Difference between revisions of "2023 AMC 12A Problems/Problem 14"
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<math>z^6=a^2+b^2</math> | <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>. | + | 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>, and these are simply the sixth roots of a positive real number. |
Adding up the solutions, we get <math>1+6=</math> <math>\boxed{\textbf{(E)} 7}</math> | Adding up the solutions, we get <math>1+6=</math> <math>\boxed{\textbf{(E)} 7}</math> | ||
+ | |||
+ | -SwordOfJustice | ||
+ | |||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | Using the fact that <math>z\bar{z}=|z|^2</math>, we rewrite our equation as <math>z^6=|z|^2</math>. Now, let <math>r</math> represent <math>|z|</math>. We know that <math>r^6 = r^2</math>; hence, we have <math>r^6-r^2=r^2(r^4-1)=0</math>. | ||
+ | |||
+ | From here, we have two cases: <math>r=0</math>, or <math>r=1</math>. In the case that <math>r=0</math>, we have <math>z^6=0</math> hence <math>z=0</math>. This gives one solution. Alternatively, if <math>r=1</math>, then we have <math>z^6=1</math>, giving <math>6</math> solutions for each root of unity. | ||
+ | |||
+ | |||
+ | Therefore, the answer is <math>6+1=</math> <math>\boxed{\textbf{(E)} 7}</math>. | ||
+ | |||
+ | |||
+ | - xHypotenuse | ||
+ | |||
+ | ==Video Solution (easy to digest) by Power Solve== | ||
+ | https://youtu.be/YXIH3UbLqK8?si=l8Ay2f0dMqSkjuQH&t=1975 | ||
==Video Solution by OmegaLearn== | ==Video Solution by OmegaLearn== |
Revision as of 20:37, 10 June 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
, 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 (easy to digest) 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.