$\begin{aligned}
& \text { }\left(\frac{1+\cos \frac{\pi}{8}-i \sin \frac{\pi}{8}}{1+\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}}\right)^8 \\
& =\left(\frac{2 \cos ^2 \frac{\pi}{16}-2 i \sin \frac{\pi}{16} \cos \frac{\pi}{16}}{2 \cos ^2 \frac{\pi}{16}+2 i \sin \frac{\pi}{16} \cos \frac{\pi}{16}}\right)^8 \\
& =\left(\frac{2 \cos \frac{\pi}{16}\left(\cos \frac{\pi}{16}-i \sin \frac{\pi}{16}\right)}{2 \cos \frac{\pi}{16}\left(\cos \frac{\pi}{16}+i \sin \frac{\pi}{16}\right)}\right)^8 \\
& =\left(\frac{\cos \frac{\pi}{16}-i \sin \frac{\pi}{16}}{\cos \frac{\pi}{16}+i \sin \frac{\pi}{16}}\right)^8 \\
& =\left(\frac{e^{-\frac{\pi}{16} i}}{e^{\frac{\pi}{16} i}}\right)^8 \\
&
\end{aligned}$
$\begin{array}{lr}
=\left(e^{-\frac{2 \pi}{16} i}\right)^8 & \text { [Using eular's theorem] } \\
=\left(e^{-8 \times \frac{2 \pi}{16} i}\right) & \text { [Using De-Moivre's theorem] }
\end{array}$
$\begin{aligned} & =e^{-\pi i} \\ & =\cos \pi-i \sin \pi \\ & =-1\end{aligned}$