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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} 📰 Games To Download On Macbook 9738342 📰 Step Into The Wii Revolution These Games Are Still Unstoppable 4384053 📰 The Greatest Classic Rock Rock Bands Every Music Lover Should Know 2647312 📰 Spain And Portugal 3071168 📰 The Heartbreak That Made Me Fall For My Other Half Did You Know 7206033 📰 Triple Jump World Record 7608936 📰 How Many Calories In A Roll From Texas Roadhouse 1359896 📰 Vectra Ai 5468123 📰 Flawless Palm Tree Drawing Tutorial You Can Complete Tonight 3746133 📰 Ps1 Emulator On Iphone 6998582 📰 You Wont Believe The Genius Move That Changed Golf Forever 9974666 📰 94 179 179 67 246 Correct 1983065 📰 Notino Shock The Hidden Secret Behind Its Wild Success That You Wont Believe 9189160 📰 Ken Schrader 7089349 📰 Mk Sparks Fireis It Just Slang Or Something Bigger 5625421