Answer
$-10 \sqrt 3 +10 \ 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:
$(4 \ cis 30^{\circ}) \cdot ( 5 \ cis 120^{\circ}) =(4)(5) \ cis (30^{\circ} +120^{\circ})= 20 \ cis 150^{\circ}$
Re-arrange as: $20 \ cis 150^{\circ}= 20 (\cos 150^{\circ}+\ i \sin 150^{\circ}) $
Next, we will write it in the rectangular form as follows:
$20 (\cos 150^{\circ}+\ i \sin 150^{\circ}) =20 (-\dfrac{\sqrt 3}{2}+i \dfrac{1}{2})= -10 \sqrt 3 +10 \ i $
