Question

对所有复数 $z$ ，函数 $f$ 定义为 $f(z) = (4 + i)z^2 + \alpha z + \gamma$ ，其中 $\alpha$ 和 $\gamma$ 是复数，且 $i^2 = -1$ 。假设 $f(1)$ 和 $f(i)$ 都是实数。 $|\alpha| + |\gamma|$ 的最小可能值是多少？

A function $f$ is defined by $f(z) = (4 + i)z^2 + \alpha z + \gamma$ for all complex numbers $z,$ where $\alpha$ and $\gamma$ are complex numbers and $i^2 = -1.$ Suppose that $f(1)$ and $f(i)$ are both real. What is the smallest possible value of $|\alpha| + |\gamma|?$

$1$

$\sqrt{2}$

$2$

$2\sqrt{2}$

$4$

答案：B

知识点：复数最优化

难度评级：1990

解答：

令 $\alpha = a + bi$ ， $\gamma = c + di$ 则 $f(1) = (4 + a + c)$ $+ (1 + b + d)i$ ，且 $f(i) = (-4 - b + c)$ $+ (-1 + a + d)i$ 。

二者都是实数，强制 $1 + b + d = 0$ 且 $-1 + a + d = 0$ ，即 $a = 1 - d$ 、 $b = -1 - d$ 。

因此当 $c = d = 0$ 时最小，值为 $\sqrt{2}$ 。 $\begin{aligned} &|\alpha| + |\gamma| \\ &= \sqrt{(1 - d)^2 + (1 + d)^2} \\ &\quad {}+ \sqrt{c^2 + d^2} \\ &= \sqrt{2 + 2d^2} \\ &\quad {}+ \sqrt{c^2 + d^2}, \end{aligned}$

因此，正确答案是 B。

Let $\alpha = a + bi$ and $\gamma = c + di.$ Then $f(1) = (4 + a + c)$ $+ (1 + b + d)i$ and $f(i) = (-4 - b + c)$ $+ (-1 + a + d)i.$

Both being real forces $1 + b + d = 0$ and $-1 + a + d = 0,$ i.e. $a = 1 - d$ and $b = -1 - d.$

Hence $\begin{aligned} &|\alpha| + |\gamma| \\ &= \sqrt{(1 - d)^2 + (1 + d)^2} \\ &\quad {}+ \sqrt{c^2 + d^2} \\ &= \sqrt{2 + 2d^2} \\ &\quad {}+ \sqrt{c^2 + d^2}, \end{aligned}$ which is smallest when $c = d = 0,$ giving $\sqrt{2}.$

Thus, the correct answer is B.

2008 AMC 12B 第 19 题

答案：B

难度评级：1990

解答：

其他年份的第 19 题