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I'm looking for a proof of this statement. I just don't know how to approach it. I recognize that $z$ has $a$ and $b$ roots of unity, but I can't seem to figure out what that tells me.

If $z \in \mathbb{C}$ satisfies $z^a = 1$ and $z^b = 1$ then $z^{gcd(a,b)} = 1$.

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    @Bill Dubuque: I deleted and rewrote, because of a typo (Dubugue!).2012-03-01

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Hint $\:$ The set of $\rm\:n\in \mathbb Z$ such that $\rm\:z^n = 1\:$ is closed under subtraction so it is closed under $\rm\:gcd$.

Recall gcds may be computed by repeated subtraction (anthyphairesis, Euclidean algorithm)

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    Very nice short conceptual solution.2012-03-01