Using the roots of $ z^5=1$, or otherwise, write $z^4+z^3+z^2+z+1$ as the product of two quadratic expressions with real coefficients.

Question

Using the roots of $z^5=1$, or otherwise, write $z^4+z^3+z^2+z+1$ as the product of two quadratic expressions with real coefficients. Hence find the exact value of the product $\cos(2\pi/5)\cos(4\pi/5)$.

Answer 1

The roots of $$z^5=1$$ are $e^{2\pi ik/5}$ where $k=\pm2,\pm1,0$

The roots of $$\dfrac{z^5-1}{z-1}=0$$ are $e^{2\pi ik/5}$ where $k=\pm2,\pm1$

$$\left(z-e^{-2\cdot2\pi i/5}\right)\left(z-e^{2\cdot2\pi i/5}\right)\left(z-e^{-1\cdot2\pi i/5}\right)\left(z-e^{1\cdot2\pi i/5}\right)=0$$

Now $$\left(z-e^{-k\cdot2\pi i/5}\right)\left(z-e^{k\cdot2\pi i/5}\right)=z^2+1-2z\cos\dfrac{2k\pi}5$$

$$\implies\left(z^2+1-2z\cos\dfrac{2\pi}5\right)\left(z^2+1-2z\cos\dfrac{2\cdot2\pi}5\right)=z^4+z^3+z^2+z+1$$

Can you take it from here?

Answer 2

Suppose $z^4+z^3+z^2+z+1 =(z^2+az+1)(z^2+bz+1) $, because that takes care of the first and last terms.

$(z^2+az+1)(z^2+bz+1) =z^4+z^3(a+b)+z^2(ab+2)+z(a+b)+1 $ so we need $a+b = 1$ and $ab+2 = 1$ or $ab = -1$.

Since $(z-u)(z-v) =z^2-z(u+v)+uv $, $a$ and $b$ are the roots of $z^2-(a+b)z+ab =z^2-z-1 $.

This will give the desired factorization.