Solution
= Solution
Use the <principal complex logarithm>, for which $\operatorname{Log}i=i\pi/2$, and set
$$
a=\frac{2\log2}{\pi},
\qquad
\boxed{z=-1-\frac{2i}{\pi}\log a}.
$$
Then the definition of <complex exponentiation> gives
$$
i^z=e^{z\operatorname{Log}i}
=e^{\log a-i\pi/2}=-ia,
$$
and therefore
$$
i^{,i^z}=e^{(-ia)i\pi/2}=e^{a\pi/2}=e^{\log2}=2.
$$
Other choices of <branch of the complex logarithm> produce further valid answers.