Question
The solution of $$2\sqrt 2 \,{x^4} = \left( {\sqrt 3 - 1} \right) + i\left( {\sqrt 3 + 1} \right)$$ is
A.
$$ \pm \left( {\cos \frac{{5\pi }}{{48}} + i\sin \frac{{5\pi }}{{48}}} \right)$$
B.
$$ \pm \left( {\cos \frac{{7\pi }}{{48}} + i\sin \frac{{7\pi }}{{48}}} \right)$$
C.
$$ \pm \left( {\cos \frac{{19\pi }}{{48}} - i\sin \frac{{19\pi }}{{48}}} \right)$$
D.
None of these
Answer :
$$ \pm \left( {\cos \frac{{5\pi }}{{48}} + i\sin \frac{{5\pi }}{{48}}} \right)$$
Solution :
$$\eqalign{
& {x^4} = \frac{{\sqrt 3 - 1}}{{2\sqrt 2 }} + i\frac{{\sqrt 3 + 1}}{{2\sqrt 2 }} = \cos \frac{{5\pi }}{{12}} + i\sin \frac{{5\pi }}{{12}} \cr
& {\text{So, }}x = \cos \left\{ {\frac{{k\pi }}{2} + \frac{{5\pi }}{{48}}} \right\} + i\sin \left\{ {\frac{{k\pi }}{2} + \frac{{5\pi }}{{48}}} \right\} \cr
& k = 0,1,2,3 \cr
& \therefore {\text{Roots are}} \cr
& \cos \frac{{5\pi }}{{48}} + i\sin \frac{{5\pi }}{{48}}{\text{ for }}k = 0 \cr
& \cos \frac{{29\pi }}{{48}} + i\sin \frac{{29\pi }}{{48}}{\text{ for }}k = 1 \cr
& \cos \frac{{53\pi }}{{48}} + i\sin \frac{{53\pi }}{{48}} = - \left( {\cos \frac{{5\pi }}{{48}} + i\sin \frac{{5\pi }}{{48}}} \right){\text{ for }}k = 2 \cr
& \cos \frac{{77\pi }}{{48}} + i\sin \frac{{77\pi }}{{48}} = - \left( {\cos \frac{{29\pi }}{{48}} + i\sin \frac{{29\pi }}{{48}}} \right){\text{ for }}k = 3 \cr} $$