Answer
$12\sqrt 3+12i$
Work Step by Step
Step 1: $[4(\cos60^{\circ}+i\sin60^{\circ})][6(\cos330^{\circ}+i\sin330^{\circ}]$
Step 2: Using the product theorem, the expression becomes
$4(6[\cos(60^{\circ}+330^{\circ})+i\sin(60^{\circ}+330^{\circ})]$
Step 3: Simplifying, $24(\cos390^{\circ}+i\sin390^{\circ})$
Step 4: Since $30^{\circ}$ and $390^{\circ}$ are coterminal angles, the expression becomes $24(\cos30^{\circ}+i\sin30^{\circ})$
Step 5: We know that $\cos30^{\circ}=\frac{\sqrt 3}{2}$ and $\sin30^{\circ}=\frac{1}{2}$
Step 6: Substituting these values in the expression, $24(\frac{\sqrt 3}{2}+\frac{1}{2}i)$
Step 7: Therefore, the final answer is $12\sqrt 3+12i$
