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Home » \( \frac43 \times 12 = 16 \text cups \). - jntua results

\( \frac43 \times 12 = 16 \text cups \). - jntua results

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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To solve \( z^4 + 4z^2 + 4 = 0 \), let us first perform a substitution...
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 \( z = -i\sqrt{2} \). Eac...
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.
Thus, the sum of the real parts is:

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