2008 AMC 12A 第 23 题

Question

方程

z^4 + 4z^3 i - 6z^2 - 4zi - i = 0

的解是复平面中一个凸多边形的顶点。该多边形面积是多少？

The solutions of the equation

z^4 + 4z^3 i - 6z^2 - 4zi - i = 0

are the vertices of a convex polygon in the complex plane. What is the area of the polygon?

2^{\frac{5}{8}}

2^{\frac{3}{4}}

2

2^{\frac{5}{4}}

2^{\frac{3}{2}}

答案：D

知识点：二项式定理复数单位根

难度评级：2240

小提示：

两边加上

1 + i

，识别一个四次幂。

Add

1 + i

to both sides to recognize a perfect fourth power

大提示：

z + i

的四个值等距分布在半径

2^{\frac{1}{8}}

的圆上，形成一个正方形。

The four values of

z + i

lie equally spaced on a circle of radius

2^{\frac{1}{8}},

forming a square

解答：

两边加上

1 + i

，左边变为

\begin{aligned} &z^4 + 4z^3 i - 6z^2 - 4zi + 1 \\ &= (z + i)^4 \end{aligned}\text{，}

所以

(z + i)^4 = 1 + i

。

令

w = z + i

。四个解等距分布在半径

|1 + i|^{\frac{1}{4}} = (2^{\frac{1}{2}})^{\frac{1}{4}} = 2^{\frac{1}{8}}

的圆上，形成正方形；减去

i

只会平移图形。

该正方形的外接圆半径为

2^{\frac{1}{8}}

，所以对角线为

2 \cdot 2^{\frac{1}{8}} = 2^{\frac{9}{8}}

，边长为

\tfrac{2^{\frac{9}{8}}}{\sqrt{2}} = 2^{\frac{5}{8}}

。

面积为

\left(2^{\frac{5}{8}}\right)^2 = 2^{\frac{5}{4}}\text{。}

所以正确答案是 D。

Adding

1 + i

to both sides, the left side becomes

\begin{aligned} &z^4 + 4z^3 i - 6z^2 - 4zi + 1 \\ &= (z + i)^4, \end{aligned}

so

(z + i)^4 = 1 + i.

The four solutions for

w = z + i

are equally spaced on a circle of radius

|1 + i|^{\frac{1}{4}} = (2^{\frac{1}{2}})^{\frac{1}{4}} = 2^{\frac{1}{8}},

and they form a square. Subtracting

i

merely translates it.

A square inscribed in a circle of radius

2^{\frac{1}{8}}

has diagonal

2 \cdot 2^{\frac{1}{8}} = 2^{\frac{9}{8}},

so its side is

\tfrac{2^{\frac{9}{8}}}{\sqrt{2}} = 2^{\frac{5}{8}}.

The area is

\left(2^{\frac{5}{8}}\right)^2 = 2^{\frac{5}{4}}.

Thus, D is the correct answer.

2008 AMC 12A 第 23 题

答案：D

难度评级：2240

小提示：

大提示：

解答：

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