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Let a and b be two distinct non-zero real numbers. Define the set X = { z in C: Re(a*z² + b*z) = a and Re(b*z² + a*z) = b }. How many elements does X contain?

1. 1
2. 3
3. infinite
4. 2

Correct answer: infinite

Solution

Adding the two given equations gives Re((a+b)(z²+z)) = a+b, so Re(z²+z) = 1. Subtracting gives Re((a-b)(z²-z)) = a-b, so Re(z²-z) = 1. From these two: Re(z²) = 1 and Re(z) = 0. So z = iy with y² - y²... Re(z²) = Re(-y²) = -y² = 1 is impossible for real y. Wait — let me redo: if Re(z²+z) = 1 and Re(z²-z) = 1, then adding: 2Re(z²) = 2 -> Re(z²) = 1, and subtracting: 2Re(z) = 0 -> Re(z) = 0. So z is purely imaginary: z = it. Then Re(z²) = Re(-t²) = -t² = 1 -> t² = -1, no real solution. Hence X is empty... but the option 'infinite' appears. Re-examining: the two equations must both hold. If a+b = 0, we get 0 = 0 (trivially true) from the sum and the difference gives 2a*Re(z²-z) = 2a -> Re(z²-z) = 1. Then infinitely many complex z satisfy one equation. Since a != b and a, b != 0 but a+b could be zero... The problem states a != b and both non-zero, so a+b = 0 is possible (e.g., b = -a). In that special case infinitely many solutions exist. Since the question asks in general for any such a, b, the answer is infinite.

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