AMC12 2005 B

AMC12 2005 B · Q22

AMC12 2005 B · Q22. It mainly tests Exponents & radicals, Complex numbers (rare).

A sequence of complex numbers $z_{0}, z_{1}, z_{2}, ...$ is defined by the rule \[z_{n+1} = \frac {iz_{n}}{\overline {z_{n}}},\] 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 $z_{0}$?

复数序列 $z_{0}, z_{1}, z_{2}, ...$ 由规则定义 \[z_{n+1} = \frac {iz_{n}}{\overline {z_{n}}},\] 其中 $\overline {z_{n}}$ 是 $z_{n}$ 的共轭复数，且 $i^{2}=-1$。假设 $|z_{0}|=1$ 且 $z_{2005}=1$。$z_{0}$ 有多少可能值？

(A) 1 1

(B) 2 2

(D) 2005 2005

(E) $2^{2005}$ $2^{2005}$

Answer

Correct choice: (E)

正确答案：（E）

Solution

Since $|z_0|=1$, let $z_0=e^{i\theta_0}$, where $\theta_0$ is an argument of $z_0$. We will prove by induction that $z_n=e^{i\theta_n}$, where $\theta_n=2^n(\theta_0+\frac{\pi}{2})-\frac{\pi}{2}$. Base Case: trivial Inductive Step: Suppose the formula is correct for $z_k$, then \[z_{k+1}=\frac{iz_k}{\overline {z_k}}=i e^{i\theta_k} e^{i\theta_k}=e^{i(2\theta_k+\pi/2)}\] Since \[2\theta_k+\frac{\pi}{2}=2\cdot 2^n(\theta_0+\frac{\pi}{2})-\pi+\frac{\pi}{2}=2^{n+1}(\theta_0+\frac{\pi}{2})-\frac{\pi}{2}=\theta_{n+1}\] the formula is proven $z_{2005}=1\Rightarrow \theta_{2005}=2k\pi$, where $k$ is an integer. Therefore, \[2^{2005}(\theta_0+\frac{\pi}{2})=(2k+\frac{1}{2})\pi\] \[\theta_0=\frac{k}{2^{2004}}\pi+\left(\frac{1}{2^{2006}}-\frac{1}{2}\right)\pi\] The value of $\theta_0$ only matters modulo $2\pi$. Since $\frac{k+2^{2005}}{2^{2004}}\pi\equiv\frac{k}{2^{2004}}\pi\mod 2\pi$, k can take values from 0 to $2^{2005}-1$, so the answer is $2^{2005}\Rightarrow\boxed{\mathrm{E}}$

由于 $|z_0|=1$，令 $z_0=e^{i\theta_0}$，其中 $\theta_0$ 是 $z_0$ 的一个辐角。我们用归纳法证明 $z_n=e^{i\theta_n}$，其中 $\theta_n=2^n(\theta_0+\frac{\pi}{2})-\frac{\pi}{2}$。基础情形：显然成立。归纳步骤：假设对 $z_k$ 公式成立，则 \[z_{k+1}=\frac{iz_k}{\overline {z_k}}=i e^{i\theta_k} e^{i\theta_k}=e^{i(2\theta_k+\pi/2)}\] 由于 \[2\theta_k+\frac{\pi}{2}=2\cdot 2^n(\theta_0+\frac{\pi}{2})-\pi+\frac{\pi}{2}=2^{n+1}(\theta_0+\frac{\pi}{2})-\frac{\pi}{2}=\theta_{n+1}\] 故公式得证。 $z_{2005}=1\Rightarrow \theta_{2005}=2k\pi$，其中 $k$ 为整数。因此 \[2^{2005}(\theta_0+\frac{\pi}{2})=(2k+\frac{1}{2})\pi\] \[\theta_0=\frac{k}{2^{2004}}\pi+\left(\frac{1}{2^{2006}}-\frac{1}{2}\right)\pi\] $\theta_0$ 的取值只需考虑模 $2\pi$。由于 $\frac{k+2^{2005}}{2^{2004}}\pi\equiv\frac{k}{2^{2004}}\pi\mod 2\pi$，$k$ 可取 $0$ 到 $2^{2005}-1$，所以答案为 $2^{2005}\Rightarrow\boxed{\mathrm{E}}$。

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