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e -1 $, we can cancel $ t + 1 $, giving: - Simpleprint

📅 March 13, 2026 👤 scraface

Mar 13, 2026

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📌 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

📌 These are distinct and purely imaginary. The real part of each is 0.

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