A) \[1+i\]
B) \[1-i\]
C) 1
D) \[-1\]
Correct Answer: D
Solution :
We have \[\frac{1+\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}}{1+\cos \frac{\pi }{8}-i\sin \frac{\pi }{8}}\] \[=\frac{2{{\cos }^{2}}\frac{\pi }{16}+i.2\sin \frac{\pi }{16}\cos \frac{\pi }{16}}{2{{\cos }^{2}}\frac{\pi }{16}-i.2\sin \frac{\pi }{16}\cos \frac{\pi }{16}}\] [using\[1+cos2\theta =2\text{ }co{{s}^{2}}\theta \] and \[\sin 2\theta =2\sin \theta \cos \theta \]] \[=\frac{\cos \frac{\pi }{16}+i\sin \frac{\pi }{16}}{\cos \frac{\pi }{16}-i\sin \frac{\pi }{16}}\] \[\therefore \] \[{{\left[ \frac{1+\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}}{1+\cos \frac{\pi }{8}-i\sin \frac{\pi }{8}} \right]}^{8}}\] \[={{\left[ \frac{\cos \frac{\pi }{16}+i\sin \frac{\pi }{16}}{\cos \frac{\pi }{16}-i\sin \frac{\pi }{16}} \right]}^{8}}\] \[=\left[ \frac{\cos \frac{8\pi }{16}+i\sin \frac{8\pi }{16}}{\cos \frac{8\pi }{16}-i\sin \frac{8\pi }{16}} \right]\] (\[\because \]By De-movers theorem) \[=\frac{\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}}{\cos \frac{\pi }{2}-i\sin \frac{\pi }{2}}=\frac{0+i}{0-i}=-1\]
You need to login to perform this action.
You will be redirected in
3 sec
