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May 16, 2026

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eq b $ implies $ heta
eq \phi $, so $ S $ is purely imaginary. However, the original questio...
Let $ S = rac{(a + b)^2 + (a - b)^2}{a^2 - b^2} = rac{2a^2 + 2b^2}{a...
S = rac{2(1 + (-1))}{1 - (-1)} = 0, ext{ but } a^2 - b^2 = 1 - (-1)...
Another example: $ a = 1 $, $ b = e^{i2\pi/3} $:
a^2 = 1, b^2 = e^{i4\pi/3}, a^2 - b^2 = 1 - e^{i4\pi/3}, a^2 + b^2 = 1...
Numerator: $ 2(1 + e^{i4\pi/3}) $, denominator: $ 1 - e^{i4\pi/3} $, s...
S = 2i \cdot rac{e^{i2\pi/3} + e^{i2\pi/3}}{e^{i2\pi/3} - e^{-i2\pi/3...
But from earlier general form $ S = rac{2(a^2 + b^2)}{a^2 - b^2} $, a...
eq b $, and $ |a| = |b| = 1 $, the only consistent value from examples...

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