Question

The expression $\left[\frac{1+sin \frac{\pi}{8}+i cos \frac{\pi}{8}}{1+ sin \frac{\pi}{8}-i cos \frac{\pi}{8}}\right]^{^{^8}} =$

• A. $1$
• B. $-1$
• C. $i$
• D. $-i$

Answer 1

Answer: (B)

Let $ 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}} $
$=(\frac{1+z}{1+\frac{1}{z}})^{8}$
$=z^{8}$
$={(sin \frac{\pi}{8} +i\,\,cos \frac{\pi}{8})^{8}}$
$=\left(cos \left(\frac{\pi}{2}-\frac{\pi}{8}\right)+i sin \left(\frac{\pi}{2}-\frac{\pi}{8}\right)\right)^{8}$
$=(cos \frac{3\pi}{8} +i sin \frac{3\pi}{8})^{8}$
$= cos 3\pi =-1$