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📅 March 1, 2026 👤 scraface

Mar 01, 2026

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Even though there are four roots? Let's double-check factorization.
Go back:
z^4 + 4z^2 + 4 = (z^2 + 2)^2 = 0
So indeed, \( (z^2 + 2)^2 = 0 \), so roots are \( z = \pm i\sqrt{2} \)...
Alternatively, interpret as sum over all **solutions** (with multiplic...
But here, the equation \( (z^2 + 2)^2 = 0 \) has roots:
z = i\sqrt{2}, i\sqrt{2}, -i\sqrt{2}, -i\sqrt{2}
Four roots total (with multiplicity 2 each), but real part of each is...
Since the real part is linear, we can sum the real parts:
\text{Re}(i\sqrt{2}) + \text{Re}(i\sqrt{2}) + \text{Re}(-i\sqrt{2}) +...

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