# Chapter 8 - Complex Numbers, Polar Equations, and Parametric Equations - Section 8.3 The Product and Quotient Theorems - 8.3 Exercises - Page 376: 18

$-6+6\sqrt{3}i$

#### 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: $12(\cos120^{\circ}+i\sin120^{\circ}) \\=12(-\frac{1}{2}+\frac{\sqrt{3}}{2}i) \\=-6+6\sqrt{3}i$

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