Answer
$12 \sqrt 3 +12 \ 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:
$(4 \ cis 60^{\circ}) \cdot ( 6 \ cis 330^{\circ}) =(4)(6) \ cis (60^{\circ} +330^{\circ})= 24 \ cis 390^{\circ}$
Re-arrange as: $24 \ cis 390^{\circ}= 24 (\cos 390^{\circ}+\ i \sin 390^{\circ}) $
Next, we will write it in the rectangular form by using calculator as follows:
$24 (\cos 390^{\circ}+\ i \sin 390^{\circ}) =24 (\dfrac{\sqrt 3}{2}+i \dfrac{1}{2})= 12 \sqrt 3 +12 \ i $
