Question

Let $P(z) = z^3 + az^2 + bz + c,$ 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 $|a + b + c|.$

Answer

Answer: 136

Concepts:complex number polynomial Vieta’s Formulas

Difficulty rating: 2410

Solution:

Write $w = x + yi$ with $x, y$ real. The sum of the roots is $4w + 12i - 4 = -a,$ which is real, so $4y + 12 = 0$ and $y = -3.$ The roots are then $x,$ $x + 6i,$ and $2x - 4 - 6i.$ Because the coefficients are real, the two non-real roots must be conjugates, so $2x - 4 = x,$ giving $x = 4.$ The roots are $4,$ $4 + 6i,$ and $4 - 6i.$

Now $\begin{aligned} 1 + a + b + c &= P(1) \\ &= (1 - 4) \\ &\quad {}\cdot \bigl(1 - (4 + 6i)\bigr) \\ &\quad {}\cdot \bigl(1 - (4 - 6i)\bigr) \\ &= (-3)(9 + 36) \\ &= -135, \end{aligned}$ so $a + b + c = -136$ and $|a + b + c| = 136.$

2010 AIME II Problem 7

Answer: 136

Difficulty rating: 2410

Solution:

Problem 7 in Other Years