Answer
$\dfrac{1}{\sqrt 2} - i \dfrac{\sqrt 3}{\sqrt 2} $ or, $ \dfrac{\sqrt 2}{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 2 (\cos (-60^{\circ}) +i \sin (-60^{\circ}) ) ....(1) $
We know that $\sin (-\theta) =- \sin \theta$ and $\cos (-\theta) =\cos \theta$
Now, $\cos (-60^{\circ})=\cos 60^{\circ}) =\dfrac{1}{2}$ and $\sin (-60^{\circ})=- \sin 60^{\circ}) = -\dfrac{ \sqrt 3}{2}$
Therefore, the equation (1) becomes:
$z =\sqrt 2(\dfrac{1}{2} -i \dfrac{\sqrt 3}{2} ) \\= \dfrac{1}{\sqrt 2} - i \dfrac{\sqrt 3}{\sqrt 2} \\= \dfrac{\sqrt 2}{2} - i \dfrac{\sqrt 6}{2}$
