Question

多项式 $f(z) = az^{2018} + bz^{2017} + cz^{2016}$ 的系数为不超过 $2019$ 的实数，且 $f\left(\frac{1 + \sqrt{3}i}{2}\right) = 2015 + 2019\sqrt{3}i$ 。求 $f(1)$ 除以 $1000$ 的余数。

The polynomial $f(z) = az^{2018} + bz^{2017} + cz^{2016}$ has real coefficients not exceeding $2019,$ and $f\left(\frac{1 + \sqrt{3}i}{2}\right) = 2015 + 2019\sqrt{3}i.$ Find the remainder when $f(1)$ is divided by $1000.$

答案

答案：53

知识点：单位根复数多项式

难度评级：2560

解答：

令 $\omega = \frac{1 + \sqrt{3}i}{2} = \cos 60^\circ + i\sin 60^\circ$ ，这是本原六次单位根。因为 $2016 = 6 \cdot 336$ ，所以 $\omega^{2016} = 1$ 、 $\omega^{2017} = \omega$ 、且 $\omega^{2018} = \omega^2 = \frac{-1 + \sqrt{3}i}{2}$ 。因此 $\begin{aligned} f(\omega) &= a\omega^2 + b\omega + c \\ &= \left(c + \frac{b - a}{2}\right) \\ &\quad {}+ \frac{(a + b)\sqrt{3}}{2}\,i. \end{aligned}$

比较虚部， $\frac{a + b}{2} = 2019$ ，所以 $a + b = 4038$ 。由于 $a \le 2019$ 且 $b \le 2019$ ，这迫使 $a = b = 2019$ 。再比较实部，得到 $c + 0 = 2015$ ，所以 $c = 2015$ 。

因此 $f(1) = a + b + c$ $= 4038 + 2015 = 6053$ ，除以 $1000$ 的余数为 $53$ 。

Let $\omega = \frac{1 + \sqrt{3}i}{2} = \cos 60^\circ + i\sin 60^\circ,$ a primitive sixth root of unity. Since $2016 = 6 \cdot 336,$ we get $\omega^{2016} = 1,$ $\omega^{2017} = \omega,$ and $\omega^{2018} = \omega^2 = \frac{-1 + \sqrt{3}i}{2}.$ Therefore $\begin{aligned} f(\omega) &= a\omega^2 + b\omega + c \\ &= \left(c + \frac{b - a}{2}\right) \\ &\quad {}+ \frac{(a + b)\sqrt{3}}{2}\,i. \end{aligned}$

Matching imaginary parts, $\frac{a + b}{2} = 2019,$ so $a + b = 4038.$ Since $a \le 2019$ and $b \le 2019,$ this forces $a = b = 2019.$ Matching real parts then gives $c + 0 = 2015,$ so $c = 2015.$

Hence $f(1) = a + b + c$ $= 4038 + 2015 = 6053,$ whose remainder upon division by $1000$ is $53.$

2019 AIME II 第 8 题

答案：53

难度评级：2560

解答：

其他年份的第 8 题