A) 1
B) -1
C) i
D) -i
Correct Answer: B
Solution :
[b] \[Let\text{ }\sin \frac{\pi }{8}+i\cos \frac{\pi }{8}=z\] \[\Rightarrow {{\left[ \frac{1+\sin \frac{\pi }{8}+i\cos \frac{\pi }{8}}{1+\sin \frac{\pi }{8}-i\cos \frac{\pi }{8}} \right]}^{8}}\] \[={{\left( \frac{1+z}{1+\frac{1}{z}} \right)}^{8}}\] \[={{z}^{8}}={{\left( \sin \frac{\pi }{8}+i\cos \frac{\pi }{8} \right)}^{8}}\] \[={{\left( \cos \left( \frac{\pi }{2}-\frac{\pi }{8} \right)+i\sin \left( \frac{\pi }{2}-\frac{\pi }{8} \right) \right)}^{8}}\] \[={{\left( \cos \frac{3\pi }{8}+i\sin \frac{3\pi }{8} \right)}^{8}}\] \[=\cos 3\pi =-1\]
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