Trigonometry (11th Edition) Clone

$\frac{-3\sqrt{2} + 3i\sqrt{6}}{\sqrt{6} + i\sqrt{2}}$ = $3i$
Now, $-3\sqrt{2} + 3i\sqrt{6} = -3\sqrt{2}(1 - \sqrt{3}i)$ where $(1 - \sqrt{3}i)$ is at $300^\circ$ with absolute value $\sqrt{1^2 + (-\sqrt{3})^2} = 2$, and $\sqrt{6} + i\sqrt{2} = \sqrt{2}(\sqrt{3} + i)$ where $(\sqrt{3} + i)$ is at $30^\circ$ with absolute value $\sqrt{(\sqrt{3})^2 + 1^2} = 2$ Therefore, $\frac{-3\sqrt{2} + 3i\sqrt{6}}{\sqrt{6} + i\sqrt{2}}$ = $\frac{-3\sqrt{2}(1 - \sqrt{3}i)}{\sqrt{2}(\sqrt{3} + i)}$ = $-3\cdot \frac{2cis300^\circ}{2cis30^\circ}$ = $-3cis(300^\circ - 30^\circ)$ (Quotient Theorem) = $-3cis270^\circ$ = $-3(0 - i)$ = $3i$