Problem source

If
$$                 
z^{4}+z^{3}+z^{2}+z+1=0\tag{*} 
$$               
and $u=z+z^{-1}$, find the possible values of $u$. Hence find the possible values of $z$. 
[Do not try to simplify your answers.]
Show that, if $z$ satisfies $(*)$, then
\[z^{5}-1=0.\]
Hence write the solutions of $(*)$ in the form $z=r(\cos\theta+i\sin\theta)$ for suitable real $r$ and $\theta$. Deduce that
\[\sin\frac{2\pi}{5}=\frac{\surd(10+2\surd 5)}{4}
\ \ \hbox{and}\ \ \cos\frac{2\pi}{5}=\frac{-1+\surd 5}{4}.\]

Solution source

\begin{align*}
&& 0 &= z^4+z^3+z^2+z+1 \\
\Rightarrow && 0 &= z^2+z+1+z^{-1}+z^{-2} \tag{$z \neq 0$} \\
&&&= \left ( z+z^{-1} \right)^2-2 + z+z^{-1} + 1 \\
&&&= u^2+u-1 \\
\Rightarrow && u &= \frac{-1 \pm \sqrt{5}}{2} \\
\Rightarrow && z+z^{-1} &= \frac{-1 \pm \sqrt{5}}{2} \\
\Rightarrow && 0 &= z^2-\left (  \frac{-1 \pm \sqrt{5}}{2}\right)z+1 \\
\Rightarrow && z &= \frac{\left (  \frac{-1 \pm \sqrt{5}}{2}\right) \pm \sqrt{\left (  \frac{-1 \pm \sqrt{5}}{2}\right)^2-4}}{2} \\
&&&= \frac{\left (  \frac{-1 \pm \sqrt{5}}{2}\right) \pm \sqrt{\frac{1+5\mp2\sqrt{5}-16}{4}}}{2} \\
&&&= \frac{\left (  \frac{-1 \pm \sqrt{5}}{2}\right) \pm \sqrt{\frac{-10\mp2\sqrt{5}-16}{4}}}{2} \\
&&&= \frac{-1\pm\sqrt{5}}{4} \pm i\frac{\sqrt{10\pm 2\sqrt{5}}}{4} 
\end{align*}

Since $z^4+z^3+z^2+z+1 = 0$ we can multiply both sides by $z-1$ to obtain $z^5-1 = 0$.

Therefore if $z = r(\cos \theta + i \sin \theta)$ we see that $z^5 = 1 \Rightarrow r^5 (\cos 5 \theta + i \sin 5 \theta) = 1 \Rightarrow r = 1, 5 \theta = 2n \pi$ ie $z = \cos \frac{2n\pi}{5} + i\sin \frac{2n \pi}{5}$.

We are looking for a solution in the first quadrant, therefore $\cos \frac{2\pi}{5} = \frac{-1 + \sqrt{5}}4$ and $\sin \frac{2\pi}{5} = \frac{\sqrt{10+2\sqrt{5}}}{4}$