2008 AMC 12B 第 19 题

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

小提示：

写

\alpha = a + bi

和

\gamma = c + di

；要求

f(1)

和

f(i)

为实数会给出两个线性条件。

Write

\alpha = a + bi

and

\gamma = c + di;

requiring

f(1)

and

f(i)

to be real gives two linear conditions

大提示：

它们推出

a = 1 - d

和

b = -1 - d

，所以

|\alpha| + |\gamma|

= \sqrt{2 + 2d^2} + \sqrt{c^2 + d^2}

；再对

c, d

最小化。

They force

a = 1 - d

and

b = -1 - d,

so

|\alpha| + |\gamma|

= \sqrt{2 + 2d^2} + \sqrt{c^2 + d^2};

minimize over

c, d

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

\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}\text{，} \end{aligned}

当

c = d = 0

时最小，值为

\sqrt{2}

。

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

小提示：

大提示：

解答：

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