Question

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

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?

$2^{5/8}$

$2^{3/4}$

$2$

$2^{5/4}$

$2^{3/2}$

答案：D

知识点：二项式定理复数单位根

难度评级：2240

解答：

两边加上 $1 + i$ ，左边变为所以 $(z + i)^4 = 1 + i$ 。 $\begin{aligned} &z^4 + 4z^3 i - 6z^2 - 4zi + 1 \\ &= (z + i)^4, \end{aligned}$

令 $w = z + i$ 。四个解等距分布在半径 $|1 + i|^{1/4} = (2^{1/2})^{1/4} = 2^{1/8}$ 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 $2^{1/8}$ ，所以对角线为 $2 \cdot 2^{1/8} = 2^{9/8}$ ，边长为 $\tfrac{2^{9/8}}{\sqrt{2}} = 2^{5/8}$ 。

面积为 $\left(2^{5/8}\right)^2 = 2^{5/4}.$

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 $|1 + i|^{1/4} = (2^{1/2})^{1/4} = 2^{1/8},$ and they form a square. Subtracting $i$ merely translates it.

A square inscribed in a circle of radius $2^{1/8}$ has diagonal $2 \cdot 2^{1/8} = 2^{9/8},$ so its side is $\tfrac{2^{9/8}}{\sqrt{2}} = 2^{5/8}.$

The area is $\left(2^{5/8}\right)^2 = 2^{5/4}.$

Thus, D is the correct answer.

2008 AMC 12A 第 23 题

答案：D

难度评级：2240

解答：

其他年份的第 23 题