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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: 📰 Is This The Ultimate Linlin One Piece Gem Click To Discover The Hype 8882623 📰 You Wont Believe What Happened When He Called The Spirit Of Mist Valley Home 5921419 📰 Agents Of Shield Cast 8267491 📰 Pc Requirements For Fortnite 4081635 📰 Barclays Us App 207258 📰 What Andrew Hussie Revealed About Fame That Will Blow Your Mind 2018261 📰 Baseball Movie Mr 3000 3156144 📰 Unlock The Ultimate Free Supermarket Simulator Experience Before Its Gone Forever 6734070 📰 Karla Mosley Husband 9552378 📰 How Many Sqm In A Sq Ft 1820255 📰 Stakataka Shocked Us All This Secret Feature Is Changing The Game Forever 1567275 📰 Shocking Escape Move How Henry Stickmin Broke Out Of Maximum Security Watch Now 6326539 📰 Speak To A Human At Verizon 5437800 📰 Nugt Stock Price Today 1108161 📰 The Lase Stock Thats Crushing Forecastswhats Stopping You From Investing Now 2685031