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

["Exploring the Expression:\n$$ S = \frac{2(a^2 + b^2)}{a^2 - b^2} $$\nfor complex numbers $ a $ and $ b $ such that $ |a| = |b| = 1 $.", "---", "### When $ a = 1 $ and $ b = i $:\nLet’s test a simple case to unpack the behavior:\n- $ a = 1 \Rightarrow a^2 = 1 $\n- $ b = i \Rightarrow b^2 = -1 $\nThen:\n$$\nS = \frac{2(1 + (-1))}{1 - (-1)} = \frac{2(0)}{2} = 0\n$$", "This suggests $ S = 0 $ in this case.", "---", "### General analysis under $ |a| = |b| = 1 $:", "Since $ |a| = 1 $, we can write $ a = e^{i\alpha} $, so $ a^2 = e^{i2\alpha} $. Similarly, $ b = e^{i\beta} $, so $ b^2 = e^{i2\beta} $, and since $ |b| = 1 $, $ b^2 = \overline{a^2} $ only if $ b = \overline{a} $, but in general $ b^2 $ is arbitrary on the unit circle.", "Thus:\n- $ a^2 + b^2 = e^{i2\alpha} + e^{i2\beta} $\n- $ a^2 - b^2 = e^{i2\alpha} - e^{i2\beta} $", "So:\n$$\nS = 2 \cdot \frac{e^{i2\alpha} + e^{i2\beta}}{e^{i2\alpha} - e^{i2\beta}}\n$$", "Multiply numerator and denominator by $ e^{-i\alpha - i\beta} $ to normalize (optional), but instead rewrite using trigonometric identities.", "Let $ \ heta = 2\alpha - 2\beta $, $ \phi = \alpha + \beta $. Then:\n- $ e^{i2\alpha} = e^{i(\phi + \ heta)} $, $ e^{i2\beta} = e^{i(\phi - \ heta)} $\nThus:\n$$\nS = 2 \cdot \frac{e^{i(\phi + \ heta)} + e^{i(\phi - \ heta)}}{e^{i(\phi + \ heta)} - e^{i(\phi - \ heta)}} = 2 \cdot \frac{2\cos(\phi + \ heta)}{2i\sin(\phi + \ heta) \cdot e^{i\phi}} = \frac{2\cos(\phi + \ heta)}{i\sin(\phi + \ heta) e^{i\phi}} = -2i \cot(\phi + \ heta) e^{-i\phi}\n$$", "This shows $ S $ is generally non-zero and complex-valued depending on $ \alpha, \beta $. However, when $ a = 1 $, $ \alpha = 0 $, and choosing $ \beta = \pi/2 $ gives $ \phi = 0 $, $ \ heta = \pi $, so $ \cot(\pi/2) = 0 $, hence $ S = 0 $, as computed.", "---", "### When does $ S = 0 $?\nFrom expression:\n$$\nS = \frac{2(a^2 + b^2)}{a^2 - b^2} = 0 \iff a^2 + b^2 = 0\n$$\nSo $ a^2 = -b^2 $. But since $ |a^2| = |b^2| = 1 $, this implies $ a^2 = -b^2 \Rightarrow b^2 = -a^2 $. But $ b^2 $ and $ a^2 $ are both on the unit circle, so this is possible only if $ b = \pm i a $.", "But earlier example: $ a = 1 $, $ b = i \Rightarrow b^2 = -1 = -a^2 $, so $ a^2 + b^2 = 0 \Rightarrow S = 0 $.\nHowever, $ a = 1 $, $ b = i $ satisfies $ |a| = |b| = 1 $, but $ S = 0 $ in this case.", "But is $ S $ always zero under $ |a| = |b| = 1 $? No — counterexample: let $ a = 1 $, $ b = 0 $ — but $ |b| = 0 <br/>\ne 1 $, invalid.", "Try $ a = 1 $, $ b = e^{i\pi/3} = \frac{1}{2} + i\frac{\sqrt{3}}{2} $, then $ b^2 = e^{i2\pi/3} $, $ a^2 = 1 $, so $ a^2 + b^2 <br/>\ne 0 $, $ a^2 - b^2 <br/>\ne 0 $, so $ S <br/>\ne 0 $.", "Hence, $ S $ is not identically zero. But specific inputs (like $ a=1, b=i $) yield $ S = 0 $.", "---", "### Alternative approach via substitution:", "Let $ z = a^2 $, $ w = b^2 $. Given $ |z| = |w| = 1 $, and $ S = \frac{2(z + w)}{z - w} $.", "Suppose $ z = 1 $, $ w = -1 $ (since $ |w| = 1 $), then $ S = \frac{2(1 - 1)}{1 - (-1)} = 0 $.\nBut if $ w = i $, $ S = \frac{2(1 + i)}{1 - i} = \frac{2(1+i)^2}{(1-i)(1+i)} = \frac{2(1 + 2i -1)}{2} = 2i <br/>\ne 0 $.", "So $ S $ varies.", "But the problem may imply a specific identity or simplification under transformation.", "Revisiting the expression:\n$$\nS = \frac{2(a^2 + b^2)}{a^2 - b^2}\n$$\nWith $ |a| = |b| = 1 $, write $ a = \frac{1}{\overline{a}} $, $ b = \frac{1}{\overline{b}} $, but $ \overline{a} = 1/a $, so $ a^2 = 1/\overline{a}^2 $, again not helpful.", "Instead, define invariants:\nLet $ u = a^2 $, $ v = b^2 $, $ |u| = |v| = 1 $. Then\n$$\nS = 2 \cdot \frac{u + v}{u - v}\n$$", "This map maps pairs $ (u,v) $ on the unit circle (with $ u <br/>\ne v $) to complex numbers via Möbius transformation. It's undefined when $ u = v $, i.e., $ a^2 = b^2 $, which avoids $ S = 0 $.", "Thus, $ S $ can be zero (e.g., $ u = 1, v = -1 $), but generically non-zero.", "---", "### Conclusion: Among valid $ a, b $ with $ |a| = |b| = 1 $,\n- $ S $ is not identically zero\n- But there exist values (like $ a = 1, b = i $) such that $ S = 0 $\n- The expression simplifies under special angles, but not generally to zero", "However, if the original intent was to analyze the maximum or minimum real part, or identify when $ S = 0 $, then $ S = 0 $ if and only if $ a^2 + b^2 = 0 $, i.e., $ b^2 = -a^2 $. With $ |a| = 1 $, this happens when $ b = \pm i a $.", "Thus, under $ |a| = |b| = 1 $,\n$$\nS = \frac{2(a^2 + b^2)}{a^2 - b^2}\n$$\nis zero when $ b = \pm i a $, and non-zero otherwise.", "---", "Bottom Line:\n- $ S $ is not universally zero\n- Specific configurations (e.g., $ a = 1, b = i $) satisfy $ S = 0 $\n- Alternative forms (substitution $ a^2 = z, b^2 = \overline{z} $) help only if $ z = -z' $, i.e., symmetry\n- Evaluating symbolically supports $ S = \frac{2(a^2 + b^2)}{a^2 - b^2} $, which vanishes under $ b^2 = -a^2 $", "For exact simplification:\n$$\n\boxed{S = 0 \quad \ ext{if and only if} \quad b^2 = -a^2}\n$$\nunder $ |a| = |b| = 1 $. For arbitrary $ a, b $ on the unit circle, $ S $ does not reduce to a constant but reflects geometric symmetry in complex arid plane.", "---", "Keywords: complex numbers, $ S = \frac{2(a^2 + b^2)}{a^2 - b^2} $, $ |a| = |b| = 1 $, evaluation at $ a=1, b=i $, identity $ a^2 + b^2 = 0 $, Möbius transformation, $ b = \pm i a $, simplification techniques."]