Difference between revisions of "2023 AMC 12A Problems/Problem 14"
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Revision as of 06:33, 24 August 2024
Contents
[hide]Problem
How many complex numbers satisfy the equation , where is the conjugate of the complex number ?
Solution 1
When , there are two conditions: either or . When , since , . . Consider the form, when , there are 6 different solutions for . Therefore, the number of complex numbers satisfying is .
~plasta
Solution 2
Let We now have and want to solve
From this, we have as a solution, which gives . If , then we divide by it, yielding
Dividing both sides by yields . Taking the magnitude of both sides tells us that , so . However, if , then , but must be real. Therefore, .
Multiplying both sides by ,
Each of the th roots of unity is a solution to this, so there are solutions.
-Benedict T (countmath 1)
Solution 3 (Rectangular Form, similar to Solution 1)
Let .
Then, our equation becomes:
Note that since every single term in the expansion contains either an or , simply setting and yields a solution.
Now, considering the other case that either or does not equal :
Multiplying both sides by (or ), we get: (since ).
Substituting back into the left hand side, we get:
Note that this will have 6 distinct, non-zero solutions since in this case, we consider that either or is not , and these are simply the sixth roots of a positive real number.
Adding up the solutions, we get
-SwordOfJustice
Solution 4
Using the fact that , we rewrite our equation as . Now, let represent . We know that ; hence, we have .
From here, we have two cases: , or . In the case that , we have hence . This gives one solution. Alternatively, if , then we have , giving solutions for each root of unity.
Therefore, the answer is .
- xHypotenuse
Video Solution by Power Solve
https://youtu.be/YXIH3UbLqK8?si=l8Ay2f0dMqSkjuQH&t=1975
Video Solution by OmegaLearn
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
2023 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
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.