Question

${\left(\frac{1+isin\frac{\pi }{8}+cos\frac{\pi }{8}}{1-isin\frac{\pi }{8}+cos\frac{\pi }{8}}\right)}^8$ equals

• A. $2^8$
• B. $0$
• C. $-1$
• D. $1$

Answer 1

Answer: (C)

${\left(\frac{1+cos\frac{\pi }{8}+isin\frac{\pi }{8}}{1+cos\frac{\pi }{8}-isin\frac{\pi }{8}}\right)}^8$
$={\left(\frac{2co{s}^2\frac{\pi }{16}+2isin\frac{\pi }{16}cos\frac{\pi }{16}}{2co{s}^2\frac{\pi }{16}-2isin\frac{\pi }{16}\cdot cos\frac{\pi }{16}}\right)}^8$
${\left[\frac{\left(cos\frac{\pi }{16}+isin\frac{\pi }{16}\right)}{\left(cos\frac{\pi }{16}-isin\frac{\pi }{16}\right)}\right]}^8$
$={\left[\left(cos\frac{\pi }{16}+isin\frac{\pi }{16}\right){\left(cos\frac{\pi }{16}-isin\frac{\pi }{16}\right)}^{-1}\right]}^8$
${\left[\left(cos\frac{\pi }{16}+isin\frac{\pi }{16}\right)\left(cos\frac{\pi }{16}+isin\frac{\pi }{16}\right)\right]}^8$
$={\left[\frac{{\left(cos\frac{\pi }{16}+isin\frac{\pi }{16}\right)}^2}{1}\right]}^8$
$=cos\pi +isin\pi =-1$ (using De-Moivre’s theorem )
Short Cut Method :
Let $z=cos\frac{\pi }{8}+isin\frac{\pi }{8}$
$\therefore \frac{1}{z}=cos\frac{\pi }{8}-isin\frac{\pi }{8}$
Now from given we have
${\left(\frac{1+cos\frac{\pi }{8}+isin\frac{\pi }{8}}{1+cos\frac{\pi }{8}-isin\frac{\pi }{8}}\right)}^8={\left(\frac{1+z}{1+z^{-1}}\right)}^8$
$={\left(\frac{\left(1+z\right)z}{\left(1+z\right)}\right)}^8=z^8={\left(cos\frac{\pi }{8}+isin\frac{\pi }{8}\right)}^8$
$=cos\pi =-1$ (using De-Moivre’s theorem )