2005 AMC 12B Problems/Problem 22
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 .
See Also
2005 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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