• Best of luck to the class of 2021 for their HSC exams. You can do it!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page

Complex Roots Of Unity Q. (1 Viewer)

kevda1st

Member
Joined
Sep 10, 2008
Messages
126
Gender
Male
HSC
2009
Prove that if: a = x+ y b=xw+yw² c=xw²+yw^4 ,
Than abc = x³+y³ and a²+b²+c²=6xy


and part ii) of this Question....
Thanks in advance


 

Trebla

Administrator
Administrator
Joined
Feb 16, 2005
Messages
7,569
Gender
Male
HSC
2006
and part ii) of this Question....
Thanks in advance


(z + √3)6 + 64 = 0
Divide by 64
[(z + √3)/2]6 + 1 = 0
Let x = (z + √3)/2
=> x6 + 1 = 0
The solutions for x are the solutions found in part (i). Therefore the solutions of z are found by the solution set of x and then substituting into z = 2x - √3
 

Timothy.Siu

Prophet 9
Joined
Aug 6, 2008
Messages
3,463
Location
Sydney
Gender
Male
HSC
2009
Prove that if: a = x+ y b=xw+yw² c=xw²+yw^4 ,
Than abc = x³+y³ and a²+b²+c²=6xy


i guess that w is a root of unity of w^3=1 then, 1+w+w^2=0
abc=(x+y)(xw+yw^2)(xw^2+yw^4)
=(x+y)(x^2w^3+xyw^5+xyw^4+y^2w^6)
=(x+y)(x^2+xy(w^2+w)+y^2)
=(x+y)(x^2-xy+y^2)=x^3+y^3

a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)
 

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

Top