Discussion
The function \(f : \mathbf{R}^{2} \to \mathbf{R}^{2}\) defined by \(f(x, y) = (2xy, x^{2} - y^{2})\) is one-one on the set :
Answer: C
\(f\) is one-one on \(\{(x, y) : \tfrac{1}{2} \lt x \lt 1, |y| \lt \tfrac{1}{4}\}\).
Write \(w = y + ix\). Then \(w^{2} = (y^{2} - x^{2}) + 2ixy\), so \(f(x, y) = (\mathrm{Im}\,w^{2}, -\mathrm{Re}\,w^{2})\): \(f\) is complex squaring in disguise. Hence \(f(x, y) = f(a, b)\) exactly when \(w^{2} = w'^{2}\) for \(w' = b + ia\), that is \(w' = \pm w\), or \((a, b) = \pm(x, y)\).
So \(f\) is one-one on a set precisely when the set contains no pair of opposite points \((x, y)\), \((-x, -y)\). The set in option C lies in the half-plane \(x \gt 0\), which contains no such pair, so \(f\) is one-one there.
The others contain opposite points: (1, 0) and \((-1, 0)\) both lie in \(y \ge 0\) and in the unit disc, and both map to (0, 1); \(\mathbf{R}^{2}\) contains every such pair.
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