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Here is the following question as is from the book:

For each of the values of n, list the elements of $\sqrt[n]{1}$ with their orders.

In my case, $n=9$. I know that when $n=4$, the elements are {$1$,$i$,$-1$,$-i$}. I drew a unit circle and on the x and y axis labeled $1,i,-1,-i$ beginnning on the x-axis and going counter-clockwise. This is how I approached it but for $n=9$:

I know that if n is a positive integer, then

$\zeta = \cos \frac{2\pi}{n} + i\sin\frac{2\pi}{n}$. So when $n=9$, I get

$\zeta = \cos \frac{2\pi}{9} + i\sin\frac{2\pi}{9}$ and then from here I conclude that the elements of $\sqrt[9]{1}$ are

$${\zeta,\zeta^2,\zeta^3,\zeta^4,\zeta^5,\zeta^6,\zeta^7,\zeta^8, 1}$$

Now when it comes to the orders of these 9 elements, I believe that for $\zeta$, we get

$o(\zeta) = 9$, but this is where I get stuck because I'm not sure of $o(\zeta^2)$. I am in headed in the right direction or am I way off track?

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    Ahhhh I see it! $l$ would be 9! – 2017-02-09
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    And then you can re-write the exponent as $(\zeta^9)^2$ – 2017-02-09
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    I deleted my comment because it was so similar to the answer below, but yes, that's it. Now for $\zeta^3$. What do you think? – 2017-02-09
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    $l=18$? It should be correct. – 2017-02-09
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    It is true that $(\zeta^3)^{18}=1$, but there are much smaller $l$ such that $(\zeta^3)^l=\zeta^{3l}=1$. – 2017-02-09
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    Ah $k=3$ and then $k=9$ and so on. – 2017-02-09
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    I mean $l$ not $k$. Sorry I was working with $k$ when discussing it with Joffan. – 2017-02-09

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You're on the right track. You just need to find the smallest $k>0$ such that $(\zeta^2)^k=1$, which implies $9\mid 2k$

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    So for the second order, $k=9$? – 2017-02-09
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    @Oliver821 You got it - and for $\zeta^3$? – 2017-02-09
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    I have $k=18$ for $o(\zeta^3)$. – 2017-02-09
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    @Oliver821 Ah, no - there is a much smaller $k$ such that $3k\mid 9$... – 2017-02-09
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    Ahh $k=3$. And for the next one $k=9$ since we have $4k|9$? – 2017-02-09
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    @Oliver821 I think you have the pattern now. – 2017-02-09
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    Yes I do. Much appreciated all of you. – 2017-02-09
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    Apologies that in my earlier comment I had the divisibility statement the wrong way round - I should have said "smaller $k$ such that $9\mid 3k$" – 2017-02-09
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    Yeah I just notice that too. Thanks for fixing it. – 2017-02-09
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Hint:

This is a (multiplicative) group isomorphic to the (additive) cyclic group $\mathbf Z/9\mathbf Z$. What are the generators of this additive group?