Mathematics - Complex Number Question with Solution | TestHub

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