Answer
$\sqrt {\dfrac{3}{2}}- \ i \ \sqrt {\dfrac{3}{2}} $ or, $\dfrac{\sqrt 6}{2} -i \ \dfrac{\sqrt 6}{2}$
Work Step by Step
The given complex number can be written in the rectangular form as follows: $r (\cos \theta +i \sin \theta) = a+i b$
Where, $a=r \cos \theta ; b =r \sin \theta $
Let us suppose that $z =\sqrt 3(\cos 315^{\circ} +i \sin 315^{\circ} ) ....(1) $
Now, $\cos 315^{\circ}=\cos (270^{\circ}+45^{\circ}) =\sin 45^{\circ}= \dfrac{\sqrt 2}{2}$ and $\sin 315^{\circ}=\sin (270^{\circ}+45^{\circ}) =- \cos 45^{\circ}= -\dfrac{\sqrt 2}{2}$
Therefore, the equation (1) becomes:
$z= \sqrt 3( \dfrac{\sqrt 2}{2} +i \dfrac{-\sqrt 2}{2})=\sqrt {\dfrac{3}{2}}- \ i \ \sqrt {\dfrac{3}{2}} $ or, $\dfrac{\sqrt 6}{2} -i \ \dfrac{\sqrt 6}{2}$
