z^2 = -2 \Rightarrow z = \pm \sqrt{-2} = \pm i\sqrt{2}
Related Articles
- w^2 + 4w + 4 = 0
- This is a quadratic in \( w \). Factoring:
- Since \( w = z^2 \), we have:
- So the solutions are \( z = i\sqrt{2} \) and \( z = -i\sqrt{2} \). 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 \( i\sqrt{2}, -i\sqrt{2} \).
- But wait — \( z^2 = -2 \) has exactly two solutions in \( \mathbb{C} \):
- z = \sqrt{2}i, \quad z = -\sqrt{2}i
