2024 AMC 12B 第 12 题

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：

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