{left[ {frac{{1 + cos (pi /8) + isin (pi /8)} | vedclass

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(A) Let $z = \frac{1 + \cos(\pi/8) + i\sin(\pi/8)}{1 + \cos(\pi/8) - i\sin(\pi/8)}$.Using the identities $1 + \cos 	heta = 2\cos^2(	heta/2)$ and $\s

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$-1$

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(A) Let $z = \frac{1 + \cos(\pi/8) + i\sin(\pi/8)}{1 + \cos(\pi/8) - i\sin(\pi/8)}$.Using the identities $1 + \cos 	heta = 2\cos^2(	heta/2)$ and $\sin 	heta = 2\sin(	heta/2)\cos(	heta/2)$:$z = \frac{2\cos^2(\pi/16) + 2i\sin(\pi/16)\cos(\pi/16)}{2\cos^2(\pi/16) - 2i\sin(\pi/16)\cos(\pi/16)}$$z = \frac{2\cos(\pi/16) [\cos(\pi/16) + i\sin(\pi/16)]}{2\cos(\pi/16) [\cos(\pi/16) - i\sin(\pi/16)]}$$z = \frac{\cos(\pi/16) + i\sin(\pi/16)}{\cos(\pi/16) - i\sin(\pi/16)}$Using Euler's formula $e^{i	heta} = \cos 	heta + i\sin 	heta$,we have $z = \frac{e^{i\pi/16}}{e^{-i\pi/16}} = e^{i\pi/8}$.Then $z^8 = (e^{i\pi/8})^8 = e^{i\pi} = \cos \pi + i\sin \pi = -1$.

${\left[ {\frac{{1 + \cos (\pi /8) + i\sin (\pi /8)}}{{1 + \cos (\pi /8) - i\sin (\pi /8)}}} \right]^8}$ is equal to

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