Roots of Unity (1 Viewer)

[shimmerz_777]

i dont remember studying them, i think that they popped up in the 1999 HSC exam paper q 7 or 8, any pointers on what they are/ how to use them cause i cant see them in my text book. thanks

 

[ianc]

Roots of unity are simply where z^{[FONT=&quot]n[/FONT]} = 1


They appear in the complex numbers topic and also in polynomials


To find the roots:

• z=r^{[FONT=&quot]1/n [/FONT]}cis[1/n(2kπ)] for all k from 0 to (n-1)
  OR
• [FONT=&quot]Use polynomial stuff[/FONT]

[FONT=&quot]
[/FONT]Just a few handy things to bear in mind when doing questions involving them:

• All the roots have a modulus of 1, hence they lie on a circle of radius 1
• They are all separated by the same argument on the Argand diagram
• If n is even, then two roots are real (ie [FONT=&quot]±[/FONT]1)
• If n is odd, then only one root is real (ie 1)
• The roots occur in conjugate pairs (eg if one root is a+bi, then the other will be a-bi)
• The roots occur in the following pattern: 1,w^{[FONT=&quot]2[/FONT]},w^{[FONT=&quot]3[/FONT]}...w^{[FONT=&quot]n-1[/FONT]} (where w is the root with the smallest arg>0)
• 1+w+w^{[FONT=&quot]2[/FONT]}w^{[FONT=&quot]3[/FONT]}+...+w^{[FONT=&quot]n-1[/FONT]} = 0

 

[shimmerz_777]

how do u prove the last result?

 

[bboyelement]

how do u prove the last result?

sum of roots i think ... where its -b/a and b should always be 0 for root of unity

 

[Riviet]

sum of roots i think ... where its -b/a and b should always be 0 for root of unity

Yep, that's the quickest and easiest way to prove it.

 

[Sober]

sum of roots i think ... where its -b/a and b should always be 0 for root of unity

Wow, I was not aware of such a simple proof, thankyou.

 

[desert fox]

And i kept adding them by number crunching...


this rocks