\mathbb Z$. Thus
$$
(i^i)^i=\exp\left[-i\left(\frac\pi2+2\pi k\right)-2\pi m\right]=-ie^{-2\pi m}.
$$
All possible <real parts> and <imaginary parts> are consequently
$$
\boxed{\Re((i^i)^i)=0,\qquad\Im((i^i)^i)=-e^{-2\pi m},\quad m\in\mathbb Z.}
$$
Choosing the principal <complex logarithm> at both stages gives the particular value $-i$. The independent logarithm choices are needed because the question asks for all possible values.
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