Complex Number Questions

A

appletini171

New Member

Mar 23, 2009

#1

Me again. Sorry

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$i)\: z^n+z^{-n}=\cos n\alpha + i\sin n\alpha + \cos (-n\alpha) + i\sin (-n\alpha)\\=\cos n\alpha + i\sin n\alpha + \cos n\alpha - i\sin n\alpha\\= 2\cos n\alpha$

$ii)\: (z+z^{-1})^6=z^6 + 6z^5z^{-1} + 15z^4z^{-2}+20z^3z^{-3}+15z^2z^{-4}+6zz^{-5}+z^{-6}\\= z^6+z^{-6} + 6(z^4+z^{-4})+15(z^2+z^{-2})+20\\= 2\cos 6\alpha + 12\cos 4\alpha + 30\cos 2\alpha + 20\\$But $(z+z^{-1})^6 = (2\cos\alpha)^6=64\cos^6\alpha\\\therefore \cos^6\alpha = \dfrac{\cos 6\alpha + 6\cos 4\alpha + 15\cos 2\alpha + 10}{32}$

$iii)\:\displaystyle\int_0^{\frac{\pi}{6}}\cos^6\alpha\,d\alpha=\dfrac{1}{32}\displaystyle\int_0^{\frac{\pi}{6}}(\cos 6\alpha + 6\cos 4\alpha + 15\cos 2\alpha + 10)\,d\alpha\\= \dfrac{1}{32}\left [\dfrac{\sin 6\alpha}{6}+\dfrac{3\sin 4\alpha}{2}+\dfrac{15\sin 2\alpha}{2}+10\alpha\right]_0^{\frac{\pi}{6}}\\= \dfrac{1}{32}\left (\dfrac{3}{2}\sin\frac{2\pi}{3}+\dfrac{15}{2}\sin\frac{\pi}{3}+\dfrac{5\pi}{3}\right )\\= \dfrac{1}{32}\left (\dfrac{9\sqrt{3}}{2}+\dfrac{5\pi}{3}\right)$