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📅 May 15, 2026 👤 scraface

May 15, 2026

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📌 To solve \( z^4 + 4z^2 + 4 = 0 \), let us first perform a substitution. Set \( w = z^2 \). Then the equation becomes:

📌 w^2 + 4w + 4 = 0

📌 This is a quadratic in \( w \). Factoring:

📌 Since \( w = z^2 \), we have:

📌 z^2 = -2 \Rightarrow z = \pm \sqrt{-2} = \pm i\sqrt{2}

📌 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

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