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support hose with zippers - Moon Smoking

📅 May 14, 2026 👤 scraface

May 14, 2026

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📌 Since $ |a| = |b| = 1 $, $ a \overline{a} = 1 $ and $ b \overline{b} = 1 $. However, observe that $ a^2 - b^2 = (a - b)(a + b) $, but no simplification arises directly. Instead, compute $ S + \overline{S} $:

📌 S + \overline{S} = rac{(a + b)^2 + (a - b)^2}{a^2 - b^2} = rac{2a^2 + 2b^2}{a^2 - b^2}.

📌 eq b $, and $ |a| = |b| = 1 $. Let $ b = \overline{a} $ (since $ |a| = 1 $), but $ a

📌 eq \overline{a} $ unless $ a $ is real, which contradicts $ a

📌 eq b $. Instead, note that $ rac{a + b}{a - b} + rac{a - b}{a + b} = rac{2(a^2 + b^2)}{a^2 - b^2} $. Let $ a = e^{i heta} $, $ b = e^{i\phi} $, then compute $ S = rac{2(e^{2i heta} + e^{2i\phi})}{e^{2i heta} - e^{2i\phi}} $. Multiply numerator and denominator by $ e^{-i heta} \overline{e^{i heta}} $:

📌 S = rac{2(e^{i heta} + e^{-i heta} + e^{i\phi} + e^{-i\phi})}{e^{i heta} - e^{-i heta} - e^{i\phi} + e^{-i\phi}}.

📌 This simplifies to $ rac{2(2\cos heta + 2\cos\phi)}{2i(\sin heta - \sin\phi)} = rac{2(\cos heta + \cos\phi)}{i(\sin heta - \sin\phi)} $, which is purely imaginary. However, the original expression simplifies directly:

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