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

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<cmath>\theta_0=\frac{k}{2^{2004}}\pi+\left(\frac{1}{2^{2006}}-\frac{1}{2}\right)\pi
 
<cmath>\theta_0=\frac{k}{2^{2004}}\pi+\left(\frac{1}{2^{2006}}-\frac{1}{2}\right)\pi
 
</cmath>
 
</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>
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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 ==
 
== Solution 2 ==
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Repeating through this recursive process, we can quickly see that
 
Repeating through this recursive process, we can quickly see that
 
<cmath>z_{2005} = ie^{i2^{2005}\theta} = i(\cos 2^{2005}\theta + i\sin 2^{2005}\theta) = 1</cmath>
 
<cmath>z_{2005} = ie^{i2^{2005}\theta} = i(\cos 2^{2005}\theta + i\sin 2^{2005}\theta) = 1</cmath>
Thus, <math>\sin 2^{2005}\theta = -1</math>. The solutions for <math>\theta</math> are <math>\frac{\frac{3\pi}{2}+2\pi k} {2^{2005}}</math> where <math>k = 0,1,2...(2^{2005}-1)</math>. Note that <math>\cos 2^{2005}\theta = 0</math> for all <math>k</math>, so the answer is <math>2^{2005}\Rightarrow\boxed{E}</math>. (Author: Patrick Yin)
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Thus, <math>\sin 2^{2005}\theta = -1</math>. The solutions for <math>\theta</math> are <math>\frac{\frac{3\pi}{2}+2\pi k} {2^{2005}}</math> where <math>k = 0,1,2...(2^{2005}-1)</math>. Note that <math>\cos 2^{2005}\theta = 0</math> for all <math>k</math>, so the answer is <math>2^{2005}\Rightarrow\boxed{\mathrm{E}}</math>. (Author: Patrick Yin)
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Quick note: the solution forgot the <math>i</math> in front of <math>z_1</math> when deriving <math>z_2</math>, so the solution is inaccurate.
  
 
== Solution 3 ==
 
== Solution 3 ==
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Solving for <math>z_n</math> here, we obtain
 
Solving for <math>z_n</math> here, we obtain
 
<cmath>z_n=\frac{\pm\sqrt{z_{n+1}}}i=-i\cdot(\pm\sqrt{z_{n+1}}).</cmath>
 
<cmath>z_n=\frac{\pm\sqrt{z_{n+1}}}i=-i\cdot(\pm\sqrt{z_{n+1}}).</cmath>
It is seen that there are two values of <math>z_n</math> which correspond to one value of <math>z_{n+1}</math>. That means that there are two possible values of <math>z_{2004}</math>, four possible values of <math>z_{2003}</math>, and so on. Therefore, there are <math>2^{2005}</math> possible values of <math>z_0</math>, giving the answer as <math>\boxed{(E)\text{ }2^{2005}}</math>.
+
It is seen that there are two values of <math>z_n</math> which correspond to one value of <math>z_{n+1}</math>. That means that there are two possible values of <math>z_{2004}</math>, four possible values of <math>z_{2003}</math>, and so on. Therefore, there are <math>2^{2005}</math> possible values of <math>z_0</math>, giving the answer as <math>\boxed{\mathrm{(E)}\text{ }2^{2005}}</math>.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 17:40, 24 January 2021

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)

Quick note: the solution forgot the $i$ in front of $z_1$ when deriving $z_2$, so the solution is inaccurate.

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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