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I need some direction in part (c) of this problem.

a) Find the 6 roots of $z^6 + 1 = 0$ (I have solved this using de Moivre's theorem)

b) Hence show that $z^6 + 1 = (z^2 + 1)(z^2 - \sqrt{3}z + 1)(z^2 + \sqrt{3}z + 1)$ (no problems here)

c) Divide both sides of this identity by $z^3$, and then let $z = cis \theta $ to show that:

$$ \cos3\theta = 4\cos\theta(\cos\theta -\cos\frac{\pi}{6})(\cos\theta -\cos\frac{5\pi}{6}) $$

I'm assuming one uses the expression in part (b) not part (a). The left hand side in part(c) can be deduced using de Moivre's theorem, it's the right hand side that confuses me.

Thanks in advance

Stephan
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1 Answers1

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$$\dfrac{ (z^2 + 1)(z^2 - \sqrt{3}z + 1)(z^2 + \sqrt{3}z + 1)}{z^3} = \dfrac{z^2+1}{z}\cdot\dfrac{z^2 - \sqrt{3}z + 1}{z}\cdot \dfrac{z^2 + \sqrt{3}z + 1}{z}$$$$ = \bigg(z+\dfrac1z\bigg)\cdot\left[\left(z+\dfrac1z\right)-\sqrt3\right]\cdot \left[\left(z+\dfrac1z\right)+\sqrt3\right]$$

Now all you're left with is adjusting some factors of $2$.

insipidintegrator
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  • Brilliant. Thanks $z^3 + z^{-3} = 2\cos3\theta$ and $ z^1 + z^{-1} = 2\cos\theta$. Thank you so much.... I just couldn't see it. Cheers. – Stephan Dec 08 '23 at 05:14