Question

Use De-Moivre's theorem to solve the equation
$2\sqrt{2}{x}^4=(\sqrt{3}-1)+i(\sqrt{3}+1)$

Answer 1

$r^2=a^2+b^2=2(3+1)$
$\therefore r=2\sqrt{2},tan\theta =\frac{b}{a}=\frac{\sqrt{3}+1}{\sqrt{3}-1}$
$=\frac{1+1/\sqrt{3}}{1-1/\sqrt{3}}=tan({45}^{\circ }+{30}^{c}irc)=tan{75}^{\circ }$
$\therefore \theta =5\pi /12$
$\therefore 2\sqrt{2}{x}^4=2\sqrt{2}(cos\frac{5\pi }{12}+isin\frac{5\pi }{12})$
$\therefore x^4=cos(2n\pi +\frac{5\pi }{12})+isincos(2n\pi +\frac{5\pi }{12})$
$\therefore x=\left[cos\right(24n+5\left)\right(\pi /12)+isin(24n+5\left)\right(\pi /12){]}^{1/4}$
$=cos\frac{24n+5}{48}\pi +isin\frac{24n+5}{48}\pi$,
where n = 0, 1, 2, 3
$\therefore x=cos\frac{r\pi }{48}+isin\frac{r\pi }{48}$
where r = 5, 29, 53 ,57