[7 points] Let $z = e^{\left(\frac{2\pi i}{n}\right)}$. Here $n \geq 2$ is a positive integer, $i^{2} = -1$ and the real number $\frac{2\pi}{n}$ can also be considered as an angle in radians.\\
(i) Show that $\displaystyle\sum_{k=0}^{n-1} z^{k} = 0$.\\
(ii) Show that $\displaystyle\sum_{k=0}^{8} \cos(40k+1)^{\circ} = 0$, i.e., $\cos(1^{\circ}) + \cos(41^{\circ}) + \cos(81^{\circ}) + \cos(121^{\circ}) + \cdots + \cos(241^{\circ}) + \cos(281^{\circ}) + \cos(321^{\circ}) = 0$.