Difference between revisions of "2005 AMC 12B Problems/Problem 22"
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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 == |
Revision as of 01:40, 22 December 2020
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)
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 .
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.