2024 AMC 12B 第 12 题

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12.

zz 是一个虚部为正、实部大于 11、且 z=2|z| = 2 的复数。在复平面中,四点 0,z,z20, z, z^2z3z^3 是一个面积为 1515 的四边形的顶点。zz 的虚部是多少?

Suppose zz is a complex number with positive imaginary part, with real part greater than 1,1, and with z=2.|z| = 2. In the complex plane, the four points 0,z,z2,0, z, z^2, and z3z^3 are the vertices of a quadrilateral with area 15.15. What is the imaginary part of z?z?

34\dfrac{3}{4}

11

43\dfrac{4}{3}

32\dfrac{3}{2}

53\dfrac{5}{3}

答案:D
知识点:复数鞋带公式
难度评级:1670
解答:

对顶点 0,z,z2,z30, z, z^2, z^3,鞋带公式给出面积 12Im(zˉz2+z2z3)=12Im((z2+z4)z)=12(z2+z4)Im(z). \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}

z=2|z| = 2 面积为 12(4+16)Im(z)=10Im(z)\tfrac12(4 + 16)\operatorname{Im}(z) = 10\operatorname{Im}(z)。令 10Im(z)=1510\operatorname{Im}(z) = 15,得 Im(z)=32\operatorname{Im}(z) = \dfrac32。又 Re(z)=494=72>1\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,z2,z30, z, z^2, z^3 the shoelace formula gives area 12Im(zˉz2+z2z3)=12Im((z2+z4)z)=12(z2+z4)Im(z). \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,|z| = 2, this is 12(4+16)Im(z)=10Im(z).\tfrac12(4 + 16)\operatorname{Im}(z) = 10\operatorname{Im}(z). Setting 10Im(z)=1510\operatorname{Im}(z) = 15 gives Im(z)=32.\operatorname{Im}(z) = \dfrac32. (Then Re(z)=494=72>1,\operatorname{Re}(z) = \sqrt{4 - \tfrac94} = \tfrac{\sqrt7}{2} \gt 1, as required.)

Thus, the correct answer is D.

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