left(frac{1+cos frac{pi}{8}-i sin frac{pi}{8} | vedclass

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(B) Let $z = \frac{1+\cos \frac{\pi}{8}-i \sin \frac{\pi}{8}}{1+\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}}$.Using the half-angle formulas $1+\cos 	heta

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$-1$

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(B) Let $z = \frac{1+\cos \frac{\pi}{8}-i \sin \frac{\pi}{8}}{1+\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}}$.Using the half-angle formulas $1+\cos 	heta = 2 \cos^2 \frac{	heta}{2}$ and $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$:$z = \frac{2 \cos^2 \frac{\pi}{16} - 2i \sin \frac{\pi}{16} \cos \frac{\pi}{16}}{2 \cos^2 \frac{\pi}{16} + 2i \sin \frac{\pi}{16} \cos \frac{\pi}{16}}$$z = \frac{2 \cos \frac{\pi}{16} (\cos \frac{\pi}{16} - i \sin \frac{\pi}{16})}{2 \cos \frac{\pi}{16} (\cos \frac{\pi}{16} + i \sin \frac{\pi}{16})}$$z = \frac{\cos \frac{\pi}{16} - i \sin \frac{\pi}{16}}{\cos \frac{\pi}{16} + i \sin \frac{\pi}{16}} = \frac{e^{-i \frac{\pi}{16}}}{e^{i \frac{\pi}{16}}} = e^{-i \frac{2\pi}{16}} = e^{-i \frac{\pi}{8}}$.Now,$z^8 = (e^{-i \frac{\pi}{8}})^8 = e^{-i \pi}$.Using Euler's formula $e^{i 	heta} = \cos 	heta + i \sin 	heta$:$e^{-i \pi} = \cos(-\pi) + i \sin(-\pi) = -1 + 0 = -1$.

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

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