Another Complex Numbers question (1 Viewer)

Hermes1 · Jul 23, 2011

if n is an integer prove:

$(1+\cos \frac{2\pi }{n}+i\sin \frac{2\pi }{n})^{n} = -2^{n}\cos ^{n}\frac{\pi }{n}$

Hermes1 · Jul 23, 2011

using double angle:

LHS:
$(1+(2\cos ^{2}\frac{\pi }{n}-1)+2i\sin \frac{\pi }{n}\cos \frac{\pi }{n})^{n}$
$((2\cos ^{2}\frac{\pi }{n}+2i\sin \frac{\pi }{n}\cos \frac{\pi }{n})^{n}$
$(2\cos \frac{\pi }{n}(\cos \frac{\pi }{n}+i\sin \frac{\pi }{n}))^{n}$
$2^{n}\cos ^{n}\frac{\pi }{n}(\cos \pi +i\sin \pi )$ (De Moivre's theorem)
$2^{n}\cos ^{n}\frac{\pi }{n}*(-1)$
= $-2^{n}\cos ^{n}\frac{\pi }{n}$ = RHS

therefore proven.

managed to solve it myself, im just leaving it up for anyone who's interested.

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Another Complex Numbers question (1 Viewer)

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