Question

设 $z$ 是一个虚部为正、实部大于 $1$ 、且 $|z| = 2$ 的复数。在复平面中，四点 $0$ ， $z$ ， $z^2$ 、 $z^3$ 是一个面积为 $15$ 的四边形的顶点。 $z$ 的虚部是多少？

Suppose $z$ is a complex number with positive imaginary part, with real part greater than $1,$ and with $|z| = 2.$ In the complex plane, the four points $0,$ $z,$ $z^2,$ and $z^3$ are the vertices of a quadrilateral with area $15.$ What is the imaginary part of $z?$

$\dfrac{3}{4}$

$1$

$\dfrac{4}{3}$

$\dfrac{3}{2}$

$\dfrac{5}{3}$

答案：D

知识点：复数鞋带公式

难度评级：1670

小提示：

四边形 $0 \to z \to z^2 \to z^3$ 的有向面积为 $\tfrac12\,\bigl|\operatorname{Im}(\bar z z^2 + \overline{z^2}\, z^3)\bigr|$ 。

The signed area of the quadrilateral $0 \to z \to z^2 \to z^3$ is $\tfrac12\,\bigl|\operatorname{Im}(\bar z z^2 + \overline{z^2}\, z^3)\bigr|$

大提示：

因为 $\bar z z^2 = |z|^2 z$ 且 $\overline{z^2} z^3 = |z|^4 z$ ，面积为 $\tfrac12(|z|^2 + |z|^4)\operatorname{Im}(z)$ ；再用 $|z| = 2$

Since $\bar z z^2 = |z|^2 z$ and $\overline{z^2} z^3 = |z|^4 z,$ the area is $\tfrac12(|z|^2 + |z|^4)\operatorname{Im}(z);$ plug in $|z| = 2$

解答：

对顶点 $0, z, z^2, z^3$ ，鞋带公式给出面积 $\begin{aligned} &\tfrac12\,\bigl|\operatorname{Im}(\bar z z^2 + \overline{z^2}\,z^3)\bigr| \\ &= \tfrac12\,\bigl|\operatorname{Im}\bigl((|z|^2 + |z|^4)z\bigr)\bigr| \\ &= \tfrac12(|z|^2 + |z|^4)\operatorname{Im}(z) \end{aligned}\text{。}$

由 $|z| = 2$ ，面积为 $\tfrac12(4 + 16)\operatorname{Im}(z) = 10\operatorname{Im}(z)$ 。令 $10\operatorname{Im}(z) = 15$ ，得 $\operatorname{Im}(z) = \dfrac32$ 。又 $\operatorname{Re}(z) = \sqrt{4 - \tfrac94} = \tfrac{\sqrt7}{2} \gt 1$ ，满足条件。

所以正确答案是 D。

For vertices $0, z, z^2, z^3$ the shoelace formula gives area $\begin{aligned} &\tfrac12\,\bigl|\operatorname{Im}(\bar z z^2 + \overline{z^2}\,z^3)\bigr| \\ &= \tfrac12\,\bigl|\operatorname{Im}\bigl((|z|^2 + |z|^4)z\bigr)\bigr| \\ &= \tfrac12(|z|^2 + |z|^4)\operatorname{Im}(z). \end{aligned}$

With $|z| = 2,$ this is $\tfrac12(4 + 16)\operatorname{Im}(z) = 10\operatorname{Im}(z).$ Setting $10\operatorname{Im}(z) = 15$ gives $\operatorname{Im}(z) = \dfrac32.$ (Then $\operatorname{Re}(z) = \sqrt{4 - \tfrac94} = \tfrac{\sqrt7}{2} \gt 1,$ as required.)

Thus, the correct answer is D.

2024 AMC 12B 第 12 题

答案：D

难度评级：1670

小提示：

大提示：

解答：

其他年份的第 12 题