But from earlier general form $ S = rac2(a^2 + b^2)a^2 - b^2 $, and $ |a| = |b| = 1 $, let $ a^2 = z $, $ b^2 = \overlinez $ (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 = rac2(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 = rac2(a^2 + b^2)a^2 - b^2 $. However, for $ |a| = |b| = 1 $, $ a^2 \overlinea^2 = 1 \Rightarrow a^2 = rac1\overlinea^2 $, but this doesn't directly help. Given $ a