Question

${\left(\begin{array}{c}\frac{1+\cos \frac{\pi }{8}-i\sin \frac{\pi }{8}}{1+\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}}\end{array}\right)}^8$

• A. 1
• B. -1
• C. 2
• D. $\frac{1}{2}$

Answer 1

The correct option is B -1

$(\frac{1+\cos \frac{\pi }{8}-i\sin \frac{\pi }{8}}{1+\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}}{)}^8=(\frac{2{\cos }^2\frac{\pi }{16}-i2\sin \frac{\pi }{16}\cos \frac{\pi }{16}}{2{\cos }^2\frac{\pi }{16}+i2\sin \frac{\pi }{16}\cos \frac{\pi }{16}}{)}^8$

$=(\frac{\cos \frac{\pi }{16}-i\sin \frac{\pi }{16}}{\cos \frac{\pi }{16}+i\sin \frac{\pi }{16}}{)}^8$

$=(\frac{e^{-i\frac{\pi }{16}}}{e^{i\frac{\pi }{16}}}{)}^8$

$=(e^{-i\frac{\pi }{8}}{)}^8$

$=e^{-i\pi }$

$=\cos \pi$

$=-1$