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：

其他年份的第 23 题