2023 AMC 12A Problems/Problem 14
Contents
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
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 |
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All AMC 12 Problems and Solutions |
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