Let $ S = rac{(a + b)^2 + (a - b)^2}{a^2 - b^2} = rac{2a^2 + 2b^2}{a^2 - b^2} $. Let $ a = 1 $, $ b = i $:
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- ight)} = -2 \cot\left( rac{ heta - \phi}{2}
- eq b $ implies $ heta
- eq \phi $, so $ S $ is purely imaginary. However, the original question may expect a real value. Re-examining:
- S = rac{2(1 + (-1))}{1 - (-1)} = 0, ext{ but } a^2 - b^2 = 1 - (-1) = 2, ext{ numerator } 2(1 + (-1)) = 0 \Rightarrow S = 0.
- 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 + e^{i4\pi/3}.
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