AMC12 2004 B

AMC12 2004 B · Q16

AMC12 2004 B · Q16. It mainly tests Complex numbers (rare).

A function $f$ is defined by $f(z) = i\overline{z}$, where $i=\sqrt{-1}$ and $\overline{z}$ is the complex conjugate of $z$. How many values of $z$ satisfy both $|z| = 5$ and $f(z) = z$?

函数 $f$ 定义为 $f(z) = i\overline{z}$，其中 $i=\sqrt{-1}$ 且 $\overline{z}$ 是 $z$ 的复共轭。满足 $|z| = 5$ 且 $f(z) = z$ 的 $z$ 有多少个值？

(A) 0 0

(B) 1 1

(D) 4 4

(E) 8 8

Answer

Correct choice: (C)

正确答案：（C）

Solution

Let $z = a+bi$, so $\overline{z} = a-bi$. By definition, $z = a+bi = f(z) = i(a-bi) = b+ai$, which implies that all solutions to $f(z) = z$ lie on the line $y=x$ on the complex plane. The graph of $|z| = 5$ is a circle centered at the origin, and there are $2 \Rightarrow \mathrm{(C)}$ intersections.

设 $z = a+bi$，则 $\overline{z} = a-bi$。由定义，$z = a+bi = f(z) = i(a-bi) = b+ai$，这意味着满足 $f(z) = z$ 的所有解在复平面上都位于直线 $y=x$ 上。$|z| = 5$ 的图像是以原点为圆心的圆，因此有 $2 \Rightarrow \mathrm{(C)}$ 个交点。

Topics

AL.019 Complex numbers (rare)