A) 0
B) 1
C) \[-1\]
D) 2
Correct Answer: C
Solution :
\[{{\left( \frac{1+\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}}{1+\cos \frac{\pi }{8}-i\sin \frac{\pi }{8}} \right)}^{8}}\] \[={{\left( \frac{2{{\cos }^{2}}\frac{\pi }{16}+i2\sin \frac{\pi }{16}\cos \frac{\pi }{16}}{2{{\cos }^{2}}\frac{\pi }{16}-i2\sin \frac{\pi }{16}.\cos \frac{\pi }{16}} \right)}^{8}}\] \[[\because \cos 2\theta =2{{\cos }^{2}}\theta -1,\sin 2\theta =2\sin \theta \cos \theta ]\] \[={{\left( \frac{\cos \frac{\pi }{16}+i\sin \frac{\pi }{16}}{\cos \frac{\pi }{16}-i\sin \frac{\pi }{16}} \right)}^{8}}\] \[={{\left( \frac{\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}}{\cos \frac{\pi }{2}-\sin \frac{\pi }{2}} \right)}^{8}}\](using De-Moiver theorem) \[=-1\]
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