Polynomials and DMT (1 Viewer)

[STx]

I think theres something wrong in the worked solutions I have to this Q.
Q:
Given that z=cosθ+isinθ

(i) Show than zⁿ+z^-n=2cos(nθ)

(ii) Hence solve the equation 4z⁴-4z³+6z²-4z+4=0

Obviously got (i), and something goes wrong i nthe simplification in (ii)

thx

 

[bboyelement]

i got z = {1-sqrt[2] + i sqrt[2sqrt[2] + 1]} /2

im so tired ... i probably made some stupid error somewhere

 

[IBODORKOS]

divide equation by z squared and use pt (1)

 

[STx]

^yep thats what I did, but the there is a mistake in the answers as they put z+(1/z)= cosθ instead of 2cosθ, so they end up getting something different. I get up to 8(cosθ)^2 - 4cosθ - 1 = 0. So cosθ= (1/4).[1±√3] (by the quadratic formula) then how do i find sinθ ? triangles?

 

[bboyelement]

uhm z+z^-1 = 2 cosθ
sub in cosθ and times z up and use quadratic again.