Answer
$-16$
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:
$(8 \ cis 210^{\circ}) \cdot ( 2 \ cis 330^{\circ}) =(8)(2) \ cis (210^{\circ} +330^{\circ})= 16 \ cis 540^{\circ}$
Re-arrange as: $16 \ cis 540^{\circ}= 16 (\cos 540^{\circ}+\ i \sin 540^{\circ}) $
Next, we will write it in the rectangular form by using calculator as follows:
$ 16 (\cos 540^{\circ}+\ i \sin 540^{\circ}) = 16 (-1 +0 i)= -16 $
