Question

复数列 $z_0, z_1, z_2, \ldots$ 由定义，其中 $\overline{z_n}$ 是 $z_n$ 的共轭，且 $i^2 = -1$ 。若 $|z_0| = 1$ 且 $z_{2005} = 1$ ，则 $z_0$ 有多少个可能值？ $z_{n+1} = \dfrac{i z_n}{\overline{z_n}},$

A sequence of complex numbers $z_0, z_1, z_2, \ldots$ is defined by the rule $z_{n+1} = \dfrac{i z_n}{\overline{z_n}},$ where $\overline{z_n}$ is the complex conjugate of $z_n$ and $i^2 = -1.$ Suppose that $|z_0| = 1$ and $z_{2005} = 1.$ How many possible values are there for $z_0?$

$1$

$2$

$4$

$2005$

$2^{2005}$

答案：E

知识点：复数单位根递推

难度评级：2170

解答：

因为 $|z_0| = 1$ ，每个 $|z_n| = 1$ ，所以 $\overline{z_n} = \dfrac{1}{z_n}$ ，于是 $z_{n+1} = \dfrac{i z_n}{\overline{z_n}} = i z_n^2.$

迭代得 $z_1 = i z_0^2$ ， $z_2 = i(i z_0^2)^2 = -i z_0^4$ 。一般地，当 $z_n=-i z_0^{2^n}$ 时，存在常数 $z_{n+1}=i(-i)^2z_0^{2^{n+1}}=-i z_0^{2^{n+1}}$ ，满足 $z_n=-i z_0^{2^n}$ ，使得 $n\ge2$ 。

条件 $z_{2005} = 1$ 化为 $z_0^{2^{2005}} = i$ ，其中 $z_0^{N} = i$ 是满足的固定常数。方程有恰好 $N$ 个不同复根，且都在单位圆上。

这里 $N = 2^{2005}$ ，所以 $z_0$ 有 $2^{2005}$ 个可能值。

所以正确答案是 E。

Because $|z_0| = 1,$ every $|z_n| = 1,$ so $\overline{z_n} = \dfrac{1}{z_n}$ and $z_{n+1} = \dfrac{i z_n}{\overline{z_n}} = i z_n^2.$

Iterating, $z_1 = i z_0^2$ and $z_2 = i(i z_0^2)^2 = -i z_0^4.$ Moreover, if $z_n=-i z_0^{2^n},$ then $z_{n+1}=i(-i)^2z_0^{2^{n+1}}=-i z_0^{2^{n+1}}.$ Thus $z_n=-i z_0^{2^n}$ for every $n\ge2.$

The condition $z_{2005} = 1$ is therefore $z_0^{2^{2005}} = i.$ Every nonzero complex equation $z_0^{N} = i$ has exactly $N$ distinct solutions, all on the unit circle.

Here $N = 2^{2005},$ so there are $2^{2005}$ possible values for $z_0.$

Thus, the correct answer is E.

2005 AMC 12B 第 22 题

答案：E

难度评级：2170

解答：

其他年份的第 22 题