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📰 S = 2i \cdot rac{e^{i2\pi/3} + e^{i2\pi/3}}{e^{i2\pi/3} - e^{-i2\pi/3}} = ext{complicated}. 📰 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 $. 📰 Mahjong Connect 2 📰 Bank Of America K Street Dc 📰 Deeply Involve Oneself Nyt 📰 How Fidelity Fidfolios Are Changing The Gamewatch How Fast Your Savings Grow 190167 📰 Viatris Stock Price 📰 Verizon Sim Karte 📰 No Sanitizer Compares These Alcohol Wipes Are Taking Over Public Spaces 7162433 📰 Oracle Account Reconciliation Manager 📰 Latest Update Best Auto Refinance Companies And The Truth Emerges 📰 Culligan Water Softening System 3642165 📰 Transformers Revenge Of The Fallen Betrayal That Redefined The Battle For Cybertron 6849752 📰 Manhunt Fixer 📰 30 Day Challenge 📰 Affirmation Affirmation 2036139 📰 Sudden Decision Pa E Docket Search By Name And Authorities Investigate