Difference between revisions of "2005 AMC 12B Problems/Problem 22"
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== Solution 4 (more accurate solution 2) == | == 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 | + | 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 | ~bomberdoodles |
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.