Question

对所有复数 $z \ne i$ ，定义 $F(z) = \frac{z + i}{z - i}$ ；对所有正整数 $n$ ，令 $z_n = F(z_{n-1})$ 。已知 $z_0 = \frac{1}{137} + i$ ，且 $z_{2002} = a + bi$ ，其中 $a$ 和 $b$ 都是实数。求 $a + b$ 。

Let $F(z) = \frac{z + i}{z - i}$ for all complex numbers $z \ne i,$ and let $z_n = F(z_{n-1})$ for all positive integers $n.$ Given that $z_0 = \frac{1}{137} + i$ and $z_{2002} = a + bi,$ where $a$ and $b$ are real numbers, find $a + b.$

答案

答案：275

知识点：复数递推找规律

难度评级：2600

解答：

将映射与自身复合，再应用一次 $F$ ，得到所以数列 $z_0, z_1, z_2, \ldots$ 的周期为 $3$ 。 $\begin{aligned} F(F(z)) &= \frac{\frac{z+i}{z-i} + i}{\frac{z+i}{z-i} - i} \\ &= \frac{(z + i) + i(z - i)}{(z + i) - i(z - i)} \\ &= \frac{(1 + i)(z + 1)}{(1 - i)(z - 1)} \\ &= i\,\frac{z + 1}{z - 1}, \end{aligned}$ $\begin{aligned} F(F(F(z))) &= \frac{i\,\frac{z+1}{z-1} + i}{i\,\frac{z+1}{z-1} - i} \\ &= \frac{(z + 1) + (z - 1)}{(z + 1) - (z - 1)} \\ &= z, \end{aligned}$

因为 $2002 = 3 \cdot 667 + 1$ ，所以 $z_{2002} = z_1 = F(z_0)$ $= \frac{z_0 + i}{z_0 - i}$ $= \frac{\frac{1}{137} + 2i}{\frac{1}{137}}$ $= 1 + 274i$ 。因此 $a + b = 1 + 274 = 275$ 。

Composing the map with itself, $\begin{aligned} F(F(z)) &= \frac{\frac{z+i}{z-i} + i}{\frac{z+i}{z-i} - i} \\ &= \frac{(z + i) + i(z - i)}{(z + i) - i(z - i)} \\ &= \frac{(1 + i)(z + 1)}{(1 - i)(z - 1)} \\ &= i\,\frac{z + 1}{z - 1}, \end{aligned}$ and applying $F$ once more gives $\begin{aligned} F(F(F(z))) &= \frac{i\,\frac{z+1}{z-1} + i}{i\,\frac{z+1}{z-1} - i} \\ &= \frac{(z + 1) + (z - 1)}{(z + 1) - (z - 1)} \\ &= z, \end{aligned}$ so the sequence $z_0, z_1, z_2, \ldots$ is periodic with period $3.$

Since $2002 = 3 \cdot 667 + 1,$ we have $z_{2002} = z_1 = F(z_0)$ $= \frac{z_0 + i}{z_0 - i}$ $= \frac{\frac{1}{137} + 2i}{\frac{1}{137}}$ $= 1 + 274i.$ Thus $a + b = 1 + 274 = 275.$

2002 AIME I 第 12 题

答案：275

难度评级：2600

解答：

其他年份的第 12 题