Question

Show that one value of
${\left(\frac{1+\sin \frac{\pi }{8}+i\cos \frac{\pi }{8}}{1+\sin \frac{\pi }{8}-i\cos \frac{\pi }{8}}\right)}^{\frac{8}{3}}$ is $-1$

Answer 1

To show:

${\left(\frac{1+\sin \frac{\pi }{8}+i\cos \frac{\pi }{8}}{1+\sin \frac{\pi }{8}-i\cos \frac{\pi }{8}}\right)}^{\frac{8}{3}}=-1$

Solution:

$\left(\frac{1+\sin \frac{\pi }{8}+i\cos \frac{\pi }{8}}{1+\sin \frac{\pi }{8}-i\cos \frac{\pi }{8}}\right)=$ $\left(\frac{1+\cos \left(\frac{\pi }{2}-\frac{\pi }{8}\right)+i\sin (\frac{\pi }{8}-\frac{\pi }{8})}{1+\cos (\frac{\pi }{2}-\frac{\pi }{8})-i\sin (\frac{\pi }{2}-\frac{\pi }{8})}\right)$

$=\frac{1+\cos \frac{3\pi }{8}+i\sin \frac{3\pi }{8}}{1+\cos \frac{3\pi }{8}-i\sin \frac{3\pi }{8}}$

$=\frac{2{\cos }^2\frac{3\pi }{16}+i2\sin \frac{3\pi }{16}\cos \frac{3\pi }{16}}{2{\cos }^2\frac{3\pi }{16}-i2\sin \frac{3\pi }{16}\cos \frac{3\pi }{16}}$

$=\frac{2\cos \frac{3\pi }{16}[\cos \frac{3\pi }{16}+i\sin \frac{3\pi }{16}]}{2\cos \frac{3\pi }{16}[\cos \frac{3\pi }{16}-i\sin \frac{3\pi }{16}]}$

$={(\cos \frac{3\pi }{16}+i\sin \frac{3\pi }{16})}^2$

$=\cos \frac{3\pi }{8}+i\sin \frac{3\pi }{8}$

Now,

${\left(\frac{1+\sin \frac{\pi }{8}+i\cos \frac{\pi }{8}}{1+\sin \frac{\pi }{8}-i\cos \frac{\pi }{8}}\right)}^{\frac{8}{3}}=$ ${(\cos \frac{3\pi }{8}+i\sin \frac{3\pi }{8})}^{\frac{8}{3}}$

$=\cos \pi +i\sin \pi$

$=-1+i\left(0\right)$

$=-1$

Hence, ${\left(\frac{1+\sin \frac{\pi }{8}+i\cos \frac{\pi }{8}}{1+\sin \frac{\pi }{8}-i\cos \frac{\pi }{8}}\right)}^{\frac{8}{3}}=-1$