2021 Fall AMC 12A Problem 15

Problem 15 of 25IntermediateAlgebra

Recall that the conjugate of the complex number

w = a + bi,

where

a

and

b

are real numbers and

i = \sqrt{-1},

is the complex number

\overline{w} = a - bi.

For any complex number

z,

let

f(z) = 4i\overline{z}.

The polynomial

P(z) = z^4 + 4z^3 + 3z^2 + 2z + 1

has four complex roots:

z_1,

z_2,

z_3,

and

z_4.

Let

\begin{aligned} &Q(z) = z^4 + Az^3 + Bz^2 \\ &\quad {}+ Cz + D \end{aligned}

be the polynomial whose roots are

f(z_1),

f(z_2),

f(z_3),

and

f(z_4),

where the coefficients

A,

B,

C,

and

D

are complex numbers. What is

B + D?

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Solution

By Vieta on

P,

\sum_{i\lt j} z_iz_j = 3

and

\prod z_j = 1,

both real, so their conjugates are also

3

and

1.

The roots of

Q

are

4i\overline{z_j}.

Then

B

is the sum of products of pairs:

B = (4i)^2 \sum_{i\lt j}\overline{z_i}\,\overline{z_j}

= -16 \cdot 3 = -48.

And

D = (4i)^4 \prod \overline{z_j} = 256 \cdot 1 = 256.

So

B + D = -48 + 256 = 208.

Thus, the correct answer is D.

More practice

Concepts: Vieta’s Formulas · complex number

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.