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📅 May 3, 2026 👤 scraface

May 03, 2026

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These are distinct and purely imaginary. The real part of each is 0.
Thus, the sum of the real parts is:
+ 0 = 0
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}

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