We have,
$$
\begin{aligned}
& \left(\frac{1+\cos \frac{\pi}{8}-i \sin \frac{\pi}{8}}{1+\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}}\right)^{12} \\
& =\left[\frac{2 \cos ^2 \frac{\pi}{16}-2 \sin \frac{\pi}{16} \cos \frac{\pi}{16} i}{2 \cos ^2 \frac{\pi}{16}+2 \sin \frac{\pi}{16} \cos \frac{\pi}{16} i}\right]^{12} \\
& =\left[\frac{\cos \frac{\pi}{16}\left(\cos \frac{\pi}{16}-i \sin \frac{\pi}{16}\right)}{\cos \frac{\pi}{16}\left(\cos \frac{\pi}{16}+i \sin \frac{\pi}{16}\right)}\right]^{12} \\
& =\left(\frac{\cos \frac{\pi}{16}-i \sin \frac{\pi}{16}}{\cos \frac{\pi}{16}+i \sin \frac{\pi}{16}}\right)^{12}=\left(\frac{e^{-i \pi / 16}}{e^{i \pi / 16}}\right)^{12} \\
& {\left[\because \cos \theta+i \sin \theta=e^{i \theta}\right]} \\
& =\left(e^{-\frac{i \pi}{8}}\right)^{12}=e^{-i 3 \pi / 2}=\cos \frac{3 \pi}{2}-i \sin \frac{3 \pi}{2} \\
& =0-i(-1)=i \\
&
\end{aligned}
$$