AoPS Wiki talk:Problem of the Day/September 11, 2011

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Solution

We put the number in Polar form: $r(cis{\theta})$. Note that $r(cis{\theta})$ is shorthand for $r(\cos{\theta}+i\sin{\theta})$.Then $(a+bi)^{2002}=r^{2002}(cis(2002\theta))=r(cis(-\theta))$. Since $r^{2002}=r$, either $r=0$ or $r=\pm1$. If $r=0$, then we have $(a,b)=(0,0)$. We can now just focus on $r=1$, since $r=-1$ gives the same solutions.

Since $2002\theta\equiv -\theta\bmod{360}$, $2003\theta=360n$ for some $n$. Each value of $n$ between 1 and 2003 gives a unique solution, so $r=1$ has $2003$ solutions for $\theta$. Thus, there are $2003+1=\boxed{2004}$ solutions; $\boxed{\mathbb E}$.

See Also

Polar form

Cosine

Sine

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