Question

If $2\sqrt{2}{x}^4=(\sqrt{3}-1)+i(\sqrt{3}+1),$ then $x=\cos \frac{1}{4}(2n\pi +k)+i\sin \frac{1}{4}(2n\pi +k);n=0,1,2,3,$ where $k=$

• A. $\frac{\pi }{12}$
• B. $\frac{5\pi }{12}$
• C. $\frac{7\pi }{12}$
• D. None of these

Answer 1

The correct option is B $\frac{5\pi }{12}$

Given $2\sqrt{2}{x}^4=(\sqrt{3}-1)+i(\sqrt{3}+1)$

$\Rightarrow x^4=\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)+i\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)$

$\Rightarrow {\left|x^4\right|}^2={\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)}^2+{\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)}^2=1$

$\therefore \left|x^4\right|=1$

and $arg\left(x^4\right)={\tan }^{-1}\left[\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)/\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)\right]$

$={\tan }^{-1}\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)={75}^0=\frac{5\pi }{12}$

$\therefore x^4=1[\cos (2n\pi +\frac{5\pi }{12})+i\sin (2n\pi +\frac{5\pi }{12})]$

Using ${(\cos \theta +i\sin \theta )}^n=\cos n\theta +i\sin n\theta$, we have

$x=\cos \frac{1}{4}(2n\pi +\frac{5\pi }{12})+i\sin \frac{1}{4}(2n\pi +\frac{5\pi }{12});n=0,1,2,3$