AMC12 2005 12B, Problem 22

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 $z_{0}$?

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 $2^{2005}\Rightarrow\fbox{2^{2005}}$

Answer: 2^{2005}