2002 AMC 12A Problem 24

Attempt Problem 24 of the 2002 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 12A solutions, or check the answer key.

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24.

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

$1001$

$1002$

$2001$

$2002$

$2004$

Answer: E

Concepts:complex number roots of unity

Difficulty rating: 2170

Solution:

Let $z = a + bi.$ The equation is $z^{2002} = \overline{z}.$ Taking magnitudes, $|z|^{2002} = |z|,$ so $|z|\big(|z|^{2001} - 1\big) = 0,$ giving $|z| = 0$ or $|z| = 1.$

If $|z| = 0,$ then $(a, b) = (0, 0),$ one solution. If $|z| = 1,$ then $\overline{z} = \dfrac1z,$ so $z^{2002} = \dfrac1z,$ i.e. $z^{2003} = 1,$ which has $2003$ distinct roots.

Altogether there are $1 + 2003 = 2004$ ordered pairs.

Thus, the correct answer is E.

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