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

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</math>
 
</math>
  
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== Solution ==
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== Solution 1 ==
 
Since <math>|z_0|=1</math>, let <math>z_0=e^{i\theta_0}</math>, where <math>\theta_0</math> is an [[argument]] of <math>z_0</math>.  
 
Since <math>|z_0|=1</math>, let <math>z_0=e^{i\theta_0}</math>, where <math>\theta_0</math> is an [[argument]] of <math>z_0</math>.  
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I will prove by induction that <math>z_n=e^{i\theta_n}</math>, where <math>\theta_n=2^n(\theta_0+\frac{\pi}{2})-\frac{\pi}{2}</math>.
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We will prove by induction that <math>z_n=e^{i\theta_n}</math>, where <math>\theta_n=2^n(\theta_0+\frac{\pi}{2})-\frac{\pi}{2}</math>.
  
 
Base Case: trivial
 
Base Case: trivial
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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>
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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{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 ==
 +
Let <math>z_0 = \cos \theta + i\sin \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(\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}</cmath>
 +
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>
 +
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)
 +
 
 +
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 ==
 +
Note that for any complex number <math>z</math>, we have <math>|z|=|\overline z|</math>. Therefore, the magnitude of <math>\frac{iz_n}{|z_n|}</math> is always <math>1</math>, meaning that all of the numbers in the sequence <math>z_k</math> are of magnitude <math>1</math>.
 +
 
 +
Another property of complex numbers is that <math>z\overline z=|z|^2</math>. For the numbers in our sequence, this means <math>z\overline z=1</math>, so <math>\overline z=z^{-1}</math>. Rewriting our recursive condition with these facts, we now have
 +
<cmath>z_{n+1}=\frac{iz_n}{z_n^{-1}}=iz_n^2.</cmath>
 +
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>
 +
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>.
 +
 
 +
== Solution 4 (more accurate solution 2) ==
 +
Let <math>z_0=e^{i\theta}</math>. Then <math>z_1=\frac{iz_0}{\overline{z_0}}=\frac{ie^{i\theta}}{e^{-i\theta}}=ie^{i2\theta}</math>, <math>z_2=\frac{iz_1}{\overline{z_1}}-e^{i2^2 \theta}</math>, <math>z_3=\frac{iz_2}{\overline{z_2}}=-ie^{i2^3 \theta}</math>, <math>z_4=\frac{iz_3}{\overline{z_3}}=e^{i2^4\theta}</math>. Now we see that every for every positive integer <math>4k</math>, <math>z_{4k}=e^{i2^{4k}\theta}</math> so <math>z_{2004}=e^{i2^{2004}\theta}</math> and <math>z_{2005}=ie^{i2^{2005}\theta}=1 \iff e^{i(2^{2005}\theta + \frac{\pi}{2})}=1</math>, which has <math>2^{2005}</math> solutions of the form <cmath>\theta=\frac{\frac{3\pi}{2}+2\pi k}{2^{2005}}</cmath> for <math>k \in \{0, 1, \dots 2^{2005}-1\}</math>. <math>\implies \boxed{\mathrm{E}}</math>
 +
 
 +
~bomberdoodles
 +
 
 +
== Video Solution ==
 +
 
 +
https://youtu.be/hKGwHUN8gQg?si=pCj35pPwVa-aaC3w
 +
 
 +
~MathProblemSolvingSkills.com
 +
 
  
 
== See Also ==
 
== See Also ==

Latest revision as of 22:50, 25 July 2024

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

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


See Also

2005 AMC 12B (Problems • Answer Key • Resources)
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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