Question

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

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

Answer 1

Answer: (B)

$-1$

$1+\cos \frac{\pi }{8}=2{\cos }^2\frac{\pi }{16}$

$\sin \frac{\pi }{8}=2\sin \frac{\pi }{16}\cos \frac{\pi }{16}$

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

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

$=\left(e^{-ix}\right)=\cos \pi -i\sin \pi$

$=-1$