Discussion
Let \(z, w\) be two complex numbers such that \(z w \in \mathbf{R}\) and \(z + w \in \mathbf{R}\). Then :
Answer: D
The answer is \(z = \bar{w}\) or \(z, w \in \mathbf{R}\).
If \(z + w = s\) and \(zw = p\) are both real, then \(z\) and \(w\) are the two roots of the real quadratic \(t^{2} - st + p = 0\). A real quadratic has either two real roots or a pair of complex conjugate roots. So either both are real, or \(z = \bar{w}\).
Each narrower option misses a case: \(z = 1, w = 2\) satisfies the conditions with \(z \ne w\) and \(z \ne \bar{w}\); \(z = i, w = -i\) satisfies them without being real; and \(z = w\) is not forced by either case.
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