Since $ |a| = |b| = 1 $, $ a \overlinea = 1 $ and $ b \overlineb = 1 $. However, observe that $ a^2 - b^2 = (a - b)(a + b) $, but no simplification arises directly. Instead, compute $ S + \overlineS $: - AMAZONAWS

📅 March 6, 2026 👤 scraface

Mar 06, 2026

$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} $:$

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