 Solution

Using the principal complex square root, $\sqrt{i}=e^{i\pi /4}$. Its two square roots differ by a sign, givingEquivalently 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.