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2021 AMC 12A Fall 第 15 题

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15.

回忆：复数 w=a+biw = a + bi（其中 aa 和 bb 为实数，i=−1i = \sqrt{-1}）的共轭是复数 w‾=a−bi\overline{w} = a - bi。对任意复数 zz，令 f(z)=4iz‾f(z) = 4i\overline{z}。多项式P(z)=z4+4z3+3z2+2z+1 P(z) = z^4 + 4z^3 + 3z^2 + 2z + 1 有四个复根：z1z_1、z2z_2、z3z_3 和 z4z_4。设Q(z)=z4+Az3+Bz2+Cz+D \begin{aligned} &Q(z) = z^4 + Az^3 + Bz^2 \\ &\quad {}+ Cz + D \end{aligned} 是以 f(z1)f(z_1)、f(z2)f(z_2)、f(z3)f(z_3) 和 f(z4)f(z_4) 为根的多项式，其中系数 AA、BB、CC 和 DD 都是复数。B+DB + D 是多少？

Recall that the conjugate of the complex number w=a+bi,w = a + bi, where aa and bb are real numbers and i=−1,i = \sqrt{-1}, is the complex number w‾=a−bi.\overline{w} = a - bi. For any complex number z,z, let f(z)=4iz‾.f(z) = 4i\overline{z}. The polynomial P(z)=z4+4z3+3z2+2z+1 P(z) = z^4 + 4z^3 + 3z^2 + 2z + 1 has four complex roots: z1,z_1, z2,z_2, z3,z_3, and z4.z_4. Let Q(z)=z4+Az3+Bz2+Cz+D \begin{aligned} &Q(z) = z^4 + Az^3 + Bz^2 \\ &\quad {}+ Cz + D \end{aligned} be the polynomial whose roots are f(z1),f(z_1), f(z2),f(z_2), f(z3),f(z_3), and f(z4),f(z_4), where the coefficients A,A, B,B, C,C, and DD are complex numbers. What is B+D?B + D?

−304-304

−208-208

12i12i

208208

304304

答案：D
知识点：韦达定理复数
难度评级：1800
小提示：

zj‾\overline{z_j} 的基本对称式是 zjz_j 的对应对称式的共轭；PP 的系数全为实数，所以这些对称和也是实数

The elementary symmetric functions of zj‾\overline{z_j} are the conjugates of those of zj;z_j; all of PP’s coefficients are real, so those symmetric sums are real

大提示：

B=(4i)2∑i<jzi‾ zj‾B = (4i)^2 \sum_{i\lt j}\overline{z_i}\,\overline{z_j}，且 D=(4i)4∏zj‾D = (4i)^4 \prod \overline{z_j}

B=(4i)2∑i<jzi‾ zj‾B = (4i)^2 \sum_{i\lt j}\overline{z_i}\,\overline{z_j} and D=(4i)4∏zj‾D = (4i)^4 \prod \overline{z_j}

解答：

对 PP，使用韦达定理，∑i<jzizj=3\sum_{i\lt j} z_iz_j = 3，且 ∏zj=1\prod z_j = 1，二者都是实数，所以它们的共轭仍为 33 和 11。

QQ 的根为 4izj‾4i\overline{z_j}。于是 BB 是两两乘积之和：B=(4i)2∑i<jzi‾ zj‾B = (4i)^2 \sum_{i\lt j}\overline{z_i}\,\overline{z_j} =−16⋅3=−48= -16 \cdot 3 = -48。且 D=(4i)4∏zj‾=256⋅1=256D = (4i)^4 \prod \overline{z_j} = 256 \cdot 1 = 256。

所以 B+D=−48+256=208B + D = -48 + 256 = 208。

所以正确答案是 D。

By Vieta on P,P, ∑i<jzizj=3\sum_{i\lt j} z_iz_j = 3 and ∏zj=1,\prod z_j = 1, both real, so their conjugates are also 33 and 1.1.

The roots of QQ are 4izj‾.4i\overline{z_j}. Then BB is the sum of products of pairs: B=(4i)2∑i<jzi‾ zj‾B = (4i)^2 \sum_{i\lt j}\overline{z_i}\,\overline{z_j} =−16⋅3=−48.= -16 \cdot 3 = -48. And D=(4i)4∏zj‾=256⋅1=256.D = (4i)^4 \prod \overline{z_j} = 256 \cdot 1 = 256.

So B+D=−48+256=208.B + D = -48 + 256 = 208.

Thus, the correct answer is D.

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