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This question already has an answer here:

$\{z\in \mathbb{C} \mid z^{60}=-1, z^k \not=-1 \text{ for } 0<k<60\}$ I tried to solve this but I've no idea on this type of questions, how to get start or in which way should I proceed. How can i do this?

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marked as duplicate by 6005, mickep, Cameron Buie, Aaron Maroja, user147263 Sep 15 '15 at 19:34

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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That are just the primitive 120th roots, so we are looking for $$\varphi(120)=\varphi(8)\cdot\varphi(3)\cdot\varphi(5)=4\cdot 2\cdot 4=32.$$

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Hint: First prove: \begin{align*} z^{60} = -1 \text{ and } z^k \ne -1 \text{ for } 0 < k < 60 &\iff z^{120} = 1 \text{ and } z^k \ne 1 \text{ for } 0 < k < 120. \end{align*} Then note that $$ \{z^{120} = 1 \text{ and } z^k \ne 1 \text{ for } 0 < k < 120\} $$ is the set of primitive $120$th roots of unity.

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