2005 AMC 12B 第 22 题

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所有题目均经美国数学协会（MAA）官方合法授权使用。

22.

复数列

z_0

，

z_1

，

z_2

，

\ldots

由

z_{n+1} = \dfrac{i z_n}{\overline{z_n}}\text{，}

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？

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

小提示：

因为

|z_n| = 1

，所以

\overline{z_n} = \dfrac{1}{z_n}

，从而

z_{n+1} = i z_n^2

。

Since

|z_n| = 1,

\overline{z_n} = \dfrac{1}{z_n},

so

z_{n+1} = i z_n^2

大提示：

迭代后

z_n

等于一个固定的常数乘以

z_0^{2^n}

；数出所得方程的根。

Iterating gives

z_n

as a fixed constant times

z_0^{2^n}

; count roots of the resulting equation

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\text{。}

迭代得

z_1 = i z_0^2

，

z_2 = i(i z_0^2)^2 = -i z_0^4

。此外，若

z_n=-i z_0^{2^n}

，则

z_{n+1}=i(-i)^2z_0^{2^{n+1}}=-i z_0^{2^{n+1}}

。于是对每个

n\ge2

都有

z_n=-i z_0^{2^n}

。

于是条件

z_{2005} = 1

就化为

z_0^{2^{2005}} = i

。而任何非零复数方程

z_0^{N} = i

恰好有

N

个不同的解，且它们都在单位圆上。

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.

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