Question

回忆：复数 $w = a + bi$ 的共轭为 $\overline{w} = a - bi$ ，其中 $a$ 和 $b$ 为实数， $i = \sqrt{-1}$ 。对任意复数 $z$ ，令 $f(z) = 4i\overline{z}$ 。多项式有四个复根： $z_1, z_2, z_3$ 和 $z_4$ 。令为根分别是 $f(z_1), f(z_2), f(z_3)$ 和 $f(z_4)$ 的多项式，其中 $A, B, C$ 和 $D$ 为复数。 $B + D$ 是多少？ $P(z) = z^4 + 4z^3 + 3z^2 + 2z + 1$ $\begin{aligned} &Q(z) = z^4 + Az^3 + Bz^2 \\ &\quad {}+ Cz + D \end{aligned}$

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 $B + D?$

$-304$

$-208$

$12i$

$208$

$304$

答案：D

知识点：韦达定理复数

难度评级：1800

解答：

对 $P$ ，使用韦达定理， $\sum_{i\lt j} z_iz_j = 3$ ，且 $\prod z_j = 1$ ，二者都是实数，所以它们的共轭仍为 $3$ 和 $1$ 。

$Q$ 的根为 $4i\overline{z_j}$ 。于是 $B$ 是两两乘积之和： $B = (4i)^2 \sum_{i\lt j}\overline{z_i}\,\overline{z_j}$ $= -16 \cdot 3 = -48$ 。且 $D = (4i)^4 \prod \overline{z_j} = 256 \cdot 1 = 256$ 。

所以 $B + D = -48 + 256 = 208$ 。

所以正确答案是 D。

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.

2021 AMC 12A Fall 第 15 题

答案：D

难度评级：1800

解答：

其他年份的第 15 题