2010 AIME II Problem 7

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

The sum of the roots,

4w + 12i - 4,

must be real; this determines the imaginary part of

w

Real coefficients force the two non-real roots to be conjugates, and

1 + a + b + c = P(1)

Write

w = x + yi

with

x

and

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.

Problem 7 in other years