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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$i$

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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 $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$,we get:$z = \frac{2 \cos^2 \frac{\pi}{16} - i (2 \sin \frac{\pi}{16} \cos \frac{\pi}{16})}{2 \cos^2 \frac{\pi}{16} + i (2 \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 \pi / 16}}{e^{i \pi / 16}} = e^{-i \pi / 8}$.Now,$z^{12} = (e^{-i \pi / 8})^{12} = e^{-i 12 \pi / 8} = e^{-i 3 \pi / 2}$.$e^{-i 3 \pi / 2} = \cos(-\frac{3 \pi}{2}) + i \sin(-\frac{3 \pi}{2}) = \cos(\frac{3 \pi}{2}) - i \sin(\frac{3 \pi}{2}) = 0 - i(1) = -i$ is incorrect,let's re-evaluate: $e^{-i 3 \pi / 2} = \cos(\frac{3 \pi}{2}) + i \sin(\frac{3 \pi}{2}) = 0 + i(1) = i$.

$\left(\frac{1+\cos \frac{\pi}{8}-i \sin \frac{\pi}{8}}{1+\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}}\right)^{12} = $

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