2005 AMC 12B Problems/Problem 22

Revision as of 01:40, 22 December 2020 by Arthy00009 (talk | contribs) (Solution 3)

Problem

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 $z_{0}$?

$\textbf{(A)}\ 1 \qquad  \textbf{(B)}\ 2 \qquad  \textbf{(C)}\ 4 \qquad  \textbf{(D)}\ 2005 \qquad  \textbf{(E)}\ 2^{2005}$

Solution 1

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 $2^{2005}\Rightarrow\boxed{\mathrm{E}}$

Solution 2

Let $z_0 = \cos \theta + i\sin \theta$. \[z_1 = \frac {iz_{0}}{\overline {z_{0}}} = \frac{i(\cos \theta + i\sin \theta)}{\cos \theta - i\sin \theta} = i(\cos \theta + i\sin \theta)^2 = i(\cos 2\theta + i\sin 2\theta) = ie^{i2\theta}\] \[z_2 = \frac {iz_{1}}{\overline {z_{1}}} = \frac{i(\cos 2\theta + i\sin 2\theta)}{\cos 2\theta - i\sin 2\theta} = i(\cos 2\theta + i\sin 2\theta)^{2} = i(\cos 2^2\theta + i\sin 2^2\theta) = ie^{i2^2\theta}\] Repeating through this recursive process, we can quickly see that \[z_{2005} = ie^{i2^{2005}\theta} = i(\cos 2^{2005}\theta + i\sin 2^{2005}\theta) = 1\] Thus, $\sin 2^{2005}\theta = -1$. The solutions for $\theta$ are $\frac{\frac{3\pi}{2}+2\pi k} {2^{2005}}$ where $k = 0,1,2...(2^{2005}-1)$. Note that $\cos 2^{2005}\theta = 0$ for all $k$, so the answer is $2^{2005}\Rightarrow\boxed{\mathrm{E}}$. (Author: Patrick Yin)

Solution 3

Note that for any complex number $z$, we have $|z|=|\overline z|$. Therefore, the magnitude of $\frac{iz_n}{|z_n|}$ is always $1$, meaning that all of the numbers in the sequence $z_k$ are of magnitude $1$.

Another property of complex numbers is that $z\overline z=|z|^2$. For the numbers in our sequence, this means $z\overline z=1$, so $\overline z=z^{-1}$. Rewriting our recursive condition with these facts, we now have \[z_{n+1}=\frac{iz_n}{z_n^{-1}}=iz_n^2.\] Solving for $z_n$ here, we obtain \[z_n=\frac{\pm\sqrt{z_{n+1}}}i=-i\cdot(\pm\sqrt{z_{n+1}}).\] It is seen that there are two values of $z_n$ which correspond to one value of $z_{n+1}$. That means that there are two possible values of $z_{2004}$, four possible values of $z_{2003}$, and so on. Therefore, there are $2^{2005}$ possible values of $z_0$, giving the answer as $\boxed{\mathrm{(E)}\text{ }2^{2005}}$.

See Also

2005 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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