Question

$\left(\frac{1+i \,sin \frac{\pi}{8}+cos \frac{\pi}{8}}{1-i\, sin \frac{\pi}{8}+cos \frac{\pi}{8}}\right)^{8}$ equals

• A. $2^8$
• B. $0$
• C. $-1$
• D. $1$

Answer 1

Answer: (C)

$\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} + 2i \,sin \frac{\pi}{16} cos \frac{\pi}{16}}{2\, cos^{2} \frac{\pi}{16} - 2i \,sin \frac{\pi}{16} \cdot cos \frac{\pi}{16}}\right)^{8} $
$\left[\frac{\left(cos \frac{\pi}{16} + i \,sin \frac{\pi}{16}\right)}{\left(cos \frac{\pi}{16} - i \,sin \frac{\pi}{16}\right)}\right]^{8} $
$= \left[\left(cos \frac{\pi}{16} + i \,sin \frac{\pi}{16}\right)\left(cos \frac{\pi}{16}-i \,sin \frac{\pi}{16}\right)^{-1}\right]^{8}$
$\left[\left(cos \frac{\pi}{16}+ i \,sin \frac{\pi}{16}\right)\left(cos \frac{\pi}{16} + i \,sin \frac{\pi}{16}\right)\right]^{8} $
$ = \left[\frac{\left(cos \frac{\pi}{16} + i\, sin \frac{\pi}{16}\right)^{2}}{1}\right]^{8}$
$= cos \,\pi + i\, sin \,\pi = -1$ (using De-Moivre’s theorem )
Short Cut Method :
Let $z = cos \frac{\pi}{8} + i\,sin \frac{\pi}{8}$
$\therefore \frac{1}{z} = cos \frac{\pi}{8} - i\,sin \frac{\pi}{8}$
Now from given we have
$\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{1+z}{1+z^{-1}}\right)^{8} $
$ = \left(\frac{\left(1+z\right)z}{\left(1+z\right)}\right)^{8} = z^{8} = \left(cos \frac{\pi}{8} + i \,sin \frac{\pi}{8}\right)^{8} $
$= cos\, \pi = -1$ (using De-Moivre’s theorem )