Answer
$\frac{2\sqrt{6} - 2i\sqrt{2}}{\sqrt{2} - i\sqrt{6}}$
= $\sqrt{3} + i$
Work Step by Step
Now, $2\sqrt{6} - 2i\sqrt{2} = 2\sqrt{2}(\sqrt{3} - i)$ where $(\sqrt{3} - i)$ is at $330^\circ$ with absolute value $\sqrt{(\sqrt{3})^2 + (-1)^2} = 2$, and $\sqrt{2} - i\sqrt{6} = \sqrt{2}(1 - i\sqrt{3})$ where $(1 - i\sqrt{3})$ is at $300^\circ$ with absolute value $\sqrt{1^2 + (-\sqrt{3})^2} = 2$
Therefore, $\frac{2\sqrt{6} - 2i\sqrt{2}}{\sqrt{2} - i\sqrt{6}}$
= $\frac{2\sqrt{2}(\sqrt{3} - i)}{\sqrt{2}(1 - i\sqrt{3})}$
= $2\cdot \frac{2cis330^\circ}{2cis300^\circ}$
= $2cis(330^\circ - 300^\circ)$ (Quotient Theorem)
= $2cis30^\circ$
= $2(\frac{\sqrt{3}}{2} + \frac{1}{2} i)$
= $\sqrt{3} + i$
