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Difference between revisions of "2002 AMC 12A Problems/Problem 24"

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</math>
 
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== Solution ==
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== 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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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>.
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 +
== Solution 2 ==
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 +
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>,
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 +
<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>
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 +
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>.
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== Solution 3 ==
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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.
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 +
Now if <math>r=0</math>, then <math>(a,b)</math> must be <math>(0,0)</math>.
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 +
However if <math>r=1</math>, we then have:
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 +
<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.
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 +
We then write:
 +
 +
<math>\cos(2002 \theta) = \cos(360-\theta)</math> which has solution of <math>\theta = \frac{360}{2003}</math>.
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 +
We can also write:
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 +
<math>\cos(2002 \theta) = \cos(720-\theta)</math> which has solution <math>\theta = \frac{720}{2003}</math>.
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 +
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.
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 +
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.
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<math>1+2003 = \boxed{2004}</math>.
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Solution by Blackhawk 9-10-17
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<math>\LaTeX</math>ed (with some edits) by PhunsukhWangdu 7/27/22
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== 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
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<cmath>z^{2003} = |z|^2 = r^{2003}e^{2023i\theta} = r^2,</cmath>
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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>.
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-Benedict T(countmath1)
  
 
== 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 19:54, 14 March 2024

Problem

Find the number of ordered pairs of real numbers $(a,b)$ such that $(a+bi)^{2002} = a-bi$.

$\text{(A) }1001 \qquad \text{(B) }1002 \qquad \text{(C) }2001 \qquad \text{(D) }2002 \qquad \text{(E) }2004$

Solution 1

Let $s=\sqrt{a^2+b^2}$ be the magnitude of $a+bi$. Then the magnitude of $(a+bi)^{2002}$ is $s^{2002}$, while the magnitude of $a-bi$ is $s$. We get that $s^{2002}=s$, hence either $s=0$ or $s=1$.

For $s=0$ we get a single solution $(a,b)=(0,0)$.

Let's now assume that $s=1$. Multiply both sides by $a+bi$. The left hand side becomes $(a+bi)^{2003}$, the right hand side becomes $(a-bi)(a+bi)=a^2 + b^2 = 1$. Hence the solutions for this case are precisely all the $2003$rd complex roots of unity, and there are $2003$ of those.

The total number of solutions is therefore $1+2003 = \boxed{2004}$.

Solution 2

As in the other solution, split the problem into when $s=0$ and when $s=1$. When $s=1$ and $a+bi=\cos\theta+i\sin\theta$,

$(a+bi)^{2002}= \cos(2002\theta)+i\sin(2002\theta)$ $=a-bi= \cos\theta-i\sin\theta= \cos(-\theta)+i\sin(-\theta)$

so we must have $2002\theta=-\theta+2\pi k$ and hence $\theta=\frac{2\pi k}{2003}$. Since $\theta$ is restricted to $[0,2\pi)$, $k$ can range from $0$ to $2002$ inclusive, which is $2002-0+1=2003$ values. Thus the total is $1+2003 = \boxed{\textbf{(E)}\ 2004}$.

Solution 3

Let $r$ be the magnitude of $a+bi$. Notice that $r$ must be either $0$ or $1$ for this to be true, as shown in the above solutions. We know this because we are taking magnitude to the $2003$rd power, and if the magnitude of $a+bi$ is larger than $1$, it will increase and if it is smaller than $1$ it will decrease. However, the magnitude on the RHS is still $r$, so this is not possible. Again, only $r=0$ and $r=1$ satisfy the equation.

Now if $r=0$, then $(a,b)$ must be $(0,0)$.

However if $r=1$, we then have:

$\cos(2002 \theta) = \cos(-\theta)$. This has solution of $\theta = 0$. This would represent the number $1+0i$, with conjugate $1-0i$. This works because the magnitude is the same and the angle is nothing anyways. We multiply angle by $2002$ through De Moivre's Theorem and also we do $-\theta$ because it is a reflection, angles therefore is negative.

We then write:

$\cos(2002 \theta) = \cos(360-\theta)$ which has solution of $\theta = \frac{360}{2003}$.

We can also write:

$\cos(2002 \theta) = \cos(720-\theta)$ which has solution $\theta = \frac{720}{2003}$.

We notice that it is simply headed upwards and the answer is of the form $\frac{720}{2003} n$, where n is some integer from $0$ to infinity, inclusive.

Well wait, it repeats itself $n=2003$, that is $360$ which is also $0$! Hence we only have $n=0$ to $2002$ as original solutions, or $2003$ solutions.

$1+2003 = \boxed{2004}$.


Solution by Blackhawk 9-10-17

$\LaTeX$ed (with some edits) by PhunsukhWangdu 7/27/22


Solution 4

Let $z = a + bi = re^{i\theta}$ where $r = |z|$ and $0\leq \theta < 2\pi.$ We want to solve $z^{2002} = \overline{z}.$ Since $z\overline{z} = |z|^2,$ we multiply by $z$ on both sides to get \[z^{2003} = |z|^2 = r^{2003}e^{2023i\theta} = r^2,\] from which we get $r\in \{0, 1\}$ (since $r\in \mathbb{R}$), immediately yielding $z = 0$ as an answer. If $r=1$ then we have $z^{2003} = 1$, which each of the $2003$rd roots of unity satisfies. Altogether, there are $1 + 2003 = \boxed{\textbf{(E)\ 2004}}$ values of $z$, each of which correspond to a unique ordered pair $(a, b)$.

-Benedict T(countmath1)

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

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