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Home » r = \fracAs = \frac{20\sqrt3}15 = \frac{4\sqrt3}3 - ToelettAPP

r = \fracAs = \frac{20\sqrt3}15 = \frac{4\sqrt3}3 - ToelettAPP

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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📌 So indeed, \( (z^2 + 2)^2 = 0 \), so roots are \( z = \pm i\sqrt{2} \), each with multiplicity 2. But the **set of distinct roots** is still two: \( i\sqrt{2}, -i\sqrt{2} \), each included twice. But the problem asks for **the sum of the real parts of all complex numbers \( z \)** satisfying the equation. Since real part is 0 for each, even with multiplicity, the sum is still \( 0 + 0 = 0 \).

📌 Alternatively, interpret as sum over all **solutions** (with multiplicity or not)? But in algebraic contexts like this, unless specified, we consider distinct roots or with multiplicity. But multiplicity doesn't affect real part — each root contributes its real part once in evaluation, but the real part function is defined per root. However, in such symmetric cases, we sum over distinct roots unless told otherwise.

📌 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 0. So sum over all roots (counting multiplicity) is \( 4 \times 0 = 0 \).

📌 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}) + \text{Re}(-i\sqrt{2}) = 0 + 0 + 0 + 0 = 0

📌 Thus, the sum is \( \boxed{0} \).

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