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

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 $a+bi=\cos\theta+\sin\theta$ , $(a+bi)^{2002}= \cos(2002\theta)+i\sin(2002\theta)= \cos\theta-i\sin\theta= \cos(-\theta)+i\sin(-\theta) = a - bi$ , 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 $1+2003 = \boxed{\textbf{(E)}\ 2004}$ .

2002 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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2002 AMC 12A Problems/Problem 24

Contents

Problem

Solution

Solution 2

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