Show that \(\sum _ { r = 1 } ^ { n } z ^ { 4 r } = \frac { z ^ { 4 n + 2 } - z ^ { 2 } } { z ^ { 2 } - z ^ { - 2 } }\), for \(z ^ { 2 } \neq z ^ { - 2 }\).
By letting \(z = \cos \theta + \mathrm { i } \sin \theta\), show that, if \(\sin 2 \theta \neq 0\),
$$\sum _ { r = 1 } ^ { n } \sin ( 4 r \theta ) = \frac { \cos 2 \theta - \cos ( 4 n + 2 ) \theta } { 2 \sin 2 \theta }$$
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