Difference between revisions of "2002 AMC 12A Problems/Problem 24"
(New page: == Problem == Find the number of ordered pairs of real numbers <math>(a,b)</math> such that <math>(a+bi)^{2002} = a-bi</math>. <math> \text{(A) }1001 \qquad \text{(B) }1002 \qquad \text{...) |
Mathboy282 (talk | contribs) (→Solution 5) |
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− | == Solution == | + | == Solution 1 == |
Let <math>s=\sqrt{a^2+b^2}</math> be the magnitude of <math>a+bi</math>. Then the magnitude of <math>(a+bi)^{2002}</math> is <math>s^{2002}</math>, while the magnitude of <math>a-bi</math> is <math>s</math>. We get that <math>s^{2002}=s</math>, hence either <math>s=0</math> or <math>s=1</math>. | Let <math>s=\sqrt{a^2+b^2}</math> be the magnitude of <math>a+bi</math>. Then the magnitude of <math>(a+bi)^{2002}</math> is <math>s^{2002}</math>, while the magnitude of <math>a-bi</math> is <math>s</math>. We get that <math>s^{2002}=s</math>, hence either <math>s=0</math> or <math>s=1</math>. | ||
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For <math>s=0</math> we get a single solution <math>(a,b)=(0,0)</math>. | For <math>s=0</math> we get a single solution <math>(a,b)=(0,0)</math>. | ||
− | Let's now assume that <math>s=1</math>. Multiply both sides by <math>a+bi</math>. The left hand side becomes <math>(a+bi)^{2003}</math>, the right hand side becomes <math>(a-bi)(a+bi)=a^2 + b^2 = 1</math>. Hence the solutions for this case are precisely all the <math>2003</math> | + | Let's now assume that <math>s=1</math>. Multiply both sides by <math>a+bi</math>. The left hand side becomes <math>(a+bi)^{2003}</math>, the right hand side becomes <math>(a-bi)(a+bi)=a^2 + b^2 = 1</math>. Hence the solutions for this case are precisely all the <math>2003</math>rd complex roots of unity, and there are <math>2003</math> of those. |
The total number of solutions is therefore <math>1+2003 = \boxed{2004}</math>. | The total number of solutions is therefore <math>1+2003 = \boxed{2004}</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | As in the other solution, split the problem into when <math>s=0</math> and when <math>s=1</math>. When <math>s=1</math> and <math>a+bi=\cos\theta+i\sin\theta</math>, | ||
+ | |||
+ | <math>(a+bi)^{2002}= \cos(2002\theta)+i\sin(2002\theta) </math> <math> =a-bi= \cos\theta-i\sin\theta= \cos(-\theta)+i\sin(-\theta)</math> | ||
+ | |||
+ | so we must have <math>2002\theta=-\theta+2\pi k</math> and hence <math>\theta=\frac{2\pi k}{2003}</math>. Since <math>\theta</math> is restricted to <math>[0,2\pi)</math>, <math>k</math> can range from <math>0</math> to <math>2002</math> inclusive, which is <math>2002-0+1=2003</math> values. Thus the total is <math>1+2003 = \boxed{\textbf{(E)}\ 2004}</math>. | ||
+ | |||
+ | == Solution 3 == | ||
+ | |||
+ | Let <math>r</math> be the magnitude of <math>a+bi</math>. Notice that <math>r</math> must be either <math>0</math> or <math>1</math> for this to be true, as shown in the above solutions. We know this because we are taking magnitude to the <math>2003</math>rd power, and if the magnitude of <math>a+bi</math> is larger than <math>1</math>, it will increase and if it is smaller than <math>1</math> it will decrease. However, the magnitude on the RHS is still <math>r</math>, so this is not possible. Again, only <math>r=0</math> and <math>r=1</math> satisfy the equation. | ||
+ | |||
+ | Now if <math>r=0</math>, then <math>(a,b)</math> must be <math>(0,0)</math>. | ||
+ | |||
+ | However if <math>r=1</math>, we then have: | ||
+ | |||
+ | <math>\cos(2002 \theta) = \cos(-\theta)</math>. This has solution of <math>\theta = 0</math>. This would represent the number <math>1+0i</math>, with conjugate <math>1-0i</math>. This works because the magnitude is the same and the angle is nothing anyways. We multiply angle by <math>2002</math> through De Moivre's Theorem and also we do <math>-\theta</math> because it is a reflection, angles therefore is negative. | ||
+ | |||
+ | We then write: | ||
+ | |||
+ | <math>\cos(2002 \theta) = \cos(360-\theta)</math> which has solution of <math>\theta = \frac{360}{2003}</math>. | ||
+ | |||
+ | We can also write: | ||
+ | |||
+ | <math>\cos(2002 \theta) = \cos(720-\theta)</math> which has solution <math>\theta = \frac{720}{2003}</math>. | ||
+ | |||
+ | We notice that it is simply headed upwards and the answer is of the form <math>\frac{720}{2003} n</math>, where n is some integer from <math>0</math> to infinity, inclusive. | ||
+ | |||
+ | Well wait, it repeats itself <math>n=2003</math>, that is <math>360</math> which is also <math>0</math>! Hence we only have <math>n=0</math> to <math>2002</math> as original solutions, or <math>2003</math> solutions. | ||
+ | |||
+ | <math>1+2003 = \boxed{2004}</math>. | ||
+ | |||
+ | |||
+ | Solution by Blackhawk 9-10-17 | ||
+ | |||
+ | <math>\LaTeX</math>ed (with some edits) by PhunsukhWangdu 7/27/22 | ||
+ | |||
+ | |||
+ | == Solution 4 == | ||
+ | Let <math>z = a + bi = re^{i\theta}</math> where <math>r = |z|</math> and <math>0\leq \theta < 2\pi.</math> We want to solve <math>z^{2002} = \overline{z}.</math> Since <math>z\overline{z} = |z|^2,</math> we multiply by <math>z</math> on both sides to get | ||
+ | <cmath>z^{2003} = |z|^2 = r^{2003}e^{2023i\theta} = r^2,</cmath> | ||
+ | from which we get <math>r\in \{0, 1\}</math> (since <math>r\in \mathbb{R}</math>), immediately yielding <math>z = 0</math> as an answer. If <math>r=1</math> then we have <math>z^{2003} = 1</math>, which each of the <math>2003</math>rd roots of unity satisfies. Altogether, there are <math>1 + 2003 = \boxed{\textbf{(E)\ 2004}}</math> values of <math>z</math>, each of which correspond to a unique ordered pair <math>(a, b)</math>. | ||
+ | |||
+ | -Benedict T(countmath1) | ||
+ | |||
+ | ==Solution 5== | ||
+ | Let <math>z=a+bi \Rightarrow z^{2022}=\overline{z}.</math> Hence, taking the magnitude of both sides, we must have: | ||
+ | <cmath>|z^{2022}|=|\overline{z}|=|z| \Rightarrow |z|=0, 1.</cmath> | ||
+ | If <math>|z|=0,</math> then <math>z=0.</math> This gives one solution. | ||
+ | |||
+ | If <math>|z|=1,</math> then we do the following. Multiply the first equation by <math>z</math> on both sides such that you obtain <math>z^{2003}=z \overline{z} =|z|^2.</math> | ||
+ | |||
+ | Now, it is clear that if <math>|z|=1,</math> we need <math>z^{2003}=1.</math> We have 2003 solutions for this, i.e. for each <math>i\in\{0,1,\cdots,2022\}</math> for <math>e^{2i\pi/2003}.</math> | ||
+ | |||
+ | A total of <math>2003+1=2004</math> solutions. | ||
+ | |||
+ | ~mathboy282 | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2002|ab=A|num-b=23|num-a=25}} | {{AMC12 box|year=2002|ab=A|num-b=23|num-a=25}} | ||
+ | {{MAA Notice}} |
Latest revision as of 02:42, 16 November 2024
Problem
Find the number of ordered pairs of real numbers such that .
Solution 1
Let be the magnitude of . Then the magnitude of is , while the magnitude of is . We get that , hence either or .
For we get a single solution .
Let's now assume that . Multiply both sides by . The left hand side becomes , the right hand side becomes . Hence the solutions for this case are precisely all the rd complex roots of unity, and there are of those.
The total number of solutions is therefore .
Solution 2
As in the other solution, split the problem into when and when . When and ,
so we must have and hence . Since is restricted to , can range from to inclusive, which is values. Thus the total is .
Solution 3
Let be the magnitude of . Notice that must be either or for this to be true, as shown in the above solutions. We know this because we are taking magnitude to the rd power, and if the magnitude of is larger than , it will increase and if it is smaller than it will decrease. However, the magnitude on the RHS is still , so this is not possible. Again, only and satisfy the equation.
Now if , then must be .
However if , we then have:
. This has solution of . This would represent the number , with conjugate . This works because the magnitude is the same and the angle is nothing anyways. We multiply angle by through De Moivre's Theorem and also we do because it is a reflection, angles therefore is negative.
We then write:
which has solution of .
We can also write:
which has solution .
We notice that it is simply headed upwards and the answer is of the form , where n is some integer from to infinity, inclusive.
Well wait, it repeats itself , that is which is also ! Hence we only have to as original solutions, or solutions.
.
Solution by Blackhawk 9-10-17
ed (with some edits) by PhunsukhWangdu 7/27/22
Solution 4
Let where and We want to solve Since we multiply by on both sides to get from which we get (since ), immediately yielding as an answer. If then we have , which each of the rd roots of unity satisfies. Altogether, there are values of , each of which correspond to a unique ordered pair .
-Benedict T(countmath1)
Solution 5
Let Hence, taking the magnitude of both sides, we must have: If then This gives one solution.
If then we do the following. Multiply the first equation by on both sides such that you obtain
Now, it is clear that if we need We have 2003 solutions for this, i.e. for each for
A total of solutions.
~mathboy282
See Also
2002 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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.