Question

函数 $f$ 定义在复数集上，且 $f(z) = (a + bi)z$ ，其中 $a$ 和 $b$ 为正数。这个函数满足：复平面中每个点的像到该点与到原点的距离相等。已知 $|a + bi| = 8$ 且 $b^2 = \frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质的正整数。求 $m + n$ 。

A function $f$ is defined on the complex numbers by $f(z) = (a + bi)z,$ where $a$ and $b$ are positive numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin. Given that $|a + bi| = 8$ and that $b^2 = \frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$

答案

答案：259

知识点：复数代数变形

难度评级：2270

解答：

条件是对所有 $z$ 都有 $|f(z) - z| = |f(z)|$ ，即 $|(a - 1 + bi)z| = |(a + bi)z|$ 。当 $z \ne 0$ 时除以 $|z|$ ，得 $|a - 1 + bi| = |a + bi|$ ，所以从而 $a = \frac{1}{2}$ 。 $(a - 1)^2 + b^2 = a^2 + b^2,$

因为 $|a + bi| = 8$ 有 $a^2 + b^2 = 64$ ，所以 $b^2 = 64 - \frac{1}{4} = \frac{255}{4}$ 。又 $\gcd(255, 4) = 1$ ，答案是 $255 + 4 = 259$ 。

The condition is $|f(z) - z| = |f(z)|$ for all $z,$ that is, $|(a - 1 + bi)z| = |(a + bi)z|.$ Dividing by $|z|$ (for $z \ne 0$ ) gives $|a - 1 + bi| = |a + bi|,$ so $(a - 1)^2 + b^2 = a^2 + b^2,$ which forces $a = \frac{1}{2}.$

Since $|a + bi| = 8,$ we have $a^2 + b^2 = 64,$ so $b^2 = 64 - \frac{1}{4} = \frac{255}{4}.$ As $\gcd(255, 4) = 1,$ the answer is $255 + 4 = 259.$

1999 AIME 第 9 题

答案：259

难度评级：2270

解答：

其他年份的第 9 题