AMC12 2008 A

AMC12 2008 A · Q23

AMC12 2008 A · Q23. It mainly tests Exponents & radicals, Complex numbers (rare).

The solutions of the equation $z^4+4z^3i-6z^2-4zi-i=0$ are the vertices of a convex polygon in the complex plane. What is the area of the polygon?

方程 $z^4+4z^3i-6z^2-4zi-i=0$ 的解在复平面中构成一个凸多边形的顶点。该多边形的面积是多少？

(A) $2^{5/8}$ $2^{5/8}$

(B) $2^{3/4}$ $2^{3/4}$

(D) $2^{5/4}$ $2^{5/4}$

(E) $2^{3/2}$ $2^{3/2}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

Looking at the coefficients, we are immediately reminded of the binomial expansion of ${\left(x+1\right)}^{4}$. Modifying this slightly, we can write the given equation as: \[{\left(z+i\right)}^{4}=1+i=2^{\frac{1}{2}}\cos \frac {\pi}{4} + 2^{\frac{1}{2}}i\sin \frac {\pi}{4}\] We can apply a translation of $-i$ and a rotation of $-\frac{\pi}{4}$ (both operations preserve area) to simplify the problem: \[z^{4}=2^{\frac{1}{2}}\] Because the roots of this equation are created by rotating $\frac{\pi}{2}$ radians successively about the origin, the quadrilateral is a square. We know that half the diagonal length of the square is ${\left(2^{\frac{1}{2}}\right)}^{\frac{1}{4}}=2^{\frac{1}{8}}$ Therefore, the area of the square is $\frac{{\left( 2 \cdot 2^{\frac{1}{8}}\right)}^2}{2}=\frac{2^{\frac{9}{4}}}{2}=2^{\frac{5}{4}} \Rightarrow D.$

观察系数，我们立刻想到 ${\left(x+1\right)}^{4}$ 的二项式展开。稍作修改，可将原方程写为： \[{\left(z+i\right)}^{4}=1+i=2^{\frac{1}{2}}\cos \frac {\pi}{4} + 2^{\frac{1}{2}}i\sin \frac {\pi}{4}\] 我们可以对图形做平移 $-i$ 与旋转 $-\frac{\pi}{4}$（这两种变换都保持面积不变）以简化问题： \[z^{4}=2^{\frac{1}{2}}\] 由于该方程的根是绕原点每次旋转 $\frac{\pi}{2}$ 弧度依次得到的，因此该四边形是一个正方形。正方形对角线的一半长度为 ${\left(2^{\frac{1}{2}}\right)}^{\frac{1}{4}}=2^{\frac{1}{8}}$。因此正方形面积为 $\frac{{\left( 2 \cdot 2^{\frac{1}{8}}\right)}^2}{2}=\frac{2^{\frac{9}{4}}}{2}=2^{\frac{5}{4}} \Rightarrow D.$

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