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,
https://www.youtube.com/watch?v=DMka35X-3WI&list=PLyhPcpM8aMvIo_foUDwmXnQClMHngjGto&index=6 (Solution by Richard Rusczyk) - AMBRIGGS
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