• Congratulations to the Class of 2024 on your results!
    Let us know how you went here
    Got a question about your uni preferences? Ask us here

polynomial (1 Viewer)

noobking

New Member
Joined
Nov 1, 2006
Messages
11
Gender
Male
HSC
2007
1. show that z^6-z^3+1= (z^2-2zcos(pi/9)+1)(z^2-2zcos(5pi/9)+1)(z^2-2zcos(7pi/9)+1)

2. show that cos(5pi/9)cos(pi/9)+cos(5pi/9)cos(7pi/9)+cos(7pi/9)cos(pi/9)=-3/4
 

haque

Member
Joined
Sep 2, 2006
Messages
426
Gender
Male
HSC
2006
It'll be better if u do it urself-but here are a few hints that will make it easier. show that z^6-z^3+1=0 are among the roots of z^9-1, from those roots u exclude the roots of z^3+1=0 . Then group these roots in the form of factors of the polynomial. Groups factors with conjugate roots and this will give u the answer once u expand. Second part is simply product of roots 2 at a time-there are two ways u could do this one-but it will become apparent after u have done part1.
 

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

Top