Which values of x and y would make the following expression represent a real number? (6 + 3i)(x + yi)

For the product to be real, its imaginary part must be 0. Multiplying gives (6 + 3i)(x + yi) = (6x − 3y) + (6y + 3x)i, so we need 6y + 3x = 0. With x = 6 and y = −3, this is 6(−3) + 3(6) = 0, so the choice x = 6, y = −3 makes the product real.

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