MedVision ad

Complex Numbers-roots of unity (1 Viewer)

Bozza555

New Member
Joined
Nov 23, 2011
Messages
29
Gender
Male
HSC
2012
Could someone explain how to work these sort of questions out please? even when i look at the answers i don't understand haha

<a href="http://www.codecogs.com/eqnedit.php?latex=if~ \omega ~ is~a~complex~cube~root~of~unity,show~that~ (1@plus;\omega )^3(1@plus;2\omega@plus;2\omega^2)=1" target="_blank"><img src="http://latex.codecogs.com/gif.latex?if~ \omega ~ is~a~complex~cube~root~of~unity,show~that~ (1+\omega )^3(1+2\omega+2\omega^2)=1" title="if~ \omega ~ is~a~complex~cube~root~of~unity,show~that~ (1+\omega )^3(1+2\omega+2\omega^2)=1" /></a>
 

nightweaver066

Well-Known Member
Joined
Jul 7, 2010
Messages
1,585
Gender
Male
HSC
2012
Since w is a complex cube root of unity, w satisfies z^3 = 1

So w^3 = 1

w^3 - 1 = 0

(w -1)(w^2 + w + 1) = 0

But w is complex, so w^2 + w + 1 = 0

Using this for that question,











Basically for these sorts of questions, keep using w^3 = 1 and w^2 + w + 1 = 0 to get a nice, simple answer.
 

Bozza555

New Member
Joined
Nov 23, 2011
Messages
29
Gender
Male
HSC
2012
ohh i get it now, i wasn't sure why w^2+w+1=0 and why it couldn't be w-1=0 but i forgot that w is complex
thanks :)
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top