(a) Show that \(\sum_{r=1}^{n} z^{2 r}=\frac{z^{2 n+1}-z}{z-z^{-1}}\), for \(z \neq 0,1,-1\).
(b) By letting \(z=\cos \theta+\mathrm{i} \sin \theta\), show that, if \(\sin \theta \neq 0\),
\[1+2 \sum_{r=1}^{n} \cos (2 r \theta)=\frac{\sin (2 n+1) \theta}{\sin \theta}\]