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.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 1 / 1C / c / Solution

Write $z=re^{i\theta}$ with $r>0$. On a fixed <branch of the complex logarithm> on which $\log z=\log r+i\theta$ and $\log\bar z=\log r-i\theta$, the equation becomes
$$
\log r+i\theta=i(\log r-i\theta)=\theta+i\log r.
$$
Its real and imaginary parts give the same condition $\log r=\theta$, or
$$
\boxed{r=e^\theta}.
$$
This is a <logarithmic spiral>. For the <principal complex logarithm>, it is the portion parametrized by $-\pi<\theta<\pi$ that avoids the branch cut.

Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 1 / 1C / d / Solution

A branch of the complex logarithm chooses the argument continuously on a domain avoiding an appropriate branch cut, making the logarithm single-valued there.


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