Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm told that $ω^n$ = $ω^{(n+3k)}$

Also

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.

share|improve this question
1  
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
1  
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
1  
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

1 Answer 1

up vote 2 down vote accepted

$\omega^{n+3k}=\omega^n\cdot(\omega^3)^k$.

share|improve this answer
    
Is this method not more complicated that the formula I was given? –  Adegoke A Apr 19 '13 at 13:24
1  
@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 $\omega^n=\omega^v$ –  Mr.ØØ7 Apr 19 '13 at 13:31

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.