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Home » Only $50 Can Deliver These Luxurious 2nd Anniversary Presents—You Won’t Believe #2! - AMAZONAWS

Only $50 Can Deliver These Luxurious 2nd Anniversary Presents—You Won’t Believe #2! - AMAZONAWS

Only $50 Can Deliver These Luxurious 2nd Anniversary Gifts—You Won’t Believe #2!

📅 March 11, 2026 👤 scraface
Mar 11, 2026
Only $50 Can Deliver These Luxurious 2nd Anniversary Presents—You Won’t Believe #2!

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📌 Thus $ n^2 = \frac{25 \cdot 9}{400} = \frac{225}{400} = \frac{9}{16} \implies n = \pm \frac{3}{4} $.
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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 \( 0 + 0 = 0 \).

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