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

Revision as of 04:47, 3 January 2025

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 $z_0 = \cos \theta + i\sin \theta = e^{i\theta}$ . $z_1 = \frac {iz_{0}}{\overline {z_{0}}} = \frac{i(\cos \theta + i\sin \theta)}{\cos \theta - i\sin \theta} = i(\cos \theta + i\sin \theta)^2 = i(\cos 2\theta + i\sin 2\theta) = ie^{i2\theta}$ $z_2 = \frac {iz_{1}}{\overline {z_{1}}} = \frac {i(ie^{i2\theta})}{-ie^{-i2\theta}} = -ie^{i4\theta}$ $z_3 = \frac {iz_{2}}{\overline {z_{2}}} = \frac {i(-ie^{i4\theta})}{ie^{-i4\theta}} = -ie^{i8\theta}$

Repeating through this recursive process, we can quickly see that for $n>1$ $z_n = -i \cdot e^{i2^n\theta}$ Plugging in $n=2005$ , we get $z_{2005} = -ie^{i2^{2005}\theta} = 1$ This means that $e^{i2^{2005}\theta}$ lies at point $-i$ on the unit circle in the complex plane, which happens when $2^{2005}\theta = \frac{3\pi}{2} + 2k\pi$ . Solving for $\theta$ , we get $\theta=\frac{\pi(2k+\frac{3}{2})}{2^{2005}}$ where $k = 0,1,2...2^{2005}-1$ . 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 $\boxed{\mathrm{(E)}\text{ }2^{2005}}$ .

Solution 4 (more accurate solution 2)

Let $z_0=e^{i\theta}$ . Then $z_1=\frac{iz_0}{\overline{z_0}}=\frac{ie^{i\theta}}{e^{-i\theta}}=ie^{i2\theta}$ , $z_2=\frac{iz_1}{\overline{z_1}}-e^{i2^2 \theta}$ , $z_3=\frac{iz_2}{\overline{z_2}}=-ie^{i2^3 \theta}$ , $z_4=\frac{iz_3}{\overline{z_3}}=e^{i2^4\theta}$ . Now we see that every for every positive integer $4k$ , $z_{4k}=e^{i2^{4k}\theta}$ so $z_{2004}=e^{i2^{2004}\theta}$ and $z_{2005}=ie^{i2^{2005}\theta}=1 \iff e^{i(2^{2005}\theta + \frac{\pi}{2})}=1$ , which has $2^{2005}$ solutions of the form $\theta=\frac{\frac{3\pi}{2}+2\pi k}{2^{2005}}$ for $k \in \{0, 1, \dots 2^{2005}-1\}$ . $\implies \boxed{\mathrm{E}}$

~bomberdoodles

Video Solution

https://youtu.be/hKGwHUN8gQg?si=pCj35pPwVa-aaC3w

~MathProblemSolvingSkills.com

@@ Line 39: / Line 39: @@
 == Solution 2 ==
-Let <math>z_0 = \cos \theta + i\sin \theta</math>.
+Let <math>z_0 = \cos \theta + i\sin \theta = e^{i\theta}</math>.
 <cmath>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}</cmath>
-<cmath>z_2 = \frac {iz_{1}}{\overline {z_{1}}} = \frac{i(ie^{i2\theta})}{-ie^{-i2\theta}} = -ie^{i4\theta}</cmath>
+<cmath>z_2 = \frac {iz_{1}}{\overline {z_{1}}} = \frac {i(ie^{i2\theta})}{-ie^{-i2\theta}} = -ie^{i4\theta}</cmath>
-Repeating through this recursive process, we can quickly see that
+<cmath>z_3 = \frac {iz_{2}}{\overline {z_{2}}} = \frac {i(-ie^{i4\theta})}{ie^{-i4\theta}} = -ie^{i8\theta}</cmath>
-<cmath>z_n = (-1)^n e^{i2^{n}\theta}</cmath>
-Plugging in <math>n=2005</math>, we get <cmath>z_{2005} = -e^{i2^{2005}\theta} = 1</cmath>
+Repeating through this recursive process, we can quickly see that for <math>n>1</math>
-This means that <math>e^{i2^{2005}\theta}</math> lies at point <math>-1</math> on the unit circle in the complex plane, which happens when <math>2^{2005}\theta = \pi + 2k\pi</math>. Solving for <math>\theta</math>, we get <math>\theta=\frac{\pi(2k+1)}{2^{2005}}</math> where <math>k = 0,1,2...2^{2005}-1</math>. So the answer is <math>2^{2005}\Rightarrow\boxed{\mathrm{E}}</math>. (Author: Patrick Yin)
+<cmath>z_n = -i \cdot e^{i2^n\theta}</cmath>
+Plugging in <math>n=2005</math>, we get <cmath>z_{2005} = -ie^{i2^{2005}\theta} = 1</cmath>
+This means that <math>e^{i2^{2005}\theta}</math> lies at point <math>-i</math> on the unit circle in the complex plane, which happens when <math>2^{2005}\theta = \frac{3\pi}{2} + 2k\pi</math>. Solving for <math>\theta</math>, we get <math>\theta=\frac{\pi(2k+\frac{3}{2})}{2^{2005}}</math> where <math>k = 0,1,2...2^{2005}-1</math>. So the answer is <math>2^{2005}\Rightarrow\boxed{\mathrm{E}}</math>. (Author: Patrick Yin)
 == Solution 3 ==

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