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2005 AMC 12B Problems/Problem 22: Difference between revisions

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== Solution ==
== Solution ==
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>.
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>.


Base Case: trivial
Inductive Step: Suppose the formula is correct for <math>z_k</math>, then
<cmath>
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)}
</cmath>
Since
<cmath>
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}
</cmath>
the formula is proven
<math>z_{2005}=1\Rightarrow \theta_{2005}=2k\pi</math>, where <math>k</math> is an integer. Therefore,
<cmath>
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
</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 only needs to take values from 0 to <math>2^{2005}-1</math>, so the answer is <math>2^{2005}\Rightarrow\boxed{E}</math>
== See Also ==
== See Also ==


{{AMC12 box|year=2005|ab=B|num-b=21|num-a=23}}
{{AMC12 box|year=2005|ab=B|num-b=21|num-a=23}}

Revision as of 00:41, 16 September 2012

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

Since $|z_0|=1$, let $z_0=e^{i\theta_0}$, where $\theta_0$ is an argument of $z_0$. I 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 only needs to take values from 0 to $2^{2005}-1$, so the answer is $2^{2005}\Rightarrow\boxed{E}$

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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