Answer
$20+20 \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:
$(8 \ cis 300^{\circ}) \cdot ( 5 \ cis 120^{\circ}) =(8)(5) \ cis (300^{\circ} +120^{\circ})= 40 \ cis 420^{\circ}$
Re-arrange as: $40 \ cis 420^{\circ}= 40 (\cos 420^{\circ}+\ i \sin 420^{\circ}) $
Next, we will write it in the rectangular form by using calculator as follows:
$ 40 (\cos 420^{\circ}+\ i \sin 420^{\circ})=40 (\dfrac{1}{2}+i \dfrac{\sqrt 3}{2})= 20+20 \sqrt 3 \ i $
