Explanation

(a)
$\left\lbrack \frac{1 + \sin{\frac{\pi}{8} + i\cos\frac{\pi}{8}}}{1 + \sin{\frac{\pi}{8} - i\cos\frac{\pi}{8}}} \right\rbrack^{n}$
$= \left\lbrack \frac{1 + \cos{\alpha + i\sin\alpha}}{1 + \cos{\alpha - i\sin\alpha}} \right\rbrack^{n}\ \ \ \ \ \ \ \ \ \left( Put\ \alpha = \frac{\pi}{2} - \frac{\pi}{8}\ \right)$
$= \left\lbrack \frac{2\cos^{2}{\frac{\alpha}{2} + 2i\sin{\frac{\alpha}{2}\cos\frac{\alpha}{2}}}}{2\cos^{2}{\frac{\alpha}{2} - 2i\sin{\frac{\alpha}{2}\cos\frac{\alpha}{2}}}} \right\rbrack^{n}$
$= \left\lbrack \frac{\cos{\frac{\alpha}{2} + i\sin\frac{\alpha}{2}}}{\cos{\frac{\alpha}{2} - i\sin\frac{\alpha}{2}}} \right\rbrack^{n}$
$= \left( e^{2i\frac{a}{2}\ } \right)^{n}$=${e^{in}}^{a}$
$= e^{in\left( \frac{3\pi}{8} \right)} = \cos{\frac{3n\pi}{8} + i\sin\frac{3n\pi}{8}}$
For $n = 4$, we get imaginary part