If $w=\cos\frac{2\pi}5+i\sin\frac{2\pi}5$ and $z=w+\frac1w$, show that $z^2+z-1=0$

Question

Let $w = \cos \frac{2\pi}{5} + i \sin \frac{2\pi}{5}$ be a 5th root of unity. Let $z = w + \frac{1}{w}$. Show that $z^2 + z - 1 = 0$.

Do I need to use de Moivre's theorem? Or other methods. An explanation is appreciated. Thanks.

Answer 1

Note $w^5-1=0$ and

$$z^2+z-1= (w+\frac1w)^2+ w+\frac1w - 1\\ = \frac1{w^2}(w^4+w^3+w^2+w+1)=0$$

that is, the sum of the five roots is zero via Vieta formula.

Answer 2

$$z = w + \frac{1}{w}\to wz=w^2+1$$now multily $z^2 + z - 1 = 0$ by $w^2$
$$w^2z^2+w^2z-w^2=0\\(wz)^2+w^2z-w^2=0\\ (w^2+1)^2+w(w^2+1)-w^2=0\\ w^4+2w^2+1+w^3+w-w^2=0\\ w^4+w^2+w^3+w+1=0\\\to \times\frac{w-1}{w-1}\\ \frac{(w-1)(w^4+w^2+w^3+w+1)}{w-1}=0\\ \frac{(w^5-1)}{w-1}=0$$