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

The correct option is B $-1$

$1+\sin \frac{\pi }{8}+i\cos \frac{\pi }{8}=1+\cos \frac{3\pi }{8}+i\sin \frac{3\pi }{8}=2{\cos }^2\frac{3\pi }{16}+i2\sin \frac{3\pi }{16}\cos \frac{3\pi }{16}=2\cos \frac{3\pi }{16}{e}^{i\frac{3\pi }{16}}$

$1+\sin \frac{\pi }{8}-i\cos \frac{\pi }{8}=1-\cos \frac{11\pi }{8}+i\sin \frac{11\pi }{8}=2{\sin }^2\frac{11\pi }{16}+i2\sin \frac{11\pi }{16}\cos \frac{11\pi }{16}=2\cos \frac{3\pi }{16}{e}^{i(\frac{\pi }{2}-\frac{11\pi }{16})}=2\cos \frac{3\pi }{16}{e}^{-i\frac{3\pi }{16}}$

$[\frac{1+\sin \frac{\pi }{8}+i\cos \frac{\pi }{8}}{1+\sin \frac{\pi }{8}-i\cos \frac{\pi }{8}}{]}^8=e^{i\frac{3\pi }{16}\times 16}=e^{i3\pi }=-1$