Difference between revisions of "2023 AMC 12A Problems/Problem 14"
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==Solution 1== | ==Solution 1== | ||
− | When <math>z^5=\overline{z}</math>, there are two conditions: either <math>z=0</math> or <math>z\neq 0</math>. When <math>z\neq 0</math>, since <math>z^5=\overline{z}</math>, <math>|z|=1</math>. <math>z^5\cdot z=z^6=\overline{z}\cdot z=|z|^2=1</math>. Consider the <math>r(\cos \theta +i\sin \theta)</math> form, when <math>z^6=1</math>, there are 6 different solutions for <math>z</math>. Therefore, the number of complex numbers satisfying <math>z^5=\bar{z}</math> is <math>\boxed{\textbf{(E)} 7}</math>. | + | When <math>z^5=\overline{z}</math>, there are two conditions: either <math>z=0</math> or <math>z\neq 0</math>. When <math>z\neq 0</math>, since <math>|z^5|=|\overline{z}|</math>, <math>|z|=1</math>. <math>z^5\cdot z=z^6=\overline{z}\cdot z=|z|^2=1</math>. Consider the <math>r(\cos \theta +i\sin \theta)</math> form, when <math>z^6=1</math>, there are 6 different solutions for <math>z</math>. Therefore, the number of complex numbers satisfying <math>z^5=\bar{z}</math> is <math>\boxed{\textbf{(E)} 7}</math>. |
~plasta | ~plasta | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Let <math>z = re^{i\theta}.</math> We now have <math>\overline{z} = re^{-i\theta},</math> and want to solve | ||
+ | |||
+ | <cmath>r^5e^{5i\theta} = re^{-i\theta}.</cmath> | ||
+ | |||
+ | From this, we have <math>r = 0</math> as a solution, which gives <math>z = 0</math>. If <math>r\neq 0</math>, then we divide by it, yielding | ||
+ | |||
+ | <cmath>r^4e^{5i\theta} = e^{-i\theta}.</cmath> | ||
+ | |||
+ | Dividing both sides by <math>e^{-i\theta}</math> yields <math>r^4e^{6i\theta} = 1</math>. | ||
+ | Taking the magnitude of both sides tells us that <math>r^4 = 1</math>, so <math>r^2 = \pm 1</math>. However, if <math>r^2 = -1</math>, then <math>r = \pm i</math>, but <math>r</math> must be real. Therefore, <math>r^2 = 1</math>. | ||
+ | |||
+ | Multiplying both sides by <math>r^2</math>, | ||
+ | |||
+ | <cmath>r^6\cdot e^{6i\theta} = z^6 = 1.</cmath> | ||
+ | |||
+ | Each of the <math>6</math>th roots of unity is a solution to this, so there are <math>6 + 1 = \boxed{\textbf{(D)}\ 7}</math> solutions. | ||
+ | |||
+ | -Benedict T (countmath 1) | ||
+ | |||
+ | ==Solution 3 (Rectangular Form, similar to Solution 1)== | ||
+ | |||
+ | Let <math>z = a+bi</math>. | ||
+ | |||
+ | Then, our equation becomes: | ||
+ | <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>, 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> | ||
+ | |||
+ | -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== | ||
https://youtu.be/rbdIrmOyczk | https://youtu.be/rbdIrmOyczk | ||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/m627Mjp3PkM | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
==See also== | ==See also== | ||
{{AMC12 box|year=2023|ab=A|num-b=13|num-a=15}} | {{AMC12 box|year=2023|ab=A|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} |
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.