Write down the five roots of the equation \(z ^ { 5 } = 1\), giving your answers in the form \(\mathrm { e } ^ { \mathrm { i } \theta }\), where \(- \pi < \theta \leqslant \pi\).
Hence find the four linear factors of
$$z ^ { 4 } + z ^ { 3 } + z ^ { 2 } + z + 1$$
Deduce that
$$z ^ { 2 } + z + 1 + z ^ { - 1 } + z ^ { - 2 } = \left( z - 2 \cos \frac { 2 \pi } { 5 } + z ^ { - 1 } \right) \left( z - 2 \cos \frac { 4 \pi } { 5 } + z ^ { - 1 } \right)$$
Use the substitution \(z + z ^ { - 1 } = w\) to show that \(\cos \frac { 2 \pi } { 5 } = \frac { \sqrt { 5 } - 1 } { 4 }\).