2004 AMC 12B 第 16 题

先试着解答 2004 AMC 12B 第 16 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2004 AMC 12B 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

16.

函数 ff 定义为 f(z)=iz‾f(z) = i\overline{z},其中 i=−1i = \sqrt{-1},z‾\overline{z} 是 zz 的复共轭。有多少个 zz 同时满足 ∣z∣=5|z| = 5 和 f(z)=zf(z) = z?
A function ff is defined by f(z)=iz‾,f(z) = i\overline{z}, where i=−1i = \sqrt{-1} and z‾\overline{z} is the complex conjugate of z.z. How many values of zz satisfy both ∣z∣=5|z| = 5 and f(z)=z?f(z) = z?
00
11
22
44
88
答案:C
知识点:复数坐标几何
小提示:
写成 z=x+iyz = x + iy,并计算 iz‾i\overline{z}。
Write z=x+iyz = x + iy and compute iz‾i\overline{z}
大提示:
f(z)=zf(z) = z 会强制 y=xy = x,这是一条直线;再与圆 ∣z∣=5|z| = 5 相交。
f(z)=zf(z) = z forces y=x,y = x, a line; intersect it with the circle ∣z∣=5|z| = 5
解答:
写 z=x+iyz = x + iy 则 f(z)=i(x−iy)=y+ixf(z) = i(x - iy) = y + ix。令 f(z)=zf(z) = z 得 y=xy = x,这是一条过原点的直线。条件 ∣z∣=5|z| = 5 是一个圆,过圆心的直线与圆相交于 22 个点。
因此正确答案是 C。
Writing z=x+iy,z = x + iy, we get f(z)=i(x−iy)=y+ix.f(z) = i(x - iy) = y + ix. Setting f(z)=zf(z) = z gives y=x,y = x, which is a line through the origin. The condition ∣z∣=5|z| = 5 is a circle, and a line through the center meets the circle in 22 points.
Thus, the correct answer is C.
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