2012 AIME I Problems/Problem 6
The complex numbers and satisfy and the imaginary part of is , for relatively prime positive integers and with Find
Substituting the first equation into the second, we find that and thus We know that because we are given the imaginary part of so we can divide by to get So, must be a nd root of unity, and thus, by De Moivre's theorem, the imaginary part of will be of the form where Note that is prime and by the conditions of the problem, so the denominator in the argument of this value will always be Thus,
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
Note: Even Richard Rusczyk said this was a low quality problem. It is a 10 second problem if you know roots of unity, but impossible to to if you don't know them.