= Solution
Directly applying the <Hadamard gate> gives
$$
H|+i\rangle=\frac12\{(1+i)|0\rangle+(1-i)|1\rangle\}=e^{i\pi/4}|-i\rangle,
$$
while
$$
H|-i\rangle=\frac12\{(1-i)|0\rangle+(1+i)|1\rangle\}=e^{-i\pi/4}|+i\rangle.
$$
Hence \b[the two states are interchanged up to the displayed <global phases>]. Their relative factors $-i$ and $+i$ are retained. They are the two <Pauli eigenstates> of the <Pauli Y gate>, and the <Hadamard interchange of Y eigenstates> is consistent with $HYH=-Y$.
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