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Question

Let P(z)=x^3+ax^2+bx+c, where a, b, and c are real. There exists a complex number w such that the three roots of P(z) are w+3i, w+9i, and 2w-4, where i^2=-1. Find a+b+c.

Accepted Answer

Let w = x + yi, where x and y are real numbers. Then the sum of the three roots is (w + 3i) + (w + 9i) + (2w - 4) = 4w - 4 + 12i = 4x + 4yi - 4 + 12i.By Vieta's formulas, the sum of the roots is -a, are real number. Hence, (4x - 4) + (4y + 12)i must be a real number, which means y = -3. Thus, the three roots are w + 3i = x, w + 9i = x + 6i, and 2w - 4 = 2x - 4 - 6i. Since the coefficients of P(z) are all real, the nonreal roots must come in conjugate pairs. Thus, x + 6i must be the conjugate of 2x - 4 - 6i, which means x = 2x - 4. Hence, x = 4, so P(z) = (z - 4)(z - 4 - 6i)(z - 4 + 6i).In particular, P(1) = (1 - 4)(1 - 4 - 6i)(1 - 4 + 6i) = -135.But P(1) = 1 + a + b + c, so a + b + c = -136.