My Math Forum Factoring the omega (cube roots of unity)
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 August 16th, 2013, 12:22 AM #1 Member   Joined: Jul 2013 Posts: 58 Thanks: 0 Factoring the omega (cube roots of unity) $(2-w)(2-w^2)(2-w^{10})(2-w^{11})= 49$ Doing L.H.S. $(4-2w^2-2w+w^3) (2-w^{10})(2-w^{11})$ $(5)-2w(1+w) (2-w^{10}) (2-w^{11})$ $(5+2w^3)(2-w^{10})(2-w^{11})$ $(7)(4-2w^{11}-2w^{10}+w^{21})$ $(7)(4-2w^3(w^8+w^7)+1)$ $(7)((4-2)(w^8+w^7)+1)$ <---- I have doubt in this line.. if (4-2) = 0, then L.H.S. = 0 ? Or am I doing wrong factoring... in short , I am not able to prove it.
 August 16th, 2013, 01:22 AM #2 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: Factoring the omega (cube roots of unity) You are close, but your last line has errors about parenthesis use. I would use brackets to be more precise. Working on your next to last line which is correct, $7 [4 - 2(w^8 + w^7) + 1]$ $7 [5 - 2(w^2 + w)]$ $7 [5 - 2(-1)]$ Back to parentheses again, hee, hi, ho ... ha-ha! $7 (5 + 2)$ $7 \cdot 7= 49$ $Q. E. D.$
 August 16th, 2013, 03:26 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,835 Thanks: 2162 As (2 - ?^10)(2 - ?^11) = (2 - ?)(2 - ?²) = (2 - ?)(3 + ?) = 6 - ? - ?² = 7, the original L.H.S. = 7² = 49.
 August 16th, 2013, 03:51 AM #4 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: Factoring the omega (cube roots of unity) Absolutely , [color=#BF0000]skipjack's[/color] approach is more elegant.
 August 16th, 2013, 08:02 PM #5 Member   Joined: Jul 2013 Posts: 58 Thanks: 0 Re: Factoring the omega (cube roots of unity) thanks agent and skipjack. I always try to factor omega using w^3 i.e. 1 but I have doubt regarding w and w^2. Since you said w and w^2 are complex conjugates to each other. w = 1 w^2 = -1 ?
 August 16th, 2013, 08:38 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,835 Thanks: 2162 No. Complex conjugates (in general) have the form x + iy and x - iy, where x and y are real. For any polynomial in ? with real coefficients, its complex conjugate can be obtained by replacing ? with ?².
August 16th, 2013, 10:47 PM   #7
Math Team

Joined: Jul 2011
From: North America, 42nd parallel

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Re: Factoring the omega (cube roots of unity)

It is informative to look at the 3 cube roots of unity in all their splendor.

$\ r_0 \= \ w^0 \ = \ 1$

$\ r_1 \= \ w^1 \ = \ - \frac{1}{2} \ + \ \frac{ \sqrt{3}}{2}i$

$\ r_2 \= \ w^2 \ = \ - \frac{1}{2} \ - \ \frac{ \sqrt{3}}{2}i$

From this you can see that w and w^2 are indeed conjugates but that will not always be the case. You can take advantage of symmetry and the fact that the roots of unity occur symmetrically on the unit circle centered at the origin of the complex plane to figure out which roots are conjugates. Look at example below where we have the seven 7th roots of unity.

Working counter clockwise from the root on the horizontal real axis let $r_0 \= \ w^0 \ = \ 1$ , then you can see that

w and w^6 are complex conjugates

w^2 and w^5 are complex conjugates

w^3 and w^4 are complex conjugates

Note* 1 can be considered as IT'S OWN COMPLEX CONJUGATE so the matter of complex conjugate pairs due to symmetry is 'tidied up'

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