Polynomials Question (from trials) (1 Viewer)

[kevda1st]

Plz help with part ii) and iii)

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[azureus88]

(i) z^9+1=(z^3+1)(z^6-z^3+1)

(ii) roots of (z^3+1) are cis(pi/3), cis(pi), cis(-pi/3)

roots of (z^9+1) are cis(pi/9), cis(pi/3), cis(5pi/9), cis(7pi/9), cis(pi), cis(-7pi/9), cis(-5pi/9), cis(-pi/3), cis(-pi/9)

so roots of (z^6-z^3+1) are cis(pi/9),cis(5pi/9), cis(7pi/9), cis(-7pi/9), cis(-5pi/9), cis(-pi/9)

using cis@+cis(-@)=2cos@ and cis@cis(-@)=1, you get the answer

(iii) equate coefficient of z^2