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I'm told that $ω^n$ = $ω^{(n+3k)}$


and $k = 0,1,2.$

How does $ω^{(-1)} = ω^{(2)}$, in this equation?

Taking $ω^{(-k)}$, $k = 1$

$= ω^{(n + 3(-1))} = ω^{(n -3)}$

Also what does $n$ equal to?

I also have this information:

More generally, $ω^n = ω^{(n + 3k)}$ for all integers n and k. Now $-1 = 2 + 3 * (-1)$, so $ω^{-1} = ω^2$.

Where did the $2$ come from?

Btw can someone please edit this, cause I don't seem to be getting it.

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Here your $\omega=\zeta_3=e^{2\pi i/3}$, which is a cube root of unity, so $\omega^3=\omega\cdot\omega^2=1$, so $\omega^2$ is the inverse $\omega^{-1}$. Is that what your concern is? – Warren Moore Apr 19 '13 at 13:21
Yeah but how does the inversing work? – Adegoke A Apr 19 '13 at 13:24
Well the inverse of $\omega$ is defined to be the element such that $\omega\cdot\omega^{-1}=1$. Since $\omega$ satisfies $\omega^3=1$, $\omega\cdot\omega^2=1$. Surely then $\omega^2$ satisfies what it means to be the inverse of $\omega$? – Warren Moore Apr 19 '13 at 13:26
Okay I get that now. But would I do it for another number apart for $-1$ – Adegoke A Apr 19 '13 at 13:27
You can do it for any number that is congruent to $-1$ modulo $3$. As you've already mentioned $\omega^n=\omega^{n+3k}$, so $\omega^{-1}=\omega^2=\omega^5=\omega^8=\cdots$ etc. As long as the power can be written as $3k-1$ for some integer $k$, then it will always be the inverse of $\omega$. – Warren Moore Apr 19 '13 at 13:29
up vote 2 down vote accepted


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Is this method not more complicated that the formula I was given? – Adegoke A Apr 19 '13 at 13:24
@AdegokeA: not if you realize that $\omega^3=1$ – Ross Millikan Apr 19 '13 at 13:28
Oh I see. But about $n$? – Adegoke A Apr 19 '13 at 13:29
@AdegokeA if $n=3u+v$ , $0\le v<3$ then $\omega^n=\omega^v$ – ABC Apr 19 '13 at 13:31

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