# 2012 AIME I Problems/Problem 6

## Problem 6

The complex numbers $z$ and $w$ satisfy $z^{13} = w,$ $w^{11} = z,$ and the imaginary part of $z$ is $\sin{\frac{m\pi}{n}}$, for relatively prime positive integers $m$ and $n$ with $m Find $n.$

## Solutions

### Solution 1

Substituting the first equation into the second, we find that $(z^{13})^{11} = z$ and thus $z^{142} = 1.$ So $z$ must be a $142$nd root of unity, and thus the imaginary part of $z$ will be $\sin{\frac{2m\pi}{142}} = \sin{\frac{m\pi}{71}}$ for some $m$ with $0 \le m < 142.$ But note that $71$ is prime and $m<71$ by the conditions of the problem, so the denominator in the argument of this value will always be $71$ and thus $n = \boxed{071.}$

### Solution 2

Note that $w^{143}=w$ and similar for $z$, and they are not equal to $0$ because the question implies the imaginary part is positive. Thus $w^{142}=z^{142}=1$, so the imaginary part of each is of the form $\sin\left(\frac{2 \pi k}{142}\right)$ where $k$ is a positive integer between $1$ and $141$ inclusive. This simplifies to $\sin\left(\frac{\pi k}{71}\right)$. Therefore, the imaginary part of $z$ is $\sin\left(\frac{\pi m}{71}\right)$, where $0 \le m < 71$. Since $071$ is prime, it is the only possible denominator, so $\boxed{n = 71}$.