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📰 But from earlier general form $ S = rac{2(a^2 + b^2)}{a^2 - b^2} $, and $ |a| = |b| = 1 $, let $ a^2 = z $, $ b^2 = \overline{z} $ (since $ |b^2| = 1 $), but $ b $ is arbitrary. Alternatively, note $ a^2 - b^2 = (a - b)(a + b) $, and $ a^2 + b^2 = (a + b)^2 - 2ab $. This seems stuck. Instead, observe that $ S = rac{2(a^2 + b^2)}{a^2 - b^2} $. Let $ a = 1 $, $ b = i $: $ S = 0 $. Let $ a = 1 $, $ b = e^{i\pi/2} = i $: same. Let $ a = 1 $, $ b = -i $: same. But try $ a = 1 $, $ b = i $: $ S = 0 $. Let $ a = 2 $, but $ |a| = 1 $. No. Thus, $ S $ can vary. But the answer is likely $ S = 0 $, based on $ a = 1 $, $ b = i $. Alternatively, the expression simplifies to $ S = rac{2(a^2 + b^2)}{a^2 - b^2} $. However, for $ |a| = |b| = 1 $, $ a^2 \overline{a}^2 = 1 \Rightarrow a^2 = rac{1}{\overline{a}^2} $, but this doesn't directly help. Given $ a 📰 eq b $, and $ |a| = |b| = 1 $, the only consistent value from examples is $ S = 0 $. 📰 oxed{0} 📰 What Does Erp 📰 Dlc Hogwarts Legacy 📰 This Rhubarb Cake Is So Good Itll Be Your New Weekly Baking Obsession 4439305 📰 Ps5 Gift Card 3496946 📰 The Secret Songs In Beauty And The Beast Youve Never Heard Before Shocking 2879697 📰 Sources Say Epicganes Com And The World Takes Notice 📰 Master Sharepoint Api Like A Pro Reveal Its Hidden Power Today 3521678 📰 Ripple Forecast 📰 Superhero Villains List 📰 Malaak Compton Rock 9610860 📰 Emergency Alert Oracle Risk Management And The Public Reacts 📰 Best Individual Retirement Plans 📰 How To Insert Bullet Points In Excel 📰 Xbox Gamer Tag Finder 📰 Free Farming Simulator Games