AMC12 2024 B

AMC12 2024 B · Q12

AMC12 2024 B · Q12. It mainly tests Complex numbers (rare), Trigonometry (basic).

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 $z$?

设 $z$ 是一个虚部为正、实部大于 $1$ 且 $|z| = 2$ 的复数。在复平面上，四个点 $0$、$z$、$z^{2}$ 和 $z^{3}$ 是四边形的顶点，该四边形的面积为 $15$。$z$ 的虚部是多少？

(A) \frac{3}{4} \frac{3}{4}

(B) 1 1

(D) \frac{3}{2} \frac{3}{2}

(E) \frac{5}{3} \frac{5}{3}

Answer

Correct choice: (D)

正确答案：（D）

Solution

By making a rough estimate of where $z$, $z^2$, and $z^3$ are on the complex plane, we can draw a pretty accurate diagram (like above.) Here, points $Z_1$, $Z_2$, and $Z_3$ lie at the coordinates of $z$, $z^2$, and $z^3$ respectively, and $O$ is the origin. We're given $|z|=2$, so $|z^2|=|z|^2=4$ and $|z^3|=|z|^3 = 8$. This gives us $OZ_1=2$, $OZ_2=4$, and $OZ_3=8$. Additionally, we know that $\angle{Z_1OZ_2}\cong\angle{Z_2OZ_3}$ (since every power of $z$ rotates around the origin by the same angle.) We set these angles equal to $\theta$. We have that [OZ1Z2Z3]=[OZ1Z2]+[OZ2Z3]=12⋅2⋅4sin⁡θ+12⋅4⋅8sin⁡θ=4sin⁡θ+16sin⁡θ=20sin⁡θ Since this is equal to $15$, we have $20\sin\theta=15$, so $\sin\theta=\frac{3}{4}$. Thus, $\text{Im}(z)=|z|\sin(\theta)=2(\frac{3}{4})=\boxed{\textbf{(D) }\frac{3}{2}}$.

通过粗略估计 $z$、$z^2$ 和 $z^3$ 在复平面上的位置，我们可以画出一个相当准确的图（如上所示）。这里，点 $Z_1$、$Z_2$ 和 $Z_3$ 分别位于 $z$、$z^2$ 和 $z^3$ 的坐标处，$O$ 是原点。已知 $|z|=2$，因此 $|z^2|=|z|^2=4$，且 $|z^3|=|z|^3 = 8$。这给出 $OZ_1=2$、$OZ_2=4$ 和 $OZ_3=8$。此外，我们知道 $\angle{Z_1OZ_2}\cong\angle{Z_2OZ_3}$（因为 $z$ 的每一幂都在原点周围旋转相同的角度）。我们设这些角度等于 $\theta$。我们有 $[OZ1Z2Z3]=[OZ1Z2]+[OZ2Z3]=\frac{1}{2}\cdot2\cdot4\sin\theta+\frac{1}{2}\cdot4\cdot8\sin\theta=4\sin\theta+16\sin\theta=20\sin\theta$ 由于这等于 $15$，我们有 $20\sin\theta=15$，所以 $\sin\theta=\frac{3}{4}$。因此，$\text{Im}(z)=|z|\sin(\theta)=2\left(\frac{3}{4}\right)=\boxed{\textbf{(D) }\frac{3}{2}}$。

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