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Home » From the third equation, $-v_2 = -4 \implies v_2 = 4\), which contradicts \(v_2 = 0\). - AMAZONAWS

From the third equation, $-v_2 = -4 \implies v_2 = 4\), which contradicts \(v_2 = 0\). - AMAZONAWS

📅 March 6, 2026 👤 scraface
Mar 06, 2026
From the third equation, $-v_2 = -4 \implies v_2 = 4\), which contradicts \(v_2 = 0\).

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📌 S = rac{2(e^{2i heta} + e^{2i\phi})}{e^{2i heta} - e^{2i\phi}} = 2 \cdot rac{e^{i heta} + e^{i\phi}}{e^{i heta} - e^{i\phi}}.
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📌 First, apply the Rational Root Theorem to test possible rational roots, which are factors of the constant term \(-4\) over factors of the leading coefficient \(2\): \( \pm 1, \pm 2, \pm 4, \pm rac{1}{2}, \pm rac{1}{4} \).
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  • The cross product \(\mathbf{v} \times \mathbf{b}\) is:
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