2021 AMC 12B Problems/Problem 18
Contents
Problem
Let be a complex number satisfying What is the value of
Solution 1
Using the fact , the equation rewrites itself as As the two quantities in the parentheses are real, both quantities must equal so
Solution 2
Let ,
By the equation given in the problem
Therefore, and
, ,
Solution 3
Let . Then . From the answer choices， we know that is real and , so . Then we have Plugging the above back to the original equation, we have So .
~Sequoia
Solution 4 (Funny Observations)
There are actually several ways to see that I present two troll ways of seeing it, and a legitimate way of checking.
Rewrite using
Symmetric in and so if is a sol, then so is
TROLL OBSERVATION #1: ALL THE ANSWERS ARE REAL. THUS, which means they must be conjugates and so
TROLL OBSERVATION #2: Note that because either solution must give the same answer! which means that
Alternatively, you can check: Let and Thus, we have and the discriminant of this must be nonnegative as is real. Thus, or which forces as claimed.
Thus, we plug in and get: ie. or which means and that's our answer since we know
- ccx09
Solution 5
Observe that all the answer choices are real. Therefore, and must be complex conjugates as this is the only way for both their sum (one of the answer choices) and their product () to be real. Thus . We will test all the answer choices, starting with . Suppose the answer is . If then and . Note that if works, then so does . It is relatively easy to see that if , then and . Thus the condition is satisfied for , and the answer is .
Video Solution by OmegaLearn (Using Complex Number Identities)
https://youtu.be/AEbMTTGEZV4 ~pi_is_3.14
Video Solution
https://youtu.be/Yw-IJvfrT_U ~MathProblemSolvingSkills.com
(includes review of complex numbers)
Video Solution by Punxsutawney Phil
See Also
2021 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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.