Difference between revisions of "2005 AMC 12B Problems/Problem 22"
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</math> | </math> | ||
− | == Solution == | + | == 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>. | ||
− | + | 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 | ||
Line 33: | Line 33: | ||
<math>z_{2005}=1\Rightarrow \theta_{2005}=2k\pi</math>, where <math>k</math> is an integer. Therefore, | <math>z_{2005}=1\Rightarrow \theta_{2005}=2k\pi</math>, where <math>k</math> is an integer. Therefore, | ||
<cmath> | <cmath> | ||
− | 2^{2005}(\theta_0+\frac{\pi}{2})=(2k+\frac{1}{2})\pi | + | 2^{2005}(\theta_0+\frac{\pi}{2})=(2k+\frac{1}{2})\pi</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 | + | 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 23:50, 25 July 2024
Contents
Problem
A sequence of complex numbers is defined by the rule
where is the complex conjugate of
and
. Suppose that
and
. How many possible values are there for
?
Solution 1
Since , let
, where
is an argument of
.
We will prove by induction that
, where
.
Base Case: trivial
Inductive Step: Suppose the formula is correct for , then
Since
the formula is proven
, where
is an integer. Therefore,
The value of
only matters modulo
. Since
, k can take values from 0 to
, so the answer is
Solution 2
Let .
Repeating through this recursive process, we can quickly see that
Thus,
. The solutions for
are
where
. Note that
for all
, so the answer is
. (Author: Patrick Yin)
Quick note: the solution forgot the in front of
when deriving
, so the solution is inaccurate.
Solution 3
Note that for any complex number , we have
. Therefore, the magnitude of
is always
, meaning that all of the numbers in the sequence
are of magnitude
.
Another property of complex numbers is that . For the numbers in our sequence, this means
, so
. Rewriting our recursive condition with these facts, we now have
Solving for
here, we obtain
It is seen that there are two values of
which correspond to one value of
. That means that there are two possible values of
, four possible values of
, and so on. Therefore, there are
possible values of
, giving the answer as
.
Solution 4 (more accurate solution 2)
Let . Then
,
,
,
. Now we see that every for every positive integer
,
so
and
, which has
solutions of the form
for
.
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.