The Student Room Group
(zneiθ)(zneiθ)=z2nzneiθzneiθ+1=z2nzn(eiθ+eiθ)+1=z2nzn(cosθ+isinθ+cosθisinθ)+1=z2n2zncosθ+1(z^{n}-e^{i\theta})(z^{n}-e^{-i\theta}) = z^{2n} - z^{n}e^{i\theta} - z^{n}e^{-i\theta} + 1 = z^{2n}-z^{n}(e^{i\theta}+e^{-i\theta}) + 1 = z^{2n}-z^{n}(\cos \theta + i\sin\theta + \cos\theta - i\sin\theta) + 1 = z^{2n}-2z^{n}\cos\theta + 1

Let n=3 and θ=π4\theta = \frac{\pi}{4} and you get z2n2zncosθ+1=z6z32+1z^{2n}-2z^{n}\cos\theta + 1 = z^{6}-z^{3}\sqrt{2}+1 which you've just shown is expressible as (zneiθ)(zneiθ)(z^{n}-e^{i\theta})(z^{n}-e^{-i\theta}) . Putting in values for n and theta gives

z6z32+1=(z3eiπ4)(z3eiπ4)z^{6}-z^{3}\sqrt{2}+1 = (z^{3}-e^{i\frac{\pi}{4}})(z^{3}-e^{-i\frac{\pi}{4}}) so z3=eiπ4z^{3} = e^{i\frac{\pi}{4}} and z3=eiπ4z^{3}=e^{-i\frac{\pi}{4}}

Work out the possible values for z and turn them into their cos + i sin forms.
Reply 2
thanks so much for all ur help :smile: