A) \[4\]
B) \[3\]
C) \[2\]
D) \[8\]
Correct Answer: A
Solution :
\[\left[ \frac{1+\sin \frac{\pi }{8}+i\,\cos \,\frac{\pi }{8}}{1+\sin \frac{\pi }{8}-i\,cos\frac{\pi }{8}} \right]\] \[=\left[ \frac{1+\cos \,\alpha +i\,\sin \alpha }{1+\cos \alpha -i\,\sin \alpha } \right]\,\,\,\left( Put\,\alpha =\frac{\pi }{2}-\frac{\pi }{8} \right)\] \[={{\left[ \frac{2{{\cos }^{2}}\frac{\alpha }{2}+2i\,\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}{2{{\cos }^{2}}\frac{\alpha }{2}-2i\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}} \right]}^{n}}\] \[=\left[ \frac{\cos \frac{\alpha }{2}+i\,\sin \frac{\alpha }{2}}{\cos \frac{\alpha }{2}-i\sin \frac{\alpha }{2}} \right]\] \[={{({{2}^{2i\frac{\alpha }{2}}})}^{n={{e}^{in\,\alpha }}}}\] \[={{e}^{in}}^{\left( \frac{3\pi }{8} \right)}=\cos \frac{3n\pi }{8}+i\,\sin \frac{3n\pi }{8}\] For \[n=4,\] we get imaginary part.
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