2002 AMC 12A Problems/Problem 24

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)

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
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