Difference between revisions of "2023 AMC 12A Problems/Problem 14"
(→Solution 3 (Rectangular Form)) |
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-Benedict T (countmath 1) | -Benedict T (countmath 1) | ||
− | ==Solution 3 (Rectangular Form)== | + | ==Solution 3 (Rectangular Form, similar to Solution 1)== |
Let <math>z = a+bi</math>. | Let <math>z = a+bi</math>. | ||
Line 37: | Line 37: | ||
Then, our equation becomes: | Then, our equation becomes: | ||
<math>(a+bi)^5=a-bi</math> | <math>(a+bi)^5=a-bi</math> | ||
+ | |||
+ | Note that since every single term in the expansion contains either an <math>a</math> or <math>b</math>, simply setting <math>a=0</math> and <math>b=0</math> yields a solution. | ||
+ | |||
+ | Now, considering the other case that either <math>a</math> or <math>b</math> does not equal <math>0</math>: | ||
+ | |||
+ | Multiplying both sides by <math>a+bi</math> (or <math>z</math>), we get: | ||
+ | <math>(a+bi)^6=a^2+b^2</math> (since <math>i^2=-1</math>). | ||
+ | |||
+ | Substituting <math>z</math> back into the left hand side, we get: | ||
+ | <math>z^6=a^2+b^2</math> | ||
+ | |||
+ | Note that this will have 6 distinct, non-zero solutions since in this case, we consider that either <math>a</math> or <math>b</math> is not <math>0</math>. | ||
+ | |||
+ | Adding up the solutions, we get <math>1+6=E)7.</math> | ||
==Video Solution by OmegaLearn== | ==Video Solution by OmegaLearn== |
Revision as of 22:14, 3 April 2024
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 .
Adding up the solutions, we get
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.