Difference between revisions of "2005 AMC 12B Problems/Problem 22"

Revision as of 00:36, 22 December 2020

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 $2^{2005}\Rightarrow\boxed{E}$ . (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 $\boxed{(E)\text{ }2^{2005}}$ .

@@ Line 36: / Line 36: @@
 <cmath>\theta_0=\frac{k}{2^{2004}}\pi+\left(\frac{1}{2^{2006}}-\frac{1}{2}\right)\pi
 </cmath>
-The value of <math>\theta_0</math> only matters [[modulo]] <math>2\pi</math>. Since <math>\frac{k+2^{2005}}{2^{2004}}\pi\equiv\frac{k}{2^{2004}}\pi\mod 2\pi</math>, k can take values from 0 to <math>2^{2005}-1</math>, so the answer is <math>2^{2005}\Rightarrow\boxed{E}</math>
+The value of <math>\theta_0</math> only matters [[modulo]] <math>2\pi</math>. Since <math>\frac{k+2^{2005}}{2^{2004}}\pi\equiv\frac{k}{2^{2004}}\pi\mod 2\pi</math>, k can take values from 0 to <math>2^{2005}-1</math>, so the answer is <math>2^{2005}\Rightarrow\boxed{\mathrm{E}}</math>
 == Solution 2 ==

2005 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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