So the solutions are \( z = i\sqrt2 \) and \( z = -i\sqrt2 \). Each appears once, but note: quadratic in \( z^2 \) gives two roots, each of multiplicity one. However, since \( z^2 = -2 \) has two complex roots, and no further multiplicity, the full set of roots is \( i\sqrt2, -i\sqrt2 \). - Abu Waleed Tea