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

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; $\boxed{\mathbb E}$ .

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

Solution

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