The smallest positive integral value of \(n\) such that \(\left[\frac{1+\sin \frac{\pi}{8}+i \cos \frac{\pi}{8}}{1+\sin \frac{\pi}{8}-i \cos \frac{\pi}{8}}\right]^{\mathrm{n}}\) \(n\) is equal to

\(\left[\frac{1+\sin \frac{\pi}{8}+i \cos \frac{\pi}{8}}{1+\sin \frac{\pi}{8}-i \cos \frac{\pi}{8}}\right]^{\mathrm{n}}\)
\(=\left[\frac{1+\cos \alpha+i \sin \alpha}{1+\cos \alpha-i \sin \alpha}\right]^{\mathrm{n}} \quad\left(\right.\) Put \(\left.\alpha=\frac{\pi}{2}-\frac{\pi}{8}\right)\)
\(=\left[\frac{2 \cos ^{2} \frac{\alpha}{2}+2 i \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{2 \cos ^{2} \frac{\alpha}{2}-2 i \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}\right]\)
\(=\left[\frac{\cos \frac{\alpha}{2}+i \sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}-i \sin \frac{\alpha}{2}}\right]^{\mathrm{n}}\)
\(=\left(e^{2 i \frac{\alpha}{2}}\right)^{n}=e^{i n \alpha}\)
\(=\mathrm{e}^{\mathrm{in}\left(\frac{3 \pi}{8}\right)}=\cos \frac{3 n \pi}{8}+\mathrm{i} \sin \frac{3 n \pi}{8}\)
For \(n=4\), we get imaginary part.