Solution
= Solution
Using the principal <complex square root>, $\sqrt i=e^{i\pi/4}$. Its two square roots differ by a sign, giving
$$
\boxed{z=e^{i\pi/8},\quad z=-e^{i\pi/8}.}
$$
Equivalently their arguments are $\pi/8$ and $9\pi/8$ modulo $2\pi$. If instead the right-hand square root is allowed to take both values, the union has four elements $e^{i(\pi/8+k\pi/2)}$, $k=0,1,2,3$, satisfying $z^4=i$. Only the two boxed values square to the principal right-hand side.