Geometric Series with Complex Numbers

A question is this type if and only if it requires summing geometric series involving z^n where z = cos θ + i sin θ to derive trigonometric identities.

2 questions · Challenging +1.2

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CAIE Further Paper 2 2024 June Q6

7 marks Challenging +1.2

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 $\sin 2 \theta \neq 0$, $$\sum _ { r = 1 } ^ { n } \sin ( 4 r \theta ) = \frac { \cos 2 \theta - \cos ( 4 n + 2 ) \theta } { 2 \sin 2 \theta }$$ \includegraphics[max width=\textwidth, alt={}, center]{27485e4a-cd34-43e3-aa92-767820a9f6f9-12_2718_35_143_2012}

CAIE Further Paper 2 2020 November Q7

7 marks Challenging +1.2

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