Answer
$20+20\sqrt 3i$
Work Step by Step
Step 1: $[8(\cos300^{\circ}+i\sin300^{\circ})][5(\cos120^{\circ}+i\sin120^{\circ}]$
Step 2: Using the product theorem, the expression becomes
$8(5)[\cos(300^{\circ}+120^{\circ})+i\sin(300^{\circ}+120^{\circ})]$
Step 3: Simplifying, $40(\cos420^{\circ}+i\sin420^{\circ})$
Step 4: Since $60^{\circ}$ and $420^{\circ}$ are coterminal angles, the expression becomes $40(\cos60^{\circ}+i\sin60^{\circ})$
Step 5: We know that $\cos60^{\circ}=\frac{1}{2}$ and $\sin60^{\circ}=\frac{\sqrt 3}{2}$
Step 6: Substituting these values in the expression, $40(\frac{1}{2}+\frac{\sqrt 3}{2}i)$
Step 7: Therefore, the final answer is $20+20\sqrt 3i$