Question

${\left[\frac{1+\cos (\pi /8)+i\sin (\pi /8)}{1+\cos (\pi /8)-i\sin (\pi /8)}\right]}^8$ is equal to

• A. -1
• B. 0
• C. 1
• D. 2

Answer 1

Answer: (A)

${\left[\frac{1+\cos (\pi /8)+i\sin (\pi /8)}{1+\cos (\pi /8)-i\sin (\pi /8)}\right]}^8={\left(\frac{2{\cos }^2\frac{\pi }{16}+i\cdot 2\sin \frac{\pi }{16}\cos \frac{\pi }{16}}{2{\cos }^2\frac{\pi }{16}-i2\cos \frac{\pi }{16}\sin \frac{\pi }{16}}\right)}^8$
$={\left(\frac{\cos \frac{\pi }{16}+i\cdot \sin \frac{\pi }{16}}{\cos \frac{\pi }{16}-i\sin \frac{\pi }{16}}\right)}^8=\cos \pi +i\sin \pi =-1$