Answer
$-\sqrt 3 - \ i$ or, $-(\sqrt 3+ i)$
Work Step by Step
The product theorem for two complex numbers is given by: $[r_1 (\cos \theta_1 +i \sin \theta_1) ] \cdot [r_2 (\cos \theta_2 +i \sin \theta_2) ]= r_1 r_2 [ \cos (\theta_1+\theta_2)+\sin (\theta_1+\theta_2)]$
This can also be re-written in the easy form as follows: $(r_1 \ cis \theta_1) \cdot ( r_2 \ cis \theta_2) = r_1 r_2 \ cis (\theta_1+\theta_2)$
Therefore, the given product becomes:
$(\sqrt 2 \ cis 300^{\circ}) \cdot ( \sqrt 2 \ cis 270^{\circ}) =(\sqrt 2)(\sqrt 2) \ cis ( 300^{\circ} + 270^{\circ})= 2 \ cis 570^{\circ} \\= 2 (\cos 570^{\circ}+\ i \sin 570^{\circ}) \\ = 2[ \cos (360^{\circ}+210^{\circ}) +i \sin (360^{\circ}+210^{\circ}]\\=2 [ \cos 210^{\circ}+i \sin 210^{\circ} ]\\= 2[ \cos (270^{\circ} -60^{\circ}) +i \sin ( 270^{\circ} -60^{\circ}] \\ = 2[ -\sin 60^{\circ} - i \cos 60^{\circ}]$
Next, we will write it in the rectangular form by using calculator as follows:
$2[ -\sin 60^{\circ} - i \cos 60^{\circ}] = 2 [-\dfrac{\sqrt 3}{2}- i \dfrac{1}{2}]= -\sqrt 3 - \ i$ or, $-(\sqrt 3+ i)$
