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

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(→‎Solution 3)
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== Solution 3 ==
 
== Solution 3 ==
  
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Notice that r=0 or r=1 for this to be true. We know this because we are taking magnitude to the 2003rd 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.
+
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 r=0, we must have (0,0) for (a,b). No exceptions.
+
Now if <math>r=0</math>, then <math>(a,b)</math> must be <math>(0,0)</math>.
  
−
However if r=1, we then have:
+
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 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 due to DeMoivre's Theorem and also we do <math>-\theta</math> because it is a reflection, angles therefore is negative.
+
<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:
 
We then write:
  
−
<math>cos(2002 \theta) = cos(360-\theta)</math> which has solution of <math>\theta = \frac{360}{2003}</math>.  
+
<math>\cos(2002 \theta) = \cos(360-\theta)</math> which has solution of <math>\theta = \frac{360}{2003}</math>.  
  
 
We can also write:
 
We can also write:
  
−
<math>cos(2002 \theta) = cos(720-\theta)</math> which has solution <math>\theta = \frac{720}{2003}</math>.
+
<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 0 to infinity inclusive.
+
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 n=2003, that is 360 which is also 0! Hence we only have n=0 to 2002 as original solutions, or 2003 solutions.
+
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>.
  
−
1+2003 = <math>\boxed{2004}</math>.
 
  
 
Solution by Blackhawk 9-10-17
 
Solution by Blackhawk 9-10-17
 +
 +
<math>\LaTeX</math>ed (with some edits) by PhunsukhWangdu 7/27/22
  
 
== See Also ==
 
== See Also ==

Revision as of 02:07, 27 July 2022

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

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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