2002 AMC 12A Problems/Problem 24

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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 $2003$ -th complex roots of unity, and there are $2003$ of those.

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

2002 AMC 12A (Problems • 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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2002 AMC 12A Problems/Problem 24

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