Answer
$-6+6\sqrt{3}i$
Work Step by Step
First, we use the division theorem to divide the absolute values and subtract the arguments:
$\frac{24(\cos150^{\circ}+i\sin150^{\circ})}{2(\cos30^{\circ}+i\sin 30^{\circ})}
\\=12(\cos (150^{\circ}-30^{\circ})+i\sin(150^{\circ}-30^{\circ}))
\\=12(\cos120^{\circ}+i\sin120^{\circ})$
Since we know that $\cos120^{\circ}=-\frac{1}{2}$ and $\sin120^{\circ}=\frac{\sqrt{3}}{2}$, we can substitute these values in the expression and simplify:
$12(\cos120^{\circ}+i\sin120^{\circ})
\\=12(-\frac{1}{2}+\frac{\sqrt{3}}{2}i)
\\=-6+6\sqrt{3}i$
