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Question Number 124595 by physicstutes last updated on 04/Dec/20 $$\mathrm{Given}\:\mathrm{that}\:\:\omega\:=\:{e}^{{i}\theta} ,\:\theta\neq\:{n}\pi,\:{n}\:\in\:\mathbb{N} \\ $$$$\mathrm{show}\:\mathrm{that}\: \\ $$$$\:\left(\mathrm{1}\right)\:\frac{\omega^{\mathrm{2}} −\mathrm{1}}{\omega}\:=\:\mathrm{2}{i}\:\mathrm{sin}\:\theta \\ $$$$\:\left(\mathrm{2}\right)\:\left(\mathrm{1}\:+\:\omega\right)^{{n}} \:=\:\mathrm{2}^{{n}} \mathrm{cos}^{{n}} \left(\frac{\mathrm{1}}{\mathrm{2}}\theta\right){e}^{\frac{\mathrm{1}}{\mathrm{2}}\left({in}\theta\right)} \\ $$ Answered by…