Answer
$$\sqrt 3 + i$$
Work Step by Step
$$\eqalign{
& \frac{{2\sqrt 6 - 2i\sqrt 2 }}{{\sqrt 2 - i\sqrt 6 }} \cr
& {\text{Convert the numerator and the denominator to trigonometric form}} \cr
& 2\sqrt 6 - 2i\sqrt 2 \cr
& r = \sqrt {{{\left( {2\sqrt 6 } \right)}^2} + {{\left( { - 2\sqrt 2 } \right)}^2}} = 4\sqrt 2 \cr
& \theta = {\tan ^{ - 1}}\left( {\frac{{ - 2\sqrt 2 }}{{2\sqrt 6 }}} \right) = - {30^ \circ } \cr
& 2\sqrt 6 - 2i\sqrt 2 = 4\sqrt 2 \left( {\cos \left( { - {{30}^ \circ }} \right) + i\sin \left( { - {{30}^ \circ }} \right)} \right) \cr
& \cr
& \sqrt 2 - i\sqrt 6 \cr
& r = \sqrt {{{\left( {\sqrt 2 } \right)}^2} + {{\left( { - \sqrt 6 } \right)}^2}} = 2\sqrt 2 \cr
& \theta = {\tan ^{ - 1}}\left( {\frac{{ - \sqrt 6 }}{{\sqrt 2 }}} \right) = - {60^ \circ } \cr
& \sqrt 2 - i\sqrt 6 = 2\sqrt 2 \left( {\cos \left( { - {{60}^ \circ }} \right) + i\sin \left( { - {{60}^ \circ }} \right)} \right) \cr
& \cr
& {\text{Then,}} \cr
& \frac{{2\sqrt 6 - 2i\sqrt 2 }}{{\sqrt 2 - i\sqrt 6 }} = \frac{{4\sqrt 2 \left( {\cos \left( { - {{30}^ \circ }} \right) + i\sin \left( { - {{30}^ \circ }} \right)} \right)}}{{2\sqrt 2 \left( {\cos \left( { - {{60}^ \circ }} \right) + i\sin \left( { - {{60}^ \circ }} \right)} \right)}} \cr
& {\text{Using the Quotient Theorem}} \cr
& = \frac{{4\sqrt 2 }}{{2\sqrt 2 }}\left[ {\cos \left( { - {{30}^ \circ } + {{60}^ \circ }} \right) + i\sin \left( { - {{30}^ \circ } + {{60}^ \circ }} \right)} \right] \cr
& {\text{Simplify}} \cr
& = 2\left[ {\cos \left( {{{30}^ \circ }} \right) + i\sin \left( {{{30}^ \circ }} \right)} \right] \cr
& {\text{Write in Rectangular form}} \cr
& = 2\left( {\frac{{\sqrt 3 }}{2} + \frac{1}{2}i} \right) \cr
& = \sqrt 3 + i \cr} $$
