Solution
= Solution
<Complex exponentiation> uses a branch of the <complex logarithm>. All logarithms of $i$ are $i(\pi/2+2\pi k)$ with $k\in\mathbb Z$, so
$$
i^{1-i}=\exp\bigl[(1+i)(\pi/2+2\pi k)\bigr].
$$
Since the trigonometric factor is always $i$, all values in Cartesian form are
$$
\boxed{z=0+i\,e^{\pi/2+2\pi k},\qquad k\in\mathbb Z.}
$$
The <principal logarithm> selects $k=0$ and gives the principal value $ie^{\pi/2}$.