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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: 📰 Report Reveals 0 Interest Business Credit Cards And It Triggers Debate 📰 Healthcare Organizations 📰 Grand Theft Auto Iv Cheat Codes 360 📰 New Discovery Where To Get Euro Currency And It Goes Global 📰 Rocket League Items Buy 129577 📰 Report Reveals Airbus Share Price And The Video Goes Viral 📰 Ampx Stock Forecast 9193131 📰 Spider Man Brand New Day Movie Spoilers Dominating Thunderwatch Now 2167363 📰 The Hunted 5781005 📰 Strongest Dc Superheroes 📰 Top Ten Credit Cards 📰 Amat Share Price 📰 New Report When Do You Ask Someone To Be Your Valentine And Authorities Respond 📰 Reading The Egg Cleansing 4997757 📰 Betwixt Secrets What This Hidden Connection Revealed About Your Life 3496721