2002 AMC 12A Problem 24

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.