Show that if \(z = \mathrm { e } ^ { \mathrm { i } \theta }\) and \(z \neq - 1\) then
$$\frac { z - 1 } { z + 1 } = \mathrm { i } \tan \frac { 1 } { 2 } \theta$$
Hence, or otherwise, show that if \(z\) is a cube root of unity then
$$\frac { z ^ { 3 } - 1 } { z ^ { 3 } + 1 } + \frac { z ^ { 2 } - 1 } { z ^ { 2 } + 1 } + \frac { z - 1 } { z + 1 } = 0$$
Hence write down three roots of the equation
$$\left( z ^ { 3 } - 1 \right) \left( z ^ { 2 } + 1 \right) ( z + 1 ) + \left( z ^ { 2 } - 1 \right) \left( z ^ { 3 } + 1 \right) ( z + 1 ) + ( z - 1 ) \left( z ^ { 3 } + 1 \right) \left( z ^ { 2 } + 1 \right) = 0$$
and find the other three roots. Give your answers in an exact form.